Ace Education Physics O'level (Ace Education Book Series) 9982913182, 9789982913188

Table of contents :
PREFACE
TABLE OF CONTENTS
MEASUREMENTS
SCALAR AND VECTOR QUANTITIES
LENGTH
TIME
MASS AND WEIGHT
VOLUME AND DENSITY
CENTRE OF GRAVITY
DISTANCE AND DISPLACEMENT
SPEED AND VELOCITY
ACCELERATION
MOTION GRAPHS
FORCE
CIRCULAR MOTION
FORCES ACTING ON SPRING
MOMENT OF A FORCE
WORK, ENERGY AND POWER
SIMPLE MACHINE
PRESSURE
THERMAL PHYSICS
TRANSFER OF THERMAL ENERGY
HEAT CAPACITY AND LATENT HEAT
WAVE MOTION
ELECTROMAGNETIC SPECTRUM
LIGHT
SOUND WAVE
MAGNETISM
ELECTROMAGNETIC INDUCTION
STATIC ELECTRICITY
CURRENT ELECTRICITY
PRACTICAL ELECTRICITY
BASIC ELECTRONICS
ATOMIC STRUCTURE
RADIOACTIVITY

Citation preview

ACE EDUCATION PHYSICS O’LEVEL

Grade 10 − 12 GCSE, GCE

NSWANA CHING’AMBU

The Authorship and Career Network [email protected] +260976008283, +260972719373

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The Authorship and Career Network ‘Impossibility Our Possibility’ The Authorship and Career Network is an organisation with a sole objective of promoting excellence in research, literacy, scholarship, education and skill development by supporting authors in Africa and beyond to publish their works.

Published for Africa by The Authorship and Career Network Indeco House, Cairo Road, Lusaka, Zambia.

Copyright © 2021 by Nswana Ching’ambu

All rights reserved. No part of this publication may be reproduced, stored or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning, or otherwise without written permission from the publisher. It is illegal to copy this book, post it to a website, or distribute it by any other means without permission.

Nswana Ching’ambu asserts the moral right as an author

First edition ISBN: 978-9982-913-18-8

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To the students Ace Education Book Series aims at giving detailed material in the simplest way to help students understand and recall information easily. The books also highlight the importance and application of each topic in real life so that students can understand why they are learning the material, how the material relates to or can be used in real life. Ace Education Physics O’level consists of 33 units. Each unit begins with the introduction and overview of the unit, and ends with the review questions and solutions. To fully benefit, students are advised to cover everything in each unit considered. Students can also learn more from our social media platforms. On these platforms, students can find additional information, ask questions or participate in helping other students. With full confidence, this book will help a number of students, not just by boosting their scores, but also to understand Physics Ordinary Level.

Ace Education

About the author Nswana Ching’ambu has a bachelor’s degree in Human Biology from the University of Zambia and Teaching Methodology from Gideon Robert University. He has been lecturing Clinical Medicine courses at Gideon Robert University for many years. He lectures Anatomy and Physiology, Medical Biochemistry and Nutrition, and previously used to lecture Cellular Pathology and Basic Microbiology. He is the founder of Ace Education and the author of Ace Education Book Series for O’Levels: Biology, Chemistry, Physics and Mathematics. +260967744388 [email protected] [email protected] [email protected]

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TABLE OF CONTENTS

UNIT 1: MEASUREMENTS……………………………………………………………………………...…..1 UNIT 2: SCALAR AND VECTOR QUANTITY……………………………………………………..….....7 UNIT 3: LENGTH……………………………………………………………………………………….…...12 UNIT 4: TIME………………………………………………………………………………………….…….18 UNIT 5: MASS AND WEIGHT……………………………………………………………………....…….23 UNIT 6: VOLUME AND DENSITY…………………………………………………………….....………30 UNIT 7: CENTRE OF GRAVITY………………………………………………………………………….37 UNIT 8: DISTANCE AND DISPLACEMENT………………………………………………….…………42 UNIT 9: SPEED AND VELOCITY……………………………………………………………….………..45 UNIT 10: ACCELERATION…………………………………………………………………….…………..49 UNIT 11: MOTION GRAPH…………………………………………………………………….………….54 UNIT 12: FORCE……………………………………………………………………………….……………64 UNIT 13: CIRCULAR MOTION……………………………………………………………..……………..73 UNIT 14: FORCE ACTING ON SPRING…………………………………………………………..……..75 UNIT 15: MOMENT OF A FORCE…………………………………………………………….…………78 UNIT 16: WORK, ENERGY AND POWER……………………………………………………..………..82 UNIT 17: SIMPLE MACHINE……………………………………………………………………………...93 UNIT 18: PRESSURE……………………………………………………………………………..………..105 UNIT 19: THERMAL PHYSICS………………………………………………………………..…………120 UNIT 20: TRANSFER OF THERMAL ENERGY……………………………………………...………..137 UNIT 21: HEAT ENERGY AND LATENT HEAT……………………………………………..………147 UNIT 22: WAVE MOTION…………………………………………………………………………..……154 UNIT 23: ELECTROMAGNETIC SPECTRUM……………………………………………………...…...161 UNIT 24: LIGHT………………………………………………………………………………………...….167 UNIT 25: SOUND WAVE…………………………………………………………………………...…….182 UNIT 26: MAGNETISM…………………………………………………………………….…………...…190 UNIT 27: ELECTROMAGNETIC INDUCTION…………………………………………………...……. 198 UNIT 28: STATIC ELECTRICITY…………………………………………………………………...….. 208 UNIT 29: CURRENT ELECTRICITY……………………………………………………….…………….214 UNIT 30: PRACTICAL ELECTRICITY……………………………………………………………….….231 UNIT 31: BASIC ELECTRONICS………………………………………………………………………...241 UNIT 32: ATOMIC STRUCTURE………………………………………………………………….……..251 UNIT 33: RADIOACTIVITY………………………………………………………………………………257

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1.0

MEASUREMENTS

Introduction Everything around us is made up of matter. This book, your pen, clothes, shoes, food, a wire, a car, an airplane, petrol, diesel, water and so on, are all matter. Some matter such as petrol, diesel and food store energy. When petrol is burnt it releases energy. How does this energy interact with matter? For instance, burning petrol in a car engine. The study of matter, energy and their interaction is called physics. In simpler terms, physics is the study of how things around us work. For example, how a car moves, an airplane flies, an electric bulb lit, a speaker works and so on. Matter has different sizes. The size of this book is different from that of a pen, cellphone, computer, car, etc. Therefore, to understand how things around us work, we need to have precise measurements of each object we study. To have exact measurements, objects have to be measured. A physical quantity is anything that can be measured. Length, time and temperature are few examples of things that can be measured and therefore are physical quantities. When we measure an object, we obtain its size. However, size alone is not sufficient. Imagine you go to buy a carpet for your room. Can it be adequate to say, “I need 5 carpets.” Likely not! Therefore, to describe a physical quantity such as length, not only the size is needed but also the unit. For example, 5 metres of carpet, 50 kilometre journey, 2 kilograms of sugar, 5 litres of cooking oil, 37 oC (degree celcius), 10 hours 30 minutes and so on.

Specific outcomes This unit covers the types of physical quantities, standard units and the importance of prefixes. By the end of this unit, you will be able to: ❖ Define physics ❖ Define the following physics terminologies: • Physical quantity • Magnitude • Unit ❖ State the seven base quantities ❖ Give examples of derived quantities ❖ Differentiate base and derived quantities ❖ Understand what SI units are ❖ State the seven SI units with their symbols ❖ Understand what prefixes are and how to use them

Measurements

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MEASUREMENTS 1.1

QUANTITY • Physics is the study of matter, energy and their interaction. • It deals with physical quantities and motions of matter. • A physical quantity is a quantity that can be measured. It is expressed by a magnitude with a suitable unit. • Magnitude is the size of a quantity, for example, 2, 150, 3, 10. • Unit is the standard measure of a quantity, for example, s (second), m (metre), kg (kilogram).

❖ TYPES OF PHYSICAL QUANTITIES 1. BASE QUANTITY • Base quantities are fundamental physical quantities. • These quantities contain only one unit. • There are seven base quantities: length, mass, temperature, time, current, amount of substance, and luminous intensity. 2. DERIVED QUANTITY • Derived quantities are quantities that consist of two or more base quantities. • Therefore, they have a combination of units. • For example, speed is a derived quantity. Speed is the total distance covered per unit time. Therefore, it consists of two base quantities; length (distance covered) and time. Types of physical quantities Examples

1.2

Derived quantities

Length Time Mass Temperature Current Amount of substance Luminous intensity

Volume Density Speed Area Force Pressure

UNITS • Unit is defined as the standard measure of a quantity. • SI units (International System of the unit) are used as standardized units of measurements. • There are seven SI units. • The table below shows SI units and their symbols. Base quantities Length Time Mass Temperature Current Amount of substance Luminous intensity

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Base quantities

Measurements

SI unit Metre Second Kilogram Kelvin Ampere Mole Candela

Symbol m s kg K A mol cd

1.3

PREFIXES • Prefixes are multiples or decimals of ten. • They are useful for expressing the size of physical quantities that are too large or small. • The table below shows the common prefixes. Prefixes Mega Kilo centi milli micro nano

Value 1 000 000 1 000 0.01 0.001 0.000 001 0.000 000 001

Form 106 103 10−2 10−3 10−6 10−9

Symbol M K c m µ n

Measurements

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REVIEW QUESTIONS 1. What is physics? A. The study of numeral values and their operation. B. The study of matter, its composition, properties and interaction to form other substances. C. The study of matter, energy and their interaction. D. The study of living things. 2. Physics as a field of science deals with A. properties of matter. B. how matter forms a living thing. C. physical quantities of matter and its motion. D. all the above. 3. Magnitude is defined as A. property of the physical quantity. B. quality of the physical quantity. C. standard measure of the physical quantity. D. size of the physical quantity. 4. Unit is defined as the A. Property of the physical quantity B. Quality of the physical quantity C. Standard measure of the physical quantity D. Size of the physical quantity 5. All physical quantities have A. Magnitude and unit B. Magnitude and size C. Quantity and quality D. Quality and unit 6. Which of the following consists of base quantities. A. Length, mass and speed B. Length, mass and time C. Time, mass and density D. Mass, volume and speed 7. The following consists of derived quantities A. Length, mass and speed B. Length, mass and time C. Time, mass and density D. Force, volume and speed 8. The SI unit of time is A. Second B. Minute C. Hour D. Day 9. The SI unit of mass is A. Gram B. Kilogram C. Metre D. Kilometre 10. The SI unit of length is A. Gram B. Kilogram C. Metre D. Kilometre 4

Measurements

11. The SI unit of temperature is A. Degree celcius B. Kelvin C. Mole D. Ampere 12. The SI unit of current is A. Candela B. Kelvin C. Mole D. Ampere 13. The SI unit of amount of substance is A. Degree celcius B. Kelvin C. Mole D. Ampere 14. The SI unit of luminous intensity is A. Candela B. Kelvin C. Mole D. Ampere 15. Which of the following value is the same as 1000 seconds? A. 10−3 ms B. 103 ms C. 10−6 ms D. 106 ms

Measurements

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SOLUTIONS 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15.

C C D C A B D A B C B D C A D

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Measurements

2.0

SCALAR AND VECTOR QUANTITIES

Introduction Imagine you walked 50 m to the right and walked back 40 m to the left. What is the total distance you covered? It’s 90 m (50 m + 40 m). How far are you positioned from the starting point? 10 m (50 m − 40 m). Why is there a difference between the first question and the second question? The first question asked only about the size of the distance covered. However, the second question asked about the size of distance and direction moved in relation to the starting point. In physics, quantities that consider only the size are called scalar quantities while quantities that consider both the size and direction are called vector quantities.

Specific outcomes This unit covers scalar and vector quantities. By the end of this unit, you will be able to: ❖ Scalar quantity: • Define a scalar quantity • Give examples of scalar quantities ❖ Vector quantity: • Define a vector quantity • Give examples of vector quantities • Differentiate scalar quantities and vector quantities • Calculate addition and subtraction of vectors

Scalar and Vector Quantities

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SCALAR AND VECTOR QUANTITIES 2.1

SCALAR QUANTITY • A scalar quantity is a quantity that only has magnitude but no direction. • Scalar quantities are indicated only by the size of the quantity. • Examples of scalar quantities are length, time, mass, volume, density, distance, speed, area, pressure, work and temperature.

2.2

VECTOR QUANTITY • A vector quantity is a quantity that has both magnitude and direction. • Vector quantities are indicated by the size and the direction of motion of the quantity. • Examples of vector quantities are displacement, velocity, acceleration, weight, gravity, and force.

2.3

ADDING VECTORS QUANTITIES • When adding vector quantities, consider both the magnitude and direction of motion of the quantity. • If both vector quantities are moving in the same direction, the resultant vector is the sum of all vectors. VR = V1 + V 2

•

where VR is the resultant vector V1 and V2 are individual vectors moving in the same direction

If vector quantities move in the opposite direction, the resultant vector is the difference between the forward and backward vectors. VR = V1 – V 2

where VR is the resultant vector V1 and V2 are individual vectors moving in the opposite direction

❖ EXAMPLES Find the resultant vector of the following forces

❖ SOLUTIONS 1. VR = V1 + V2 = 1N + 3N = 4N 3.

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VR = V1 + V2 − V3 = 3N + 1N − 3N = 1N

Scalar and Vector Quantities

2. VR = V1 − V2 = 1N − 3N = −2N

REVIEW QUESTIONS 1. What is a scalar quantity? A. A quantity that can be measured B. A quantity that has both magnitude and direction C. A quantity that has a unit D. A quantity that only has magnitude but no direction 2. What is a vector quantity? A. A quantity that can be measured B. A quantity that has both magnitude and direction C. A quantity that has a unit D. A quantity that only has magnitude but no direction 3. A quantity that also considers the direction of motion of an object A. Vector quantity B. Scalar quantity C. Magnitude D. Unit 4. Which of the following is a scalar quantity? A. Velocity B. Gravity C. Speed D. Force 5. Which of the following is a scalar quantity? A. Acceleration B. Distance C. Displacement D. Force 6. Which of the following is a vector quantity? A. Gravity B. Length C. Speed D. Mass 7. Which of the following is a vector quantity? A. Volume B. Area C. Mass D. Weight 8. Two forces 10 N and 15 N are pushing an object in the same direction. What is the resultant force acting on an object? A. 25 N B. 5 N C. −5 N D. 10 N 9. A 120N force and a 300 N force are acting on an object in the same direction. What is the resultant force? A. 180 N B. 200 N C. 420 N D. 550 N

Scalar and Vector Quantities

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10. A forces 50 N and 30 N are acting on the same object. A 50 N force is pulling the object to the left while a 30 N force is pulling an object to the right. What is the resultant force and direction of motion? A. 80 N to the right B. 80 N to the left C. 20 N to the right D. 20 N to the left 11. A bag of cement weighing 500 N is acted upon by an upward force of 720 N. What is the resultant force and direction of motion? A. 220 N upward B. 220 N downward C. 1220 N upward D. 1220 N downward

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Scalar and Vector Quantities

SOLUTIONS 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

D B A C B A D A C D A

Scalar and Vector Quantities

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3.0

LENGTH

Introduction “How far is the journey?”, “How tall is the building?”, “How thick is the hair?”. What is common in all these questions? The distance between two points. The first is about the distance from the start to the end of the journey. The second is about the distance from the ground to the top of the building. The third is about the distance between one side of the hair to the other. The distance between two points is called length. This unit cover instruments commonly used to measure the length in physics and how to use them.

Specific outcomes By the end of this unit, you will be able to: ❖ Define length ❖ State the SI unit of length, and other units ❖ Know the instrument used for measuring length ❖ Measuring tape and metre rule: • Know how to read the metre rule • State the causes of errors ❖ Vernier calipers: • Know the parts of the vernier calipers • Know how to read the vernier calipers ❖ Micrometer screw gauge: • Know the parts of the micrometer screw gauge • Know how to read the micrometer screw gauge

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Length

LENGTH 3.1

LENGTH • Length is defined as the measure of distance between two points. • SI unit: metre (m); other common units used to measure length are centimetre (cm), kilometre (km). • Instruments for measuring length. 1. Measuring tape • A measuring tape measures long lengths. • It is 1mm accuracy. • Example of length measured; length of the classroom, width or height of the house. 2. Metre rule • A metre rule measures medium lengths. • It is 1mm accuracy. • Example of length measured; length of the book, chair, desk. 3. Vernier caliper • A vernier caliper measures short lengths. • It is 0.1mm accuracy. • Example of length measured; width of the pen, the diameter of the water tap. 4. Micrometer screw gauge • Micrometer crew gauge measures very short lengths. • It is 0.001 mm accuracy. • Example of length measured; diameter of the hair, the thickness of a razor blade. Correct reading: 12 cm Wrong reading:

3.2

MEASURING TAPE AND METRE RULE Wrong reading: 13 cm 11 cm • How to use a metre rule 1. Put the zero mark at the end of an object. 2. Read the mark at the other end of an object. • Causes of reading errors 5 10 20 0 15 1. Zero error: the value error due to not placing the object at the zero mark of a ruler. 2. Parallax error: the value error due to the position of the eye when reading. The eye must be vertically placed above the mark of the reading scale.

3.3

VERNIER CALIPER • A vernier caliper consists of the main scale, vernier scale and two jaws. • Object to be measured is put between jaws. • The outer jaw is used to measure external diameters such as the diameter of a solid metal rod. • The inner jaw is used to measure internal diameters such as the internal diameter of the water tap.

Length

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❖

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HOW TO READ A VERNIER CALIPER i. Read the main scale by reading the mark on the main scale before the vernier scale's zero mark. ii. Read the vernier scale by reading the mark on the vernier scale in line with the main scale. Then, multiply the value of the vernier scale by 0.01cm. iii. Add the main scale and vernier scale.

Length

3.4

MICROMETER SCREW GAUGE • A micrometer crew gauge consists of a ratchet, thimble, sleeve, frame, spindle and anvil. • The object to be measured is gripped between the anvil and the spindle using the ratchet.

❖ HOW TO READ A MICROMETER SCREW GAUGE i. Read the main scale by reading the mark on the sleeve before the thimble. ii. Read the circular scale by reading the mark in line with the horizontal line of the main scale. Then, multiply the value by 0.01 mm. iii. Add the main scale and the circular scale.

Length

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REVIEW QUESTIONS 1. Which of the following devices can be used to measure the thickness of a banknote? A. Micrometer screw gauge B. Vernier caliper C. Metre rule D. Measuring tape 2. Vernier caliper measures the A. Diameter B. Thickness C. Depth D. All the above 3. Inner jaws of vernier caliper are used to measure A. Thickness B. Internal depth C. Outer diameter D. Internal diameter 4. Which instruments can be used to accurately measure a piece of wire about 1m long and 4mm in diameter. A. Metre rule for length; vernier caliper for diameter B. Metre rule for length; micrometer screw gauge for diameter C. Metre for both length and diameter D. Vernier caliper for length; micrometer screw gauge for diameter 5. The figure below shows the reading on a vernier caliper.

A. What is the reading on? i. main scale ii. vernier scale B. What is the overall reading of a vernier caliper? C. After zeroing the instrument it read +0.02cm. What is the actual reading of the object measured? 6. The figure shows the reading of a device used to measure the thickness of the material.

A. What is the name of a device? B. What is the thickness of the material? C. What are some of the precautions to be taken when using this device?

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Length

SOLUTIONS 1. 2. 3. 4.

A D D B

5. A. i. 6.3cm ii. 0.05cm B. 6.35cm C. 6.33cm (6.35cm − 0.02cm) 6. A. Micrometer screw gauge B. 4.11mm C. Precautions • The device should be dry and clean • The device should be zeroed • Avoid over tightening the thimble

Length

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4.0

TIME

Introduction “The event will commence in the afternoon and end in the evening.” What physical quantity can be used to measure the period of occurrence of events? It’s time! Time is a physical quantity used to tell the progression of events. How was time measured in ancient times? Different methods were used. The commonest was the position of the sun and phases of the moon. Sometimes inventions were also used such as the hourglass. This instrument consisted of two rounded glass bulbs that were separated by a narrow neck. It contained particles of sand within it. When turned upside down, a measured amount of sand would drop from the top of the glass to the bottom in a specific period of time. In the 17th century, a pendulum clock was invented using the pendulum law discovered by Galileo Galilei. A pendulum is an object hanging from a fixed point that swings back and forth. According to the pendulum law, it takes the same amount of time to make one complete sing; that is, swing back and forth.

Specific outcomes This unit introduces the simple pendulum. It covers the terminologies used to describe the pendulum and calculations. By the end of this unit, you will be able to: ❖ State the SI unit of time ❖ State the instruments used to measure time ❖ Simple pendulum: • Label the parts of the simple pendulum • Define oscillation • Define period, and work out period calculations • Define frequency, and work out frequency calculations • State factors that affect the period of the pendulum

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Time

TIME 4.1

TIME • SI unit of time: second (s). • Other units: minute, hour, day, month, year, century. • Instruments used to measure time: • Stopwatch • Clock • Pendulum

4.2

SIMPLE PENDULUM • It consists of a bob attached by a string hanging from a supporter. • The length of the pendulum (l) is measured from the supporter to the middle of a bob. 1. OSCILLATION • Oscillation of the pendulum is the swing of a bob back and forth. • One complete oscillation of a simple pendulum is the swing of a bob forth and back to its starting point. For example, a swing from point A through B to point C and back to point A is one complete oscillation. 2. PERIOD (T) • Period is the time taken by a pendulum to complete one oscillation. • The formula for calculating the period of the pendulum.

T=

t n

where T = Period (in seconds) t = time taken for n oscillations n = number of oscillations

3. FREQUENCY (f ) • Frequency is the number of oscillations completed in one second. • SI unit: Hertz (Hz). • The formula for calculating the frequency of the pendulum.

f =

1 T

where f = frequency (in Hz) T = Period (in seconds)

4. FACTORS THAT AFFECT PERIOD OF PENDULUM i. Length of the pendulum. The length of the pendulum is directly proportional to the period. The longer the length the higher the period. ii. Acceleration due to gravity. Gravity is inversely proportional to the period. Therefore, greater gravity decreases the period of the pendulum. ➢ Note: Mass of bob does not affect the period of a pendulum.

Time

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REVIEW QUESTIONS 1. The measure of the number of oscillations per second is called A. Period B. Amplitude C. Frequency D. Vibration 2. A motion of a pendulum repeated over and over again is called A. Frequency B. Time C. Oscillation D. Period 3. The amount of time required for one oscillation is called A. Frequency B. Period C. Hertz D. Amplitude 4. If the same pendulum is performed on earth and the moon, its period on the moon will A. Increase B. Decrease C. Stay the same D. Swing forever 5. If the mass of a bob of a pendulum is increased, the period will A. Increase B. Decrease C. Stay the same D. Become zero 6. If the length of the pendulum increases, the period will A. Increase B. Decreases C. Stay the same D. Swing forever 7. A pendulum makes one complete swing in 1.8 seconds. What is its frequency A. 0.56 Hz B. 0.59 Hz C. 1.8 Hz D. 2.3 Hz 8. A pendulum swings back and forth 5 times in 4 seconds. What is its period? A. 0.50 seconds B. 0.80 seconds C. 1.25 seconds D. 1.35 second

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Time

9. The diagram shows a pendulum oscillating.

A. It takes 1.5s to move from A to C and back to B. What is the period of the pendulum? B. What is its frequency? C. How long does it take for 5 times complete oscillation?

Time

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SOLUTIONS 1. 2. 3. 4. 5. 6. 7. 8.

C C B A C A A B

9. A. T = =

t

T=? t = 1.5 s

n 1.5 0.75

= 2s B.

f= =

1 T 1

n=

3 4

= 0.75 (three quarter oscillation) f=? T = 2s

2

= 0.5 Hz C.

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Time

t=T×n =2×5 = 10s

T = 2s n=5 t=?

5.0

MASS AND WEIGHT

Introduction Objects differ in heaviness. The heaviness of this book is different from that of the pen, phone, car, airplane, etc. What physical quantities could be used to measure the amount of heaviness? It is mass and weight! What is the difference between mass and weight? What instruments are used to measure the mass and weight of an object? What is the relationship between mass and weight? This unit covers these questions.

Specific outcomes By the end of this unit, you will be able to: ❖ Mass: • Define mass • Write the SI unit of mass correctly • Convert from kilogram to gram and from gram to kilogram • State the instruments used to measure the mass of an object • Label the parts of the beam balance and how to read it ❖ Weight: • Define weight • Know the SI unit of weight • State the instruments used to measure weight ❖ Differentiate mass and weight ❖ State the relationship between mass and weight ❖ Work out weight, mass calculations using W = mg

Mass and Weight

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MASS AND WEIGHT 5.1

MASS • Mass is defined as the amount of matter in an object. • The mass of an object does not change with location. For example, the mass of an object on earth and the moon is the same. • SI unit of mass: kilogram (kg) • Other units: gram, tonne. 1kg = 1000g 1000kg = 1 tonne • Instruments for measuring mass: • Triple beam balance • Electronic pan balance 1. TRIPLE BEAM BALANCE • A triple beam balance consists of a pan, base, beams, three riders and a balance pointer. • It measures mass to 0.1 gram accuracy.

❖ HOW TO USE A TRIPLE BEAM BALANCE i. Clean the pan. ii. Move all riders to the far left. iii. Place an object to be measured on the pan. iv. Start with the largest rider. Move the rider to the right into each groove. When the pointer drops below the zero mark, move the rider one groove backward. v. Slide the middle rider like in step iv. vi. Slide the smallest rider until the pointer is in line with the zero mark. vii. Add the values of all three riders.

5.2

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WEIGHT • Weight is defined as the force of gravity acting on an object. • The weight of an object changes with location. It increases with an increase in gravity and decreases with a decrease in gravity. The weight of an object on earth is different from its weight on the moon. • SI unit of weight is Newton (N). • Weight is measured by a spring balance. • The diagram on the next page shows a spring balance.

Mass and Weight

❖

DIFFERENCES BETWEEN MASS AND WEIGHT Mass Mass of an object is the same everywhere Mass can never be zero Mass is a scalar quantity Measured by beam balance Measured in kilogram

5.3

Weight Can change with change in gravity With zero gravity, weight is zero Weight is a vector quantity Measured by a spring balance Measured in Newton

THE RELATIONSHIP BETWEEN MASS AND WEIGHT • The weight of an object is directly proportional to its mass. • Therefore, as the mass of an object increases its weight increases, provided that gravity is constant. • The formula for calculating weight. W = mg

where W = weight (in Newton) m = mass (in kg) g = acceleration due to gravity (≈ 10m/s2)

❖

EXAMPLE What is the weight of an 80kg man: A. on earth? B. on moon? C. in space?

❖

SOLUTION A. W = mg = 80 × 10 = 800 N

earth’s gravity = 10 m/s2

Mass and Weight

25

B.

W = mg = 80 × 1.67 = 133.6 N

C.

W = mg = 80 × 0 =0N

26

Mass and Weight

moon’s gravity = 1.67 m/s2

space gravity = 0 m/s2

REVIEW QUESTIONS 1. Mass is defined as A. The amount of space occupied by an object B. The amount of matter in an object C. The force of gravity acting on an object D. All the above 2. Weight is defined as A. The amount of space occupied by an object B. The amount of matter in an object C. The force of gravity acting on an object D. All the above 3. The SI unit of mass A. Metre B. Newton C. Gram D. Kilogram 4. The SI unit of weight is ……….. A. Metre B. Newton C. Gram D. Kilogram 5. Which of the following conversion is correct? A. 10kg = 1000g B. 1kg = 1000 tonne C. 100g = 0.1kg D. 100kg = 10g 6. Which statement correctly describes the mass of an object? A. The amount of space taken up by an object B. The amount of matter in an object C. The amount of substance in an object D. The pull of gravity on the object 7. What statement correctly describes the weight of an object? A. The amount of space taken up by an object B. The amount of matter in an object C. The amount of substance in an object D. The pull of gravity on the object 8. What is the mass of a 5kg steel block on earth? A. 5kg B. 50kg C. 8.35kg D. 0kg 9. A 5kg steel block was taken to the moon. What is the mass of the block on the moon? A. 5kg B. 50kg C. 8.35kg D. 0kg 10. A 5kg steel block was taken in space. What is the mass of the block in space? A. 5kg B. 50kg C. 8.35kg D. 0kg Mass and Weight

27

11. What is the weight of a 5kg steel block on earth? A. 5kg B. 50kg C. 8.35kg D. 0kg 12. A 5kg steel block was taken to the moon. What is the weight of the block on the moon? A. 5kg B. 50kg C. 8.35kg D. 0kg 13. A 5kg steel block was taken in space. What is the weight of the block in space? A. 5kg B. 50kg C. 8.35kg D. 0kg 14. What is the mass of a 15000 N vehicle? 15. Calculate the weight of a 50g book.

28

Mass and Weight

SOLUTIONS 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13.

B C D B C B D A A A B C D

14. m = =

w g 15000

m=? W = 15 000N g = 10 N/kg

10

= 1500 kg

15. W = mg = 0.05 × 10 = 0.5 N

W=? m = 50g ÷ 1000g = 0.05kg g = 10 N/kg

Mass and Weight

29

6.0

VOLUME AND DENSITY

Introduction All objects occupy some amount of space. Whether placed on a table or held in your hands, this book has occupied some amount of space. Smaller objects tend to occupy a smaller amount of space than larger objects. For example, the space occupied by a toy car is smaller than the space occupied by a real car. What physical quantity is used to measure the amount of space occupied by an object? It is called volume! This unit covers how to measure the volume of objects. It also includes the relationship between density, mass and volume.

Specific outcomes By the end of this unit, you will be able to: ❖ Volume: • Define volume • State the SI unit of volume • Measure the volume of: • Regular solid objects • Irregular solid objects • Liquid substances ❖ Density: • Define density • State the SI unit of density m • Work out density, mass, volume calculations using P = • •

30

Define relative density Calculate the relative density of the substance

Volume and Density

V

VOLUME AND DENSITY 6.1

VOLUME • Volume is defined as the amount of space occupied by an object. • SI unit of volume: cubic metre (m3). ❖ HOW TO MEASURE VOLUME OF 1. REGULAR SOLID • The volume of regular solid objects is obtained by using a formula for calculating the volume of that object. • For example, the volume of a rectangular metal rod is calculated using the formula for calculating rectangular objects. • The common volume of the regular solid object calculated in physics are rectangular objects such as a rectangular metal rod, rectangular wood and rectangular rock; and cylindrical objects such as a cylindrical metal rod, cylindrical wood and cylindrical instruments. ❖ FORMULA FOR RECTANGULAR OBJECTS V=l×b×h

where V = volume of an object l = length of an object b = breadth of an object h = height of an object

❖ FORMULA CYLINDRICAL OBJECTS V = πr2 h

where V = volume of an object π = 3.142 r = radius of an object h = height of an object

Volume and Density

31

2.

IRREGULAR SOLIDS • To measure the volume of irregular solid objects, the displacement method is used. • Some objects that can be measured using this method include irregular metal and irregular rock. • Steps involved in the displacement method. i. Put water in a cylinder and take the initial reading. ii. Lower an object in a cylinder and take the final reading. iii. The volume of an object equals the volume of water displaced. Vo = Vf − Vi

where Vo = volume of an object Vf = final volume Vi = initial volume

Volume of water without an irregular solid (Vi) = 12cm3

3.

32

Volume of water with an irregular solid (Vi) = 23cm3

Vo = Vf − Vi = 23 − 12 = 11cm3

VOLUME OF LIQUID SUBSTANCES • The volume of liquid substances is measured by a measuring cylinder. • Steps involved: i. Check that the measuring cylinder is on a flat surface. ii. Pour the liquid to be measured in a cylinder. iii. Read the scale at the meniscus. ➢ For the concave meniscus, read a volume from the bottom of the meniscus. Water displays a concave meniscus in a cylinder. ➢ For the convex meniscus, read a volume from the top of the meniscus. Mercury displays a convex meniscus in a cylinder.

Volume and Density

6.2

DENSITY • Density is defined as mass per unit volume. • It depends on the arrangement of particles and the state of an object. For example, solids have particles closely packed therefore have a relatively higher density. • SI unit of density: kg/m3 or g/cm3. • The formula for calculating density is given as: P=

6.3

m V

where P = density (in kg/m3 or g/cm3) m = mass (in kg or g) V = volume (in m3 or cm3)

RELATIVE DENSITY • Relative density is defined as the ratio of the density of a substance to the density of water. • The density of water is 1 g/cm3. Any substance with a density greater than 1 g/cm3 or a relative density greater than one will sink in water. Any substance with a density less than 1 g/cm3 or a relative density less than one will float on water. • Relative density has no unit as it is a ratio of the same units. Therefore they get cancelled. • The formula for calculating relative density. Relative density =

density of substance density of water

=

mass of any volume of substance mass of equal volume of water

Volume and Density

33

REVIEW QUESTIONS 1. Which statement correctly describes the volume of an object? A. The amount of space taken up by an object B. The amount of matter in an object C. The amount of substance in an object D. The pull of gravity on the object 2. Which of the following is a method used to measure the volume of an irregular solid object? A. Calculate the volume of an object mathematically B. Weigh an object on a beam balance C. Use displacement method D. All the above 3. Which of the following is a method used to measure the volume of water? A. Pour water in a cylinder and get the reading B. Pour water in a cylinder and weigh it C. Pour water in a cylinder and calculate the volume of a cylinder D. None of the above 4. Which of the following is a method used to the volume of a rectangular metal? A. Calculate volume by using V = πr2h B. Calculate volume by using V = lbh C. Calculate volume by using A = πr2 D. Calculate volume by using A = lb 5. The following is true about a substance with a relative density of less than 1. The substance will A. Sink in water B. Suspended in C. Heavier than water D. Lighter than water 6. What is the volume of a metal bar which is 4cm long, 1 cm wide and 2 cm high? 7. A cylinder contained 20 ml of water. After lowering a stone in it water raised to 55 ml. What is the volume of a stone? 8. A 10cm by 10cm by 10 cm wooden cube weighs 900g. A. What is the density of a cube? B. If it emerged in water, would it sink or float given that the density of water is 1g/cm3? 9. A 32g stone is placed in a cylinder containing 50ml of water. Water raised to 67ml. calculate the density of a stone 10. 200g mercury is poured in a cylinder to the 40ml mark. What is the density of mercury? 11. Students performed an experiment to determine the relative density of cooking oil. They used a known volume of a glass bottle. After weighing the bottle, bottle with water and bottle with cooking oil, they obtained the following results: Mass of empty bottle = 150g Mass of bottle filled with water = 900g Mass of bottle filled with cooking oil = 825g A. What is the mass of: i. Water? ii. cooking oil? B. Calculate the relative density of cooking oil. C. If cooking oil is mixed with water, would it sink or float?

34

Volume and Density

SOLUTIONS 1. 2. 3. 4. 5.

A C A B D

6. V = lbh =4×1×2 = 8 cm3 7. Vstone = Vf × Vi = 55ml − 20ml = 35ml 8. A.

Vcube = lbh = 10cm × 10cm × 10cm = 1000cm3 Pcube = =

m V 900 1000

= 0.9g/cm3 B.

Relative density of cube = =

density of cube density of water 0.9g/cm3

1g/cm3 = 0.9 As the relative density of the wooden cube is less than 1, it will float on water 9. Vstone = Vf – Vi = 67ml – 50ml = 17ml

P= =

m V 32g 17ml

= 1.88g/cm3

10. P =

200g 40cm3

= 5g/cm3

Volume and Density

35

11. A.

i. Mass of water = mass of bottle filled with water – mass of empty bottle = 900g – 150g = 750g ii. Mass of cooking oil = mass of bottle filled with cooking oil – mass of empty bottle = 825g – 150g = 675g

B.

Relative density of cooking oil = =

mass of cooking oil mass of equal volume of water 675g 750g

= 0.9

36

Volume and Density

7.0

CENTRE OF GRAVITY

Introduction Have you ever tried to balance a pen on your finger? Try it! Place your pen or pencil on your finger. Move it until it cannot fall sideways without supporting it. The point at which the pen balances on your finger is the centre of gravity of that pen. Usually, it balances near to or at the midpoint of the pen. The centre of gravity is the point on an object where the weight of an object acts. What is the importance of knowing the centre of gravity of an object? Try this! Put your pen or pencil on the table. Lift it on one end. What happens? Likely it rotates at the axis as you lift it. Put it back on the table and lift it again, this time at the centre of gravity, the point where it balanced on your finger. What happens? Likely it will be raised without rotating about the axis. Therefore, if an object is lifted or pushed at its centre of gravity, its whole points will move as if they are a single point. The object will not rotate about any axis. However, if an object is lifted or pushed at any point other than the centre of gravity, it will rotate about the axis. The stability of an object is also affected by the centre of gravity. Therefore, knowing the centre of gravity of an object helps to know if the object is stable or unstable. Then, how can you increase the stability of an object? This unit covers this question and many more.

Specific outcomes By the end of this unit, you will be able to: ❖ Centre of gravity: • Define centre of gravity • Find the centre of gravity of lamina objects ❖ Stability: • Define stability • State how to increase the stability of an object • Describe the three states of stability

Centre of Gravity

37

CENTRE OF GRAVITY 7.1

CENTRE OF GRAVITY • The centre of gravity of an object is defined as the imaginary point in an object where its whole weight acts. • How to find the centre of gravity of a lamina object; for example, a lamina paper: i. Make at least three holes on different edges of the paper. ii. Attach strings to each hole. iii. Hang the paper with one string. iv. Hang a plumb line in a hole of step iii. and mark the line it passes through. v. Repeat steps iii. and iv. to get more lines. vi. Find the intersection of the three lines. Their intersection point is the centre of gravity.

7.2

STABILITY • Stability is defined as the ability of an object to return to its original position after being slightly displaced. • Stability is affected by the position of the centre of gravity. • When the centre of gravity is low, the object is more stable and less likely to tip over when tilted. • Objects with a high centre of gravity are less stable and easily tip over when tilted. ❖ HOW TO INCREASE THE STABILITY OF AN OBJECT i. Lower the centre of gravity. This can be done by making an object short. ii. Increase the base area. ❖ EXAMPLES Which of the following is more stable, and why?

38

Centre of Gravity

❖

SOLUTIONS 1. A is more stable because it is short therefore has a low centre of gravity than B.

2.

7.3

B is more stable because it has a larger base area.

STATE OF STABILITY • There are three states of stability. 1. STABLE EQUILIBRIUM • An object is in stable equilibrium if it returns to its original position when slightly displaced. • For example, a book on the table.

2. UNSTABLE EQUILIBRIUM • An object in unstable equilibrium does not return to its original position when slightly displaced. • For example, an erect pencil on the table.

3. NEUTRAL EQUILIBRIUM • An object is in neutral equilibrium if it remains in the displaced position when it is displaced. • For example, a ball on flat ground.

Centre of Gravity

39

REVIEW QUESTIONS 1. The famous Leaning Tower of Pisa doesn’t topple over because its centre of gravity A. Stabilized by its structure B. Very high up in the building C. In the same place as its centre of mass D. Kept within its base 2. If an object is in stable equilibrium, any displacement A. Raise its centre of gravity B. Lower its centre of gravity C. Increase its mass D. Decrease its mass 3. If an object is in unstable equilibrium, any displacement A. Raise its centre of gravity B. Lower its centre of gravity C. Increase its mass D. Decrease its mass 4. A fish suspended in water is in what kind of equilibrium? A. Stable B. Unstable C. Neutral D. None of the above 5. A picture hangs on a wall from a wire that is passed over a supporting nail. The picture is in stable equilibrium because A. Its centre of gravity is directly below the supporting nail B. Any slight push will raise its centre of gravity C. After any small push it returns to its original position D. All the above 6. Indicate the following figure as stable, unstable or neutral

40

Centre of Gravity

SOLUTIONS 1. 2. 3. 4. 5. 6.

D A B C D A. Stable B. Neutral C. Unstable

Centre of Gravity

41

8.0

DISTANCE AND DISPLACEMENT

Introduction David’s house is located 100 metres from the market. He walks to the market and comes back home. What is the total length he took? Definitely it is 200 metres. How far is he from the house after the trip to the market? 0 metres. What is the difference between the two questions? The first question asked about the total distance David covered. The second question asked about the displacement of David from his house. What is the difference between distance and displacement? This unit answers this question.

Specific outcomes By the end of this unit, you will be able to: ❖ Distance: • Define distance • State the units of distance • Convert m to km, km to m, m to cm, cm to m ❖ Displacement: • Define displacement • State the units of displacement ❖ Differentiate distance and displacement ❖ Work out distance and displacement calculations

42

Distance and Displacement

DISTANCE AND DISPLACEMENT 8.1

DISTANCE • Distance is defined as the total length taken between two points. • It is a scalar quantity. • SI unit of distance: metre (m); other units are kilometre (km), centimetre (cm), mile (mi): 1m = 100cm 1km = 1000m 1mi = 1.6km

8.2

DISPLACEMENT • Displacement is defined as the shortest length taken by an object in a specific direction. • It is a vector quantity. • SI unit: metre (m).

❖

•

If ABCD is a path taken from point A to D, then: • distance = ABCD, • displacement = AD.

•

If a car moves from point L through M, N, O to point P, then: • distance covered = 3km + 1km + 2km + 1km = 7km, • displacement = 2km south.

DIFFERENCES BETWEEN DISTANCE AND DISPLACEMENT Distance Total length taken between two points It is a scalar quantity Magnitude depends on path covered Always positive Can never be zero

Displacement Shortest length taken between two points It is a vector quantity Magnitude not depend on path covered Can be positive or negative Can be zero

Distance and Displacement

43

❖

EXAMPLES 1. A car moves 7km north then 5km east. What is the distance covered? What displacement is covered? 2. A race track is 800m long. If a racer does one lap, what distance and displacement are covered by a racer? 3. Below is the path taken by David. What is the distance and displacement moved?

❖

SOLUTIONS 1. Distance = 7km + 5km = 11km Displacement = 7km north and 5km east 2. Distance = 800m Displacement = 0m 3. Distance = 1km + 3km = 4km Displacement = 1km − 3km = −2km east (or 2km west)

44

Distance and Displacement

9.0

SPEED AND VELOCITY

Introduction “The car is moving at 500 km/hr”. “The car is moving at 50 km/hr south”. What is the difference between the two statements? The first statement is describing how fast the car is moving. The second statement is describing how fast the car is moving in a particular direction. In terms of physics, the first statement is about the speed of the car while the second statement is about the velocity of the car. In everyday life, speed and velocity are used interchangeably. However, they are different terms. Speed describes how fast an object is moving regardless of the direction. It can never be zero or negative for a moving object. Velocity is used to describe how fast an object is changing its position. It has a reference point. When an object moves forward from the reference point, its velocity is positive. Suppose the object turns and moves toward the reference point. As it crosses the reference point, its velocity is zero. If it moves back beyond the reference point, its velocity becomes negative. Therefore, unlike speed which is always positive for a moving object, velocity can be positive, zero or negative for a moving object.

Specific outcomes This unit covers speed and velocity and how to calculate them. By the end of this unit, you will be able to: ❖ Speed: • Define speed • State the units of speed •

Work out speed, distance, time calculations using S =

❖ Velocity: • Define velocity • State the units of velocity •

d t

Work out velocity, displacement, time calculations using V =

❖ Differentiate distance and displacement

d t

Speed and Velocity

45

SPEED AND VELOCITY 9.1

SPEED • Speed is defined as the distance covered per unit time. • It is a scalar quantity. • SI unit of speed is m/s or km/hr. • The formula for calculating speed is given as: S=

d t

where S = distance (in m/s or km/h) d = distance covered (in m or hr) t = time taken (in s or hr)

Average speed =

total distance traveled total time taken

❖ EXAMPLES 1. A car travels 500m in 50s. What is the speed of a car? 2. A travelled at 70 km/hr in 5 hr. What distance was covered? 3. A girl runs for a distance of 140m at a speed of 7m/s. How long does it take her to cover this distance? 4. A truck travels 60km/hr in the first hour and 80km/hr in the second hour. What is the average speed of a truck? ❖ SOLUTIONS d 1. S = t 500m = 50s = 10m/s

S=? d = 500m t = 50s

2. d = S × t = 70km/h × 5hr = 350 km

3. t = =

d S 140m

S = 70km/hr d=? t = 5hr

S = 7m/s d = 140m t=?

7m/s = 20s

4. Total distance covered = 60 + 80 = 140km/hr Total time taken = 1hr + 1hr = 2hr total distance travelled 140 Average speed = = = 70km/hr total time taken 2

46

Speed and Velocity

9.2

VELOCITY • Velocity is defined as the rate of change of displacement per unit time. • It is a vector quantity. • SI unit of velocity is m/s or km/hr. • The formula for calculating velocity. V=

d t

where V = velocity (in m/s or km/hr) d = displacement (in m or km) t = time (s or hr)

Average velocity =

• • •

resultant displacement time taken

Two bicycles traveling at 7km/hr, bicycle X toward north and Y toward east, have same speed but different velocity. The speed of bicycle X and Y is 7km/hr. The velocity of bicycle X is 7km/hr north, the velocity of bicycle Y is 7km/hr east.

❖ DIFFERENCES BETWEEN SPEED AND VELOCITY Speed How fast an object is moving It is a scalar quantity Has only a magnitude Can never be zero Can never be negative

Velocity How fast an object is changing position It is a vector quantity Has both magnitude and direction Can be zero Can be positive or negative

❖ EXAMPLES 1. A bird travels to the right a displacement of 35 metres in 5 seconds. What is the bird’s velocity? 2. A car travels south at 40m/s for 18 seconds. What is the car’s displacement? 3. How long will it take an elevator to travel 25 metres if it is traveling at a constant velocity of 5m/s? 4. A car travels along a straight road to the east 50 metres in 3 seconds, then goes to the west 150 metres in 7 seconds. Find the average speed and velocity of a car? ❖ SOLUTIONS d 1. V = t 35 = 5 = 7m/s 2.

d=V×t = 40 × 18 = 720m

V=? d = 35m t = 5s

V = 40m/s d=? t = 18s

Speed and Velocity

47

3.

t= =

d

V = 5m/s d = 25m t=?

V 25

5 = 5s

4.

Total distance covered = 50 + 150 = 200m Resultant displacement = 50 − 150 = −100m Total time taken = 7 + 3 = 10s Average speed = =

total distance covered total time taken 200

10 = 20m/s

Average velocity =

resultant velocity total time taken 100

= − 10 = −10m/s

48

Speed and Velocity

10

ACCELERATION

Introduction "How fast is the car speeding up?" This statement describes the acceleration of the car. Acceleration is the rate of change of velocity. It is a vector quantity; therefore, it can be positive, zero or negative. An object moving at a constant speed in a circular path is changing its position, hence, it is acceleration. This unit covers acceleration and how to calculate acceleration problems. It also includes the motion of a free falling object.

Specific outcomes By the end of this unit, you will be able to: ❖ Acceleration: • Define acceleration • State the units of acceleration •

Work out acceleration, velocity, time calculations using a =

❖ Uniform acceleration: • Master uniform acceleration formulae • Work out uniform acceleration calculations ❖ Free falling objects: • Describe the motion of a free falling object • Work out free falling calculations

v

u 𝑡

Acceleration

49

ACCELERATION 10.1 ACCELERATION • Acceleration is defined as the rate of change of velocity per unit time. • It describes how fast an object is changing its motion. • For example, a car is increasing its motion such that in 1 second it is moving at 10m/s, in 2 seconds at 20m/s, in 3 seconds at 30m/s. Therefore, it can be said that a car is accelerating at 10m/s per second or written as 20m/s2. • Acceleration is a vector quantity. • Unit of acceleration: m/s2. • The formula for calculating acceleration

• •

u

v

a =

𝒕

where a = acceleration (in m/s2) v = final velocity (in m/s) u = initial velocity (in m/s) t = time (in sec)

If the final velocity is larger than the initial velocity, it is positive acceleration. Positive acceleration indicates that an object is speeding up. If the final velocity is smaller than the initial velocity, it is negative acceleration and indicates that an object is slowing down. Negative acceleration is also called deceleration or retardation. It is indicated by a negative sign.

❖ EXAMPLES 1. Using an example of a car changing its motion above, find the acceleration of the car using the formula. 2. A train from rest increases its velocity to 70m/s in 5 seconds. What is the train’s acceleration? 3. Nelson is driving a car at 40m/s when he saw a child in front of him. He applied brakes and came to stop in 4 seconds. What was the acceleration of Winnie’s car? ❖ SOLUTIONS 1. a = = =

u

v

𝑡 20

10 1

10 1

= 10m/s

2.

a = =

2

u

v

𝑡 70 0 5

= 14m/s2

50

Acceleration

a=? v = 20m/s u = 10m/s t =2−1 = 1sec

OR

a = = =

v

u

𝑡 30 20 1 10 1

a=? v = 30m/s u = 20m/s t=3−2 = 1sec

= 10m/s2

u = 0m/s, object from rest has zero initial velocity v = 70m/s t = 5s a=?

3.

a = =

u

v

u = 40m/s v = 0m/s, object going to rest has zero final velocity t = 4s a=?

𝑡 0

=−

40 4 40 4

= −10m/s2

10.2 UNIFORM ACCELERATION • If an object is moving with uniform acceleration; that is constant acceleration, four equations can be used to solve its motion. 1.

v = u + at

2.

s = (u + v)t

3. 4.

𝟏 2

where v = final velocity (in m/s) u = initial velocity (in m/s) s = distance travelled (in m)

𝟏

s = ut + a𝒕𝟐 𝟐

t = time taken (in s) a = acceleration (in m/s2)

v 2 = u 2 + 2as

❖ EXAMPLES 1. A jet accelerates uniformly from 15m/s to 400m/s in 35 seconds. A. Find the acceleration of the jet at this period. B. How far has the jet travelled? 2. A car is moving on a highway at 20m/s. The driver wishes to pass another car and speed up to 30m/s over a time of 4 seconds. A. What is the acceleration over the 4 seconds? B. How much distance did the car travel in 4 seconds? 3. A car from rest accelerates at 4m/s2 in 7 seconds. Find its final velocity. 4. A bus moving at 30m/s slows at a rate of 4m/s2. A. How long does it take for the bus to stop? B. How far does it travel while breaking? ❖ SOLUTIONS 1. A. v = u + at 400 = 15 + a(35) 35a = 400 − 15 35a = 385 a = 11m/s2

B.

1

s = (u + v)t 2 1

s = (15 + 400) × 35 2

s = 7262.5m

u = 15m/s v = 400m/s t = 35s a=?

u = 15m/s v = 400m/s t = 35s s=?

Acceleration

51

2.

A.

v = u + at 30 = 20 + a(4) 4a = 30 − 20 4a = 10 a = 2.5m/s2

u = 20m/s v = 30m/s t = 4s a=?

1

B. s = ut + at 2 2

s = 20 × 4 +

1 2

× 2.5 × 4 2

s = 80 + 20

u = 20m/s t = 4s a = 2.5m/s2 s=?

s = 100m

3.

v = u + at v=0+4×7 v = 28m/s

4.

A.

v = u + at t= t= t=

v

u 𝑎

0

20

−𝟒 −𝟐𝟎

u = 0m/s, object from rest has zero initial velocity v=? a = 4m/s2 t = 7s

u = 20m/s v = 0m/s, object going to rest has zero final velocity a = −4m/s, deceleration has a negative sign t=?

4

t = 5s B. v 2 = u 2 + 2as 0 2 = 20 2 + 2 × (−4)s 8s = 400 s = 50m

u = 20m/s v = 0m/s, object going to rest has zero final velocity a = −4m/s, deceleration has a negative sign s=?

10.3 FREE FALLING OBJECTS • A free falling object falls with a constant acceleration. • The acceleration of an object is due to the earth’s gravity, which is about 10m/s2. This acceleration is called acceleration due to gravity. • Therefore, the acceleration of a free falling object is equal to acceleration due to gravity ≈ 10m/s2. • If an object is dropped from a tall building, it accelerates uniformly at 10m/s2. However, as it starts from rest its initial velocity is 0m/s. • If an object is thrown up, it decelerates uniformly at −10m/s2 until it stops momentarily. At that point, its velocity is 0m/s. Then it falls back, accelerating with acceleration due to gravity.

52

Acceleration

❖

EXAMPLES 1. A ball is dropped off from a tall building. A. What is its velocity after 2 seconds? B. How far does it reach in 2 seconds? C. If the building is 25m, what will be its velocity as it just impacts the ground? 2. A stone is thrown up with a velocity of 30m/s. A. How far will the stone reach? B. How long does it take for the stone to reach the maximum height? C. How long does it take for the stone to return?

❖

SOLUTIONS 1. A. v = u + at v = 0 + 10 × 2 v = 20m/s

1

B. s = ut +

u = 0m/s, starting from rest v=? a = 10m/s2, acceleration due to gravity t = 2s

at 2

2

s=0×2+ s = 20m

1 2

× 10 × 2 2

C. v 2 = u 2 + 2as v 2 = 0 + 2 × 10 × 25 v 2 = 500 √v 2 = √500 v = 22.4m/s 2.

A.

v 2 = u 2 + 2as s= s= s=

v2

u2 2a

02

30 2

2 ( 10) 900

u = 0m/s, starting from rest a = 10m/s2, acceleration due to gravity t = 2s s=?

u = 0m/s, starting from rest v=? a = 10m/s2, acceleration due to gravity s = 25m

u = 30m/s v = 0m/s, object thrown upward has zero final velocity at its maximum height a = −10m/s2, acceleration due to gravity, negative if object is moving against gravity s=?

20

s = 45m B.

v = u + at t= t=

u

v 𝑎 0

30 10

t = 3s C.

T = 2t T=2×3 T = 6s

u = 30m/s v = 0m/s, object thrown upward has zero final velocity at its maximum height a = −10m/s2, acceleration due to gravity, negative if object is moving against gravity t=? Time taken to return = time taken ascending + time taken descending Time taken ascending = time taken descending; therefore, Therefore, time taken to return = 2 × time taken ascending Acceleration

53

11

MOTION GRAPHS

Introduction Graphs are important in physics. They help us understand the data and the relationship between quantities. There are two types of motion graph; distance (or displacement) − time graph, and speed (or velocity) − time graph. Distance − time graph describes the relation between distance covered by an object and time taken. Speed − time graph describes the relation between speed of an object and time taken. This unit covers these two types of motion graphs.

Specific outcomes By the end of this unit, you will be able to: ❖ Distance (displacement) – time graph: • Determine the feature of the distance – time graph • Draw a distance – time graph from the data given • Calculate speed from the distance – time graph given ❖ Speed (velocity) – time graph: • Determine feature of the speed – time graph • Draw a speed – time graph from the data given • Calculate acceleration and distance from the speed – time graph given

54

Motion Graphs

MOTION GRAPHS 11.1 DISTANCE (DISPLACEMENT) – TIME GRAPH • Distance (or displacement) – time graph describes the distance covered by an object per unit time. • Features on the distance-time graph: 1. Gradient (slope line) = speed 2. Horizontal (flat) line = stop 3. Curved line = acceleration

❖ EXAMPLES 1. A car travels 60km for 2 hours. It stops at a lay-by for 30 minutes. Then it continues at a steady speed of 40km/hr for 1 hour. Draw a distance-time graph.

Distance covered after stopping d=S×t = 40km × 1hr = 40km

2. Steve walks at a uniform speed of 2m/s for 30 seconds. He stops for 2 seconds, then walks back from where he came from for 20 seconds. Draw a distance-time graph.

Distance covered in first 30s d=S×t = 2 × 30 = 60m

Motion Graphs

55

3.

Gilbert walks at a uniform speed of 3m/s for 20 seconds. He stops for 20 seconds, then continues to walk for another 30 seconds at 2m/s. Draw a distance-time graph. Distance covered in first 20s d=S×t = 3 × 20 = 60m Distance covered in last 20s d=S×t = 2 × 30 = 60m

11.2 SPEED (VELOCITY) – TIME GRAPH • Speed (or velocity) – time graph describes the speed of an object with time taken. • Features on speed-time graph: 1. Gradient (slope line) = constant acceleration ➢ Slant upward = acceleration ➢ Slant downward = deceleration 2. Horizontal (flat) line = constant speed, zero acceleration 3. Curved line = changing acceleration 4. Area of a graph = distance covered ➢ To find the distance covered by an object on a speed-time graph, calculate the area of the graph using the following formulae.

i.

Triangle area =

1 2

bh

ii. Rectangle area = bh

iii. Trapezium area =

56

Motion Graphs

1 2

(a + b)h

From the graph, total distance = total area Total area = area A (triangle) + area B (rectangle) + area C (trapezium) + area D (rectangle) + area E (triangle)

❖

EXAMPLE 1. A motorcyclist accelerates from rest to 5m/s2 for 8 seconds, then travels at a constant speed for 10 seconds. He then decelerates at 2m/s2 for 5 seconds and continues at a constant for 5 seconds. A. Draw a speed-time graph. B. Find the total distance covered by the motorcyclist.

❖

SOLUTION 1. A.

First phase t = 8s v = u + at v=0+5×8 v = 40m/s

Second phase v = 40m/s t = 8 + 10 t = 18s

Third phase t = 18 + 4 t = 22s v = u + at v = 40 + (−2)5 v = 40 − 10 v = 30m/s

Fourth phase v = 30m/s t = 23 + 5 t = 28s

Motion Graphs

57

B.

Total distance = total area = area A + Area B + area C + area D Area A =

1 2

× 8 × 40 = 160m

Area B = 10 × 40 = 400m Area C =

1 2

(40 + 30) × 4 = 140m

Area D = 5 × 30 = 150m Total distance = 160 + 400 + 140 + 150 = 850m

58

Motion Graphs

REVIEW QUESTIONS 1. A car reversed 10m backward then moved 30m forward. What distance is covered by the car? A. 10m B. 20m C. 30m D. 40m 2. A car reversed 10m backward then moved 30m forward. What is the displacement of a car? A. 10m B. 20m C. 30m D. 40m 3. Distance moved per unit time is called A. Displacement B. Velocity C. Speed D. Acceleration 4. Velocity is defined as the change of displacement per A. Speed B. Distance C. Acceleration D. Time 5. A soccer ball takes 7 seconds to roll 21m. What is its average speed? A. 2m/s B. 3m/s C. 4m/s D. 5m/s 6. A storm is moving 30km/h west. This information is about the storm’s A. Acceleration B. Constant speed C. Velocity D. Displacement 7. Acceleration involves the change in A. Distance travelled per time B. Speed of an object in motion C. Displacement in a particular direction D. Velocity per time 8. A train starts from rest and uniformly accelerates to 15m/s in 25 seconds. How far does the train travel during this time? A. 186m B. 186.5m C. 187m D. 187.5m 9. Kennedy moves 50m in 20 seconds. If he continues, how far will he move in the next 2 minutes? A. 200m B. 300m C. 400m D. 500m

Motion Graphs

59

10. When an object is at rest, what is its speed? A. 4m/s B. 0.5m/s C. 10m/s D. 0m/s 11. The area under a velocity-time graph gives A. Distance B. Time C. Acceleration D. Speed 12. A bus traveling at 10m/s speeds up at a constant rate of 5m/s2 for 4 seconds. A. What velocity does it reach? B. How far does it travel in this period? C. If it continues with this accelerates, what velocity does it reach in 10 seconds? 13. A car accelerates uniformly from rest for 30 seconds with an acceleration of 2m/s. It then travels at a constant speed for 2 minutes before slowing down with a uniform deceleration to come to rest in further 10 seconds. A. What is the maximum speed of the car? B. Sketch a speed-time graph from the about information 14. A taxi driver saw a child from afar crossing the road and applied emergency brakes. During the driver’s reaction time the taxi travelled at a steady speed and then slowed down until it stopped. The graph shows the speed of a taxi.

A. B. C. D. E.

60

What was the driver’s reaction time? Calculate the distance covered during the reaction time. How long did it take the taxi to stop once brakes were applied? Find the deceleration of the taxi. How far did the taxi move after brakes were applied?

Motion Graphs

15. The figure below shows the motion of a bus.

A. Calculate the acceleration of the bus in the first 6 seconds. B. What is the total distance travelled by bus in 10 seconds?

Motion Graphs

61

SOLUTIONS 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

D B C D B C D D C D A

12. A.

B.

v = u + at v = 10 + 5 × 4 v = 30m/s

s = ut +

1 2

at 2

s = 10 × 4 + s = 40 + 40 s = 80m C.

13. A.

1 2

× 5 × 42

u = 10m/s a = 5m/s2 t = 4s s=?

v = u + at v = 10 + 5 × 10 v = 70m/s

v=? u = 10m/s a = 5m/s2 t = 10s

v = u + at v = 0 + 2 × 30 v = 60m/s

u = 0m/s, object from rest has zero initial velocity v=? a = 4m/s2 t = 7s

B.

62

v=? u = 10m/s a = 5m/s2 t = 4s

Motion Graphs

14. A. 1 second B. Distance = area covered by the 1 second = 30 × 1 = 30m C. 3 seconds D.

Deceleration = =

u

v

𝑡 0

30

4

1

= −10m/s E.

u = 30m/s v = 0m/s, taxi going to rest t = 3s a=?

Distance = area covered from 1s to 4s =

1 2

× 30 × 3

= 45m

15. A. a = a=

u

v

𝑡 30 0 6

a = 5m/s2 B.

u = 0m/s, from rest v = 30m/s t = 6s a=?

Distance covered in first 6 seconds d=

1 2

Distance in first 6 sec = area of triangle 1

× 30 × 6

= bh 2

d = 90m Distance covered from 6s to 10s d = 30 × (10 − 6) d = 30 × 4 d = 120m

Distance from 6s to 10s = area of rectangle = bh

Total distance covered = 90m + 120m = 210m Total distance covered in 10sec can also be calculated using 1 the trapezium formula; d = 2 (a + b)h Where; a = 10 − 6 = 4 b = 10 h = 30

Motion Graphs

63

12

FORCE

Introduction What causes a car to move? A car engine consists of cylinders and pistons. First, a mixture of air and fuel is introduced into the cylinder. Next, the burning of air and fuel mixture produces a force that pushes the piston up and down, which rotates the crankshaft. Finally, the crankshaft causes the wheels of the car to turn. Force is simply a pull or push. It causes an object to move, stop or change direction. An object at rest will remain at rest unless acted upon by a force. An object in motion will continue in a straight path unless acted upon by a force. This unit covers Newton’s law of motion.

Specific outcomes By the end of this unit, you will be able ❖ Force: • Define force • State the units of force • State the effects of force on objects • Work out addition and subtraction of forces moving in the same and opposite direction ❖ Newton’s first law of motion: • State Newton’s first law of motion • Define inertia ❖ Newton’s second law of motion: • State Newton’s second law of motion • State the relationship between force, mass and acceleration • Work out force, mass, acceleration calculations using F = ma • State the relationship between weight, mass and acceleration • Perform weight, mass, acceleration calculations using W = mg ❖ Friction: • Define friction • State the advantages of friction • State the disadvantages of friction ❖ Terminal velocity: • Define terminal velocity • Explain what causes terminal velocity • Draw the terminal velocity graph ❖ Newton’s third law of motion: • State Newton’s third law of motion

64

Force

FORCE 12.1 FORCE • Force is defined as a pull or push. • It is a vector quantity, therefore has both magnitude and direction. • SI unit of force is Newton (N). • Force is measured by a spring balance. ❖ EFFECTS OF FORCE ON OBJECT 1. Force changes the direction of a moving object. 2. Force increase or decrease the motion of an object. ❖ ADDING FORCES • When two or more forces act on an object in a straight line, the resultant force can be found by adding or subtracting individual forces. • Addition of forces; if individual forces are moving an object toward the same direction. FR = F1 + F2

where FR is the resultant vector F1 and F2 are individual forces acting toward the same direction

• Subtraction of forces; if individual forces are moving an object toward the opposite direction. FR = F1 – F2

where FR is the resultant vector F1 and F2 are individual vectors acting toward the opposite direction

• If the resultant force is 0N, forces are balanced and an object will be at rest. • If the resultant force is greater than 0N, forces are unbalanced and an object will be in motion, moving toward the direction of the larger forces.

Force

65

❖

EXAMPLES Find the resultant forces of the following

12.2 NEWTON’S FIRST LAW OF MOTION • Newton’s first law of motion states that if all forces are balanced on an object ➢ if it is at rest, it will continue to be at rest ➢ if it is moving, it will keep moving at a constant speed in a straight line • Newton First law is based on inertia (Latin term meaning ‘laziness’), the objects tendency to resist change in its motion • Therefore, a moving object has a tendency to keep moving in a particular direction and at a constant speed. An object at rest, as well, has a tendency to remain at rest • That’s why we tend to move forward when breaks are suddenly applied, as the body in a moving car has a tendency to continue in motion • Object’s inertia depends on its mass. The larger the mass, the greater its inertia, the more difficult it is to move it or stop it • It is easier to push a book than a refrigerator because a refrigerator has greater inertia. It is easier to stop a smaller vehicle such as a motorbike than a larger vehicle such as a truck as a larger mass, hence greater inertia

12.3 NEWTON’S SECOND LAW OF MOTION • Newton’s second law of motion states that if forces acting on an object are unbalanced, they produce an acceleration in the direction of a force • An object’s acceleration is directly proportional to the net force and inversely proportional to the mass • The larger the unbalanced force, the larger the acceleration of an object • The larger the mass of an object, the greater its inertia, therefore the lower the acceleration • The relationship between force, mass and acceleration is as shown below F = ma

where F = force (in Newton, N) m = mass (in kg) a = acceleration (in m/s2)

❖ EXAMPLES 1. A car of mass 2000kg accelerates at 4m/s2. What is its force? 2. A force of 40N was applied to a football to accelerate at 20m/s2. What is the mass of a football? 3. If a 65kg swimmer pushes off a wall with a force of 300N, at what rate will the swimmer accelerate from the wall?

66

Force

❖

SOLUTIONS 1. F = ma F = 2000 × 4 F = 8000N

2. m = m=

F

F=? m = 2000kg a = 4m/s2 F = 40N m=? a = 20m/s2

𝑎 40 20

m = 2kg 3. a = a=

F

F = 300N m = 65 a=?

m 300 65

a = 4.62m/s2

12.4 WEIGHT AND GRAVITY • The weight of an object is the same as the force of gravity acting on an object pulling it toward the earth’s centre. • Therefore, weight (W) = force (F) = mass × acceleration; where acceleration = gravity ≈ 10m/s2 • The formula for calculating weight is given as: W = mg

where W = weight (in Newton, N) m = mass (in kg) g = acceleration due to gravity (≈10m/s2)

❖ EXAMPLES 1. What is the weight of a 25kg bag of mealie meal? 2. What is the mass of a person who weighs 700N? ❖ SOLUTIONS 1. W = mg = 25 × 10 = 250N

2. m = =

W g 700

W=? m = 25kg g = 10m/s2 W = 700N m=? g = 10m/s2

10

= 70kg

Force

67

12.5 FRICTION • Friction is defined as the force which opposes motion when objects slide over each other. • For a moving object, friction act in the direction opposite to the object’s direction of motion. • Greater friction exists on rough surfaces. The rougher the surface, the greater the amount of friction.

❖ ADVANTAGE OF FRICTION 1. We can hold objects in our hands. 2. Vehicles can move and stop on the road. 3. We walk on the road. ❖ DISADVANTAGE OF FRICTION 1. Causes tear and wear. 2. Causes energy loss. 3. Produce heat in the machine.

12.6 TERMINAL VELOCITY • Terminal velocity is the maximum velocity reached by a freely falling object in the fluid. • The fluid can be air or liquid. • Terminal velocity occurs when the downward falling force (weight) equals the upward resisting force (air resistance), resulting in a zero net force. • At terminal velocity, an object has zero acceleration and move at a constant speed. • For a freely falling object, the object accelerates downward at the start due to the force of gravity. At this point, there is no air resistance. • As the object is falling downward, air resistance gradually increases. As a result, the acceleration of an object is reduced. • Finally, the force of gravity pulling an object downward equals air resistance pushing it upward. • At that point, an object ceases to accelerate but continues moving at a maximum speed reached called terminal velocity.

68

Force

Terminal velocity graph

12.6 NEWTON’S THIRD LAW OF MOTION • Newton's third law of motion states that for every action, there is an opposite but equal reaction. • According to this law, forces always occur in pairs. Thus, if object A exerts a force on object B, then object B must exert a force equal in magnitude but in the opposite direction back to object A. • For example, the action force of jumping off the boat produces a reaction force that pushes it backward. • The forward action force on the bullet of a fired gun produces a backward reaction force on the gun.

Force

69

REVIEW QUESTIONS 1. Force A. Is a pull or push B. Is a product of mass and acceleration C. Friction is a type of force D. All the above 2. Weight is a force that depends on the mass and A. Volume B. Acceleration C. Gravity D. Velocity 3. According to Newton's first law A. Object in motion has a tendency to continue moving B. Object in motion has a tendency to stop C. Unbalanced forces produce an acceleration on the object D. A forward force has an equal backward force 4. According to Newton's third law A. Object in motion has a tendency to continue moving B. Object in motion has a tendency to stop C. Unbalanced forces produce an acceleration on the object D. A forward force has an equal backward force 5. According to Newton's second law A. Object in motion has a tendency to continue moving B. Object in motion has a tendency to stop C. Unbalanced forces produce an acceleration on the object D. A forward force has an equal backward force 6. A 25N force and a 30N force act on an object in the opposite direction. What is the net force on the object? A. 0N B. 5N C. 50N D. 55N 7. A 25N force and a 30N force act on an object in the same direction. What is the net force on the object? A. 0N B. 5N C. 50N D. 55N 8. A 10N force act on the object toward the north and another 3N force act on the same object toward the south. The object will A. Be stationary B. Move toward south C. Move toward north D. Move toward east 9. Objects falling through air experience a type of friction called A. Inertia B. Air resistance C. Terminal velocity D. Gravity

70

Force

10. The tendency of an object to resist change in its motion is called A. Force B. Friction C. Mass D. Inertia 11. The force that one object’s surface exerts on another when the two rub against each other is called A. Inertia B. Gravity C. Friction D. Mass 12. What causes a moving object to change its direction? A. Acceleration B. Velocity C. Gravity D. Force 13. The figure shows a graph for a 0.035g falling raindrop from clouds until it strikes the ground.

A. B. C. D. E.

What is the weight of the raindrop? What is the terminal velocity of the raindrop? At what time did the raindrop reach the terminal velocity? What distance was covered while traveling with the terminal velocity? Describe in terms of forces acting on the raindrop why the gradient changed.

Force

71

SOLUTIONS 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

D C A D C B D C B D C D

13. A.

W = mg W = 0.000035kg × 10m/s W = 0.00035 W = 3.5 × 10−4 N

W=? m = 0.035g ÷ 1000g = 0.000035kg g = 10m/s2

B. 50m/s C. 6 second D. Distance traveling with terminal velocity equal area from 6s to 11s d = 50m/s × 5s d = 250m E. Initially, the raindrop’s weight is greater than air resistance. Therefore it was accelerating. Air resistance gradually increased until it was equal to the weight of the raindrop in the sixth second. With weight being equal to air resistance, the raindrop ceased accelerating and travelled at a constant velocity called terminal velocity shown by a horizontal line.

72

Force

13

CIRCULAR MOTION

Introduction You tie a stone to a rope and swing it in a circle. What force will be acting on the stone as it is moving in a circular path? What keeps the moon orbiting the earth? An object moving in a circular path is acted upon by three forces: centripetal force, inertia, and centrifugal force. Centrifugal force is only present when the objectin circular motion is attached to the centre by physical means such as a stone attached by a rope. Objects not physically linked to the centre, such as the moon or a satellite orbiting the earth, are kept in motion only by centripetal force and inertia. What is the centripetal force, inertia and centrifugal force? How does each of these forces keep the object in circular motion? This unit answers these questions.

Specific outcomes By the end of this unit, you will be able to: ❖ Forces affecting circular motion: • Explain how centripetal force affects circular motion • Explain how inertia affects circular motion • Explain how centrifugal force affects circular motion ❖ Draw a diagram of an object in a circular motion, including all the forces acting upon it

Circular Motion

73

FORCES ACTING ON CIRCULAR MOTION 13.1 FORCES ACTING ON CIRCULAR MOTION 1. CENTRIPETAL FORCE • Centripetal force is a force acting on an object that tends to pull it toward the centre of a circular path. • It is perpendicular to the motion of an object. • For example; centripetal force (force of gravity of the earth) acts on the moon orbiting the earth, pulls it toward the centre of the earth. 2. INERTIA • According to Newton first law of motion, an object in motion will remain in motion with the same speed and direction. • In a circular motion, at any instantaneous point an object has the tendency to remain moving in a straight line while centripetal force pulling it toward the centre. This combination produces a circular motion. • An object in circular motion travelling with a constant speed is continuously changing its direction, therefore it is accelerating. 3. CENTRIFUGAL FORCE • When an object in circular motion is attached to the centre by a string, the object tends to pull the centre toward it. • This force is due to the inertia of the object in a circular motion to move in a straight line, therefore it tends to pull the centre away. • If the centrifugal force is greater than the centripetal force, the centre would be displaced toward the object.

74

Circular Motion

14

FORCES ACTING ON SPRING

Introduction When force is applied to a spring, it stretches or is compressed. When force is removed, it returns to its original position. So what governs the behaviour of springs? It is Hooke’s law. This law states that the force applied to spring is directly proportional to its extension, provided it does not pass its elastic limit. Hooke’s law is also true to elastic objects such as rubber bands.

Specific outcomes This unit covers Hooke’s law. By the end of this unit, you will be able to: ❖ State Hooke’s law ❖ Draw a graph to support Hooke’s law ❖ Work out extension, load, mass calculations using;

Extension Load

= constant

Force Acting on Spring

75

FORCE ACTING ON SPRING 14.1 HOOKE’S LAW • Hooke's law explains the relationship between the force stretching (or load) the spring and the amount of spring extension. • Hooke’s law states that the force (or load) applied to spring is directly proportional to its extension, provided it does not pass the elastic limit. • The formula for calculating constant is given as:

Extension Load • •

= Constant

In other words, the larger the load applied to the spring, the more it will stretch until it reaches its maximum extension. If the load is plotted against extension the result produces a straight line.

Note: Load = Weight = mg where; m = mass g = gravity = 10N/kg ❖ EXAMPLES 1. A spring of length 50cm is extended by 4cm when a load of 0.8kg is suspended at the end. How far would it be extended if a load of 0.16kg is applied? 2. A 500g extended the spring by 6cm. How many grams would extend the spring to 2cm?

❖ SOLUTIONS 1.

Constant = =

Extension

4

Load

8 = 0.5

Extension = constant × load = 0.5 × 16 = 8cm

76

Force Acting on Spring

Extension = 4cm Load = mg = 0.8 × 10 = 8N Constant = ?

Extension = ? Load = mg = 0.16 × 10 = 16N Constant = 0.5

2.

Constant = =

Extension 6

Load

5

= 1.2

Extension

Load = =

Constant 2 1.2

Extension = 6cm Load = mg =

500g 1000

× 10

= 5N Constant = ?

Extension = 2cm Load = ? Constant = 1.2

= 1.67N

m= =

L g 1.67 10

Load = Weight = mass × gravity = 1.67N Mass = ? Gravity = 10m/s2

= 0.167kg = 167g

Force Acting on Spring

77

15

MOMENT OF A FORCE

Introduction What happens when an object is supported on one end and force acts on the other end? The object rotates. This phenomenon is called the moment of a force or torque. The moment of a force is defined as the product of force and perpendicular distance from its point of action. The point of action is the pivot, where an object rotates from. For example, the door opens by applying a perpendicular force on the door handle. A nut is turned by applying a perpendicular force on a spanner. The longer the spanner, the less the force required to turn the nut. Therefore, the size of the moment of a force depends on the force applied and the perpendicular distance from where force is applied to the pivot.

Specific outcomes This unit covers the moment of a force. By the end of this unit, you will be able to: ❖ Moment of a force: • State the moment of a force • State the SI unit of moment of a force • State the applications of the moment of a force • Work out moment of a force, force, distance calculations using M = Fd ❖ Principle of moment: • State the principle of moment • Work out the principle of moment calculations using F1d1 = F2d2

78

Moment of a Force

MOMENT OF A FORCE 15.1 MOMENT OF A FORCE • Moment of a force is defined as the turning effect of a force about a pivot. • It is also defined as the product of force and perpendicular distance from its point of action. • The moment of a force depends on the size of a force applied and how far it is applied from the pivot. The larger the force applied or the longer the distance is applied from the pivot, the greater the turning effect. • SI unit of moment of a force: Newton metre (Nm). • The formula for calculating moment of a force is given as: M = Fd

where M = moment of a force ( in Nm) F = force (in N) d = distance (in m)

❖ APPLICATION OF MOMENT OF A FORCE i. Force is applied at the right angle when opening a door. ii. A door handle is placed at a distance from the door hinges. ii. It is easier to turn a nut with a long spanner. ❖ EXAMPLES 1. Harry opens a door using the handle placed 0.5m away from the door hinges. If he applies a force of 6N, what is the moment of force? 2. What is the moment of force of a spanner when it is pushed with a force 15N, 30cm from the nut? 3. What is the distance when the moment of force is 12Nm? d

4. If the moment is 30Nm, calculate the force applied.

Moment of a Force

79

❖

SOLUTIONS 1. M = Fd M = 6 × 0.5 M = 3Nm

2. M = Fd M = 15 × 0.3 M = 4.5Nm

3. d = d=

M=? F = 6N d = 0.5m

M=? F = 15N d = 30cm ÷ 100cm = 0.3m

M

M = 12Nm F = 30N d=?

F 12 30

d = 0.4m

4. F = F=

M

M = 18Nm F=? d = 0.6m

d 18 0.6

F = 50N

15.2 PRINCIPLE OF MOMENT • Principle of moment states that for an object to be at equilibrium, the sum of clockwise moments must be equal to the sum of anticlockwise moments. • At equilibrium, total clockwise moments = total anticlockwise moments. • The formula for calculating principle of moment is given as: F1d1 = F2d2

where F1 = clockwise forces (in N) F2 = anticlockwise forces (in Nm) d1 = clockwise distance (in m) d2 = anticlockwise distance (in m)

❖ EXAMPLES 1. Calculate distance x needed to balance the seesaw below.

80

Moment of a Force

2.

Calculate force K needed to balance the following:

❖

SOLUTIONS 1. F1d1 = F2d2 d2 = x=

F1d1 F2 5 × 70

F1 = 70N d1 = 5m F2 = 50N d2 = x = ?

50

x = 7m

2. F1d1 = F2d2 + F3d3 8×2=2×1+K×4 16 = 2 + 4K 4K = 14 K = 3.5N

F1 = 8N d1 = 2m F2 = 2N d2 = 1m F3 = K = ? d3 = 4m

Moment of a Force

81

16

WORK, ENERGY AND POWER

Introduction Imagine a boy pushing against a wall for hours. He is tired and sweating. Is he doing work? In physics, he is not doing any work! Work is defined as the product of force and distance moved in the direction of a force. So as long as the wall has not moved, the boy’s work done is zero. For work to be done, energy is needed. Energy is the capacity to do work. An object with a lot of energy can do lots of work. When work is done, energy is converted from one form to another. For example, if you drop your pen, work is done on the pen. As the pen falls, the potential energy of the pen is converted to kinetic energy. Work is done because the pen moves a distance from your hand to the ground in the direction of the force; gravity, acting on the pen. Power is the rate of doing work. An object that can do a lot of work in a short period of time is said to be powerful. For example, a car that can cover a long distance in a short period of time is powerful. Same as a man who can do a lot of work in a short period time is said to be powerful.

Specific outcomes This unit covers work, energy and power. By the end of this unit, you will be able to: ❖ Work: • Define work • State the SI unit of work • Work out work, force, distance calculations using W = Fd ❖ Energy: • Define energy • State the SI unit of energy • Define gravitational potential energy (GPE) • Work out GPE, mass, height calculations using GPE = mgh • Define kinetic energy (KE) 1 • Work out kinetic energy, mass, velocity calculations using KE = 2 mv 2 • State other forms of energy • Explain the conversation of energy • Define renewable energy, and give examples of renewable energy • Define non-renewable energy and gives examples of non-renewable energy ❖ Power: • Define power • State the units of power •

82

Work out power, work, energy, time calculations using P =

Work, Energy and Power

W t

WORK, ENERGY AND POWER 16.1 WORK • Work, also called work done, is defined as the product of force and distance moved in the direction of force. • SI unit of work done is Joule (J). • The formula for calculating work done is given as: W = Fd

• • •

where W = work done (in J) F = force (in N) d = distance moved in direction of force (in m)

For work to be done, the force applied must move the object from one point to another. If force is applied but an object has not moved, no work is done. So, for example, if you have pushed a wall for 2hr but it has not moved or fallen, what is lost is energy, but no work is done. To do work, energy is required.

❖ EXAMPLE 1. How much work is done when a 100N force moves a block 25m? 2. Calculate work done when an object of mass 25kg is lifted by 1m. (g = 10m/s2) 3. What mass requires 2kJ of work to lift it by 20m? (g = 10m/s2) 4. Calculate the distance moved by a force of 620N when doing 31kJ. ❖ SOLUTIONS 1. W = Fd W = 100 × 25 W = 2500 W = 2.5kJ

2. F = W = mg F = 25 × 10 F = 250N W = Fd W = 250 × 1 W = 250J

W=? F = 100N d = 25m

Force = Weight F=? m = 25kg g = 10m/s2 W=? F = 250N d = 1m

Work, Energy and Power

83

3.

F= =

W

W = 2kJ × 1000 = 2000J F=? d = 20m

d 2000 20

= 100N F = W = mg m= =

Force = Weight F = 100N m=? g = 10m/s2

F

g 100 10

= 10kg

4.

d= =

W

W = 31kJ × 1000 = 31000J F = 620N d=?

F 31000 620

= 50m

16.2 ENERGY • Energy is defined as the capacity to do work. • SI unit of energy is Joule (J) ❖ TYPES OF ENERGY 1. POTENTIAL ENERGY • Potential energy is defined as the stored energy due to the position of an object. • Gravitational potential energy (GPE) is the potential energy of an object due to the height of an object from the ground. • The object at the surface of the earth has zero GPE. The higher the object from the earth’s surface, the greater its GPE. • The formula for calculating GPE is given as: GPE = mgh

where GPE = gravitational potential energy (in J) m = mass (in kg) h = height (in m) g = gravity (10m/s2)

2. KINETIC ENERGY • Kinetic energy is defined as the energy possessed by an object in motion. • The formula for calculating kinetic energy is given as: KE =

84

Work, Energy and Power

1 2

mv 2

Where KE = kinetic energy (in J) m = mass (in kg) v = velocity (in m/s)

3.

ELECTRICAL ENERGY • Electrical energy is defined as the energy stored or moving charges.

4.

HEAT ENERGY • Heat energy is defined as the energy due to moving particles in an object.

❖

CONSERVATION OF ENERGY • Conservation of energy states that energy can be changed from one form to another but cannot be created or destroyed • Suppose an object is lifted to height h, its gravitational potential energy (GPE) = mgh. When an object is released and falls, its gravitational potential energy is changed to kinetic energy. • Formula: GPE = KE

❖

where GPE = gravitational potential energy KE = kinetic energy

SOURCES OF ENERGY 1. RENEWABLE SOURCES: Renewable sources are sources that can be reused or replaced. The following are examples of renewable sources: • Sun (solar) energy: energy radiating from the sun. • Wind energy: generation of electricity by conversion of the kinetic energy of wind into electrical energy. • Hydro energy: generation of electricity by the conversion of the potential energy of water into kinetic energy, then to electrical energy. • Geothermal energy: the use of heat energy stored inside the earth to produce electrical energy. • Biomass energy: generation of energy from organic materials (plants and animals). 2. NON – RENEWABLE SOURCES: Non-renewable sources are sources that cannot be replaced once used. The following are examples of non-renewable sources: • Fossil fuel • Coal • Natural gas • Nuclear energy Work, Energy and Power

85

❖

EXAMPLES 1. A 3kg cat is lifted 2m into the air. How much GPE does it gain? (g = 10m/s2) 2. Nkisu runs up a staircase of 10m high and gains 6500J. What is her mass? (g = 10m/s2) 3. A huge block of mass 120kg is lifted and gains 36kJ. How high was it raised? (g = 10m/s2) 4. A bow can shoot a 0.05kg arrow at a speed of 20m/s. What is the kinetic energy of the arrow? 5. A car has a kinetic energy of 675kJ and travelling at 30m/s. What is the mass of a car? 6. What is the velocity of the 12000kg wind turbine blade with the kinetic energy of 1350kJ?

❖

SOLUTIONS 1. GPE = mgh = 3 × 10 × 2 = 60J

2. GPE = mgh m= =

GPE hg 6500

GPE = ? m = 3kg h = 2m g = 10m/s2

GPE = 6500J m=? h = 10m g = 10m/s2

10 × 10

= 65kg 3. GPE = mgh GPE h= mg =

36000 120 × 10

GPE = 36kJ = 36000J m = 120kg h=? g = 10m/s2

= 30m 1

4. KE = =

2 1 2

mv 2 × 0.05 × 20 2

KE = ? m = 0.05kg v = 20m/s

= 10J

5. KE = m= =

1

mv 2

2 2KE

v2 2 × 675000 30 2

= 1500kg

86

Work, Energy and Power

KE = 675kJ = 675000J m=? v = 30m/s

6.

KE =

1 2

KE = 1350kJ = 1350000J m = 12000kg v=?

mv 2

v=√ =√

2KE m 2 × 1350000 12000

= √225 = 15m/s

16.3 POWER • Power is defined as the rate of doing work. • It is also defined as the rate of transferring energy from one form to another. • SI unit of power is Watt (W). • The formula for calculating power is given as: P= P=

W

where P = power (in W) W = work done (in J) E = energy transferred (in j) t = time taken (in s)

t E t

❖ EXAMPLES 1. A man lifted a 50kg bag of cement by m in 4 seconds. (g = 10m/s2) A. Calculate work done. B. Calculate power. 2. A car has a mass of 1100kg. Its speed changes from zero to 30m/s in 15s. A. Calculate the change in kinetic energy of the car. B. Calculate the average power generated by the engine. ❖ SOLUTIONS 1. A. F = W = mg F = 50 × 10 F = 500N W = Fd W = 500 × 2 W = 1000J

B. P = P=

W t 1000 4

Force = Weight F=? m = 50kg g = 10m/s2 W=? F = 500N d = 2m

P=? W = 1000J t = 4s

P = 250W

Work, Energy and Power

87

2.

A. Initial KE KE = =

1 2 1 2

mv

2

Initial KE = ? m = 1100kg v = 0m/s

× 1100 × 0 2

= 0J

Final KE KE = =

1 2 1 2

mv 2

Final KE = ? m = 1100kg v = 0m/s

× 1100 × 30 2

= 495000J Change in KE = Final KE − Initial KE = 495000 − 0 = 495000J

B. P = P=

E

=

Change in KE

t time taken 495000 15

P = 33000W

88

Work, Energy and Power

P=? E = Change in KE = 495000J t = 15s

REVIEW QUESTIONS 1. A type of energy possessed by a moving object is called A. Potential energy B. Kinetic energy C. Heat energy D. Electrical energy

.

2. A type of energy possessed by a stationary object is called A. Potential energy B. Kinetic energy C. Heat energy D. Electrical energy 3. All are true except. Which sentence is not true? A. Joule is a unit of work done. B. Work done is a product of force and distance. C. Is always greater than zero if the distance is zero. D. Force moves in the direction of distance. 4. After energy changes from one form to another, you always end up with the same total amount of energy. Which of the following laws explain this rule? A. Law of energy change. B. Law of conservation of energy. C. Law of kinetic energy. D. Law of potential energy. 5. The ability to do work is called A. Energy B. Work done C. Power D. Force 6. Which of the following is the rate at which work is done? A. Energy B. Weight C. Force D. Power 7. How much work is done by a machine in lifting a mass of 1kg through a height of 4m? A. 2.5J B. 25J C. 40J D. 60J 8. What is the kinetic energy of a 20kg object moving at 5m/s? A. 500J B. 250J C. 100J D. 2000J 9. A man weighs 700N, runs up a staircase of height 3m in 3s. How much useful power is needed to do this work? A. 700W B. 2100W C. 233W D. 4200W

Work, Energy and Power

89

10. The amount of work done when a 60kg person jumps 1m into the air in 0.5s is A. 30J B. 120J C. 300J D. 600J

.

11. The figure shows a police car of mass 2000kg moving on a level road. The driving force of the car is 5000N and the frictional force is 1000N.

A. Calculate the acceleration of the car, B. If the car started from rest, calculate; i. the kinetic energy gained by the car in 15s, ii. the power developed by the car engine in 15s. 12. The figure shows a soccer ball of mass 450g on top of the hill. (take g = 10m/s2)

A. Calculate the potential energy possessed by a ball while on top of the hill (at point A). B. What is the kinetic energy of the ball as it reaches point B? C. Find its velocity as it reaches point B.

90

Work, Energy and Power

SOLUTIONS B A C B A D C B A D

.

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

11. A. Resultant force = Driving force − frictional force Resultant force = 5000 − 1000N Resultant force = 4000N F = ma a= =

F = 4000N m = 2000kg a=?

F m 4000 2000

= 2m/s2 B. i. Velocity in 15s v = u + at = 0 + 2 × 15 = 30m/s KE = =

1 2 1 2

mv 2 × 2000 × 30 2

v=? u = 0m/s, from rest a = 2m/s2 t = 15s KE = ? m = 2000kg v = 30m/s

= 900000J

ii. P = =

E t 900000 15

P=? E = KE = 900000J t = 15s

= 60000W

Work, Energy and Power

91

PE = ? m = 450g ÷ 1000 = 0.45kg h = 14m g = 10m/s2

.

12. A. PE = mgh = 0.45 × 10 × 14 = 63J

B. 63J. All kinetic energy is changed to kinetic energy

C. KE =

1 2

mv 2

v=√ =√

2KE m 2 × 63 0.45

= 16.7m/s

92

Work, Energy and Power

KE = 63J m = 0.45kg v=?

17

SIMPLE MACHINE

Introduction Every day we perform many works. For example, we cut fabrics, vegetables, lift heavy objects, climb stairs and so on. To achieve these tasks, we use energy. To reduce the amount of energy spent, simple machines are used. For example, we can cut a piece of fabric with our hands, but it can require a lot of energy. To ease the work, we use a razor blade or scissors. A razor blade reduces the input force needed to cut a piece of fabric. Therefore, simple machines such as a razor blade make life easier. .

Specific outcomes This unit covers the types of simple machines and the terms used to describe the effectiveness of a simple machine. By the end of this unit, you will be able to: ❖ Simple machine: • Define simple machine, effort and load • Describe the following simple machines with examples: • Lever • Inclined plane • Pulley • Wheel and axle • Screw • Wedge ❖ Mechanical advantage: • Describe mechanical advantage (MA) of a machine • Calculate the mechanical advantage of a machine ❖ Velocity ratio: • Describe the velocity ratio (VR) of a simple machine • Calculate the velocity ratio of a machine ❖ Efficiency: • Define efficiency of a machine • Calculate the efficiency of a machine .

Simple Machine

93

SIMPLE MACHINE 17.1 SIMPLE MACHINE • A simple machine is a device that makes life easier. • These devices multiply the force we apply to the machine, making everyday work easier to perform. • The force applied to a machine is called input force or effort. • The force produced or overcome by a machine is called output force or load.

.

17.2 TYPES OF SIMPLE MACHINES 1. LEVER • A lever is a bar used for raising an object. • Levers have a fulcrum, a point where a lever turn. • A load is placed on one end and force is applied on another end. • Examples of levers include scissors, brooms, pliers, bottle opener, wheelbarrow.

2. INCLINED PLANE • An inclined plane is a flat surface with one end raised higher than the other. • It is used for raising or lowering a heavy load. • Examples of inclined planes include a staircase, slope road, playground slides, and ramp.

3. PULLEY • A simple machine with one or more grooved wheels connected by a rope. • Pulleys are used to lift or lower heavy objects. • An object is tied to one end of the rope and force is applied to the other end of the rope. • Examples of pulleys include elevators, water wells, and window blinds.

94

Simple Machine

4.

WHEEL AND AXLE • Wheel and axle is a simple machine consisting of a larger wheel attached to a smaller axle. • These two rotate together, so the force applied to the axle is transferred to the wheel and vice versa. • Examples of wheel and axle include doorknob, water wheel, and wheelbarrow.

5.

SCREW • A screw is a specialised type of inclined plane wrapped around a cylinder. • It converts circular motion into a linear force. • A screw is used to hold objects together and lift heavy objects. • Examples of screws include drills, car jacks, bolts and nuts, bottle tops, circular staircase.

6.

WEDGE • Wedge is a type of inclined plane. • It is wide on one end and tapered on the other end. • Wedges are used to separate objects by cutting, piercing and splitting. • Examples of wedges include a pickaxe, knife, shovel, needle and nail.

.

Simple Machine

95

17.3 MECHANICAL ADVANTAGE • Mechanical advantage is defined as the ratio of the load to the effort. • It indicates the number of times a simple machine multiplies the input force. • Simple machines are designed to lift or move a larger load with a smaller effort to save the input force. • If the mechanical advantage (MA) is greater than one (MA > 1), the load is greater than the effort (L > E). The larger the MA is, the smaller the effort to be applied to the machine. • Friction affects the mechanical advantage of the machine. The greater the friction is, the lower the MA of the machine, as more effort is applied to overcome friction. • Ideal mechanical advantage (IMA) is the multiplication of force that would be achieved in the absence of friction. • Actual mechanical advantage (AMA) is the multiplication of force achieved in the presence of friction. • The formula for calculating mechanical advantage is given as:

.

MA =

L

where MA = mechanical advantage L = load E = effort

E

17.4 VELOCITY RATIO • Velocity ratio is defined as the ratio of the distance moved by the effort to the distance moved by the load. • The formula for calculating velocity ratio is given as: VR =

DE

where VR = velocity ratio DE = distance moved by effort DL = distance moved by load

DL

17.5 EFFICIENCY • Efficiency is defined as the ratio of useful work done by the machine to the total work input into the machine. • The formula for calculating the efficiency of a simple machine is given as: Efficiency =

Work output work input

× 100%

Or Efficiency =

• •

96

MA VR

× 100%

The ideal machine should produce 100% efficiency. However, in the real world some input work is used to overcome friction. Therefore, efficiency is always ≤ 100%. The efficiency of a machine can be maximised by reducing friction through lubrication and smoothening of surfaces.

Simple Machine

EXAMPLES 1. The figure below shows a lever.

.

❖

A. What is the mechanical advantage of the lever? B. What is the velocity ratio? C. Calculate the efficiency of the lever. 2. The figure below shows an inclined plane.

A. B. C. D.

What is the mechanical advantage of the inclined plane? What is the velocity ratio of the inclined plane? What is the efficiency of the inclined plane? Calculate the angle of inclination θ.

3. A tower crane lifts a 400kg block 60m off the ground with 1000N as shown below.

A. What is the mechanical advantage? B. What is the velocity ratio? C. What is efficiency?

Simple Machine

97

SOLUTIONS 1. A. MA = =

L

MA = ? L = 80N E = 40N

E 80 40

.

❖

=2 DE

B. VR =

DE = distance from pivot to effort DL = distance from pivot to load

DL 60

=

10

=6 C. Efficiency = =

2 6

MA VR

Efficiency = ? MA = 2 VR = 6

× 100%

× 100%

= 33.3% 2. A. MA = =

L

MA = ? L = 50N E = 10N

E 50 10

=5 B. VR = =

DE

DE = length of the slope DL = height

DL 3 0.5

=6 C. Efficiency = =

MA VR 5

× 100%

× 100%

6

Efficiency = ? MA = 5 VR = 6

= 83.3% D. sin θ = sin θ = sin θ =

opposite hypotenuse height length of the slope 0.5 3 0.5 ) 3

θ = sin−𝟏 ( θ = 9.6o

98

Simple Machine

Sine function definition

Use a scientific calculator to find the sin inverse

A. Load = Weight W = mg = 400 × 10 = 4000N MA = =

Weight = mass (m) × acceleration due to gravity (g) acceleration due to gravity = 10 N/kg

.

3.

L

The easiest way to found MA is to count the number of ropes lifting the load in the pulley system. In the above pulley system, ropes lifting the load are 4, therefore MA = 4

E 4000 1000

=4 B. VR = =

DE

VR = ? DE = 60m DL = 10m

DL 60 10

=6 C. Efficiency = =

MA VR 4 6

× 100%

× 100%

= 66.7%

Simple Machine

99

REVIEW QUESTIONS

.

1. How are machines helpful? A. They make work bearable B. They eliminate work C. They provide effort for work D. All the above 2. Force exerted on a machine is called A. Load B. Effort C. Out force D. Power 3. Force exerted by a machine is called A. Input B. Effort C. Load D. Work

4. A simple machine consisting of a rigid bar that rotates about a pivot is called A. Crew B. Pulley C. Inclined plane D. Lever 5. A simple machine consisting of a sloped surface is called A. Wedge B. Lever C. Inclined plane D. Wheel and axle 6. A simple machine with a rope wrapped around a wheel groove is called A. Pulley B. Wheel and axle C. Crew D. Wedge 7. A simple machine consisting of two circular objects of different sizes that are attached and rotate together is called A. Lever B. Wedge C. Wheel and axle D. Pulley 8. A simple machine that is an inclined plane wrapped around a cylinder is called A. Wheel and axle B. Pulley C. Lever D. Crew 9. A simple machine that is an inclined plane with wide and tapered ends is called A. Wedge B. Pulley C. Crew D. Lever

100

Simple Machine

10. Which of the following simple machines is a lever? A. Car jack B. Door knob C. Elevator D. Bottle top opener

.

11. Which of the following simple machines is a wedge? A. Window blinds B. Knife C. Bolts and nuts D. Wheelbarrow 12. A ramp is an example of A. Wheel and axle B. Lever C. Inclined crane D. Pulley 13. Which of the following is not a simple machine? A. Needle B. Comb C. Shoes D. Razor 14. The number of times the input force is multiplied by a machine is called A. Velocity ratio B. Mechanical advantage C. Efficiency D. None of the above 15. The ratio of the distance moved by the effort to the distance moved by the load is called A. Velocity ratio B. Mechanical advantage C. Efficiency D. None of the above 16. The ratio of work out by the machine to the work input expressed as a percentage is known as A. Velocity ratio B. Mechanical advantage C. Efficiency D. None of the above 17. The figure below shows a 40kg block dragged up a hill of length 30m and height 5m using a 200N force.

A. B. C. D.

Calculate the velocity ratio. Calculate the mechanical advantage. What is the efficiency of the hill? Calculate the angle of inclination θ.

Simple Machine

101

18. The figure below shows a pulley system that lifts an 8000N block 50m off the ground with 2000N.

. A. B. C. D. E.

102

Name pulleys L and M. What is the purpose of pulley L? Calculate the mechanical advantage of the pulley system. Calculate the velocity ratio of the pulley system. What is the efficiency of the pulley system?

Simple Machine

SOLUTIONS A B C D C A C D A D B C C B A C

.

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16.

17. A. VR = =

DE

DE = length of the slope DL = height

DL 30 10

=3 B. Load = Weight W = mg = 40 × 10 = 400N MA = =

Load = ? m = 40 kg g = 10N/kg

L

MA = ? L = 400N E = 200N

E 400 200

=2 C. Efficiency =

MA

=

VR 2 3

× 100%

× 100%

Efficiency = ? MA = 2 VR = 3

= 66.7% D. sin θ = sin θ = sin θ =

opposite hypotenuse height

opposite = height = 10m hypotenuse = length of the slope = 30m

length of the slope 10 30 10

θ = sin−𝟏 (30) θ = 19.5o

Use a scientific calculator to find the sin inverse

Simple Machine

103

18. A. Pulleys L = fixed pulleys Pulleys M = movable pulleys

.

B. Fixed pulleys change the direction of the load C. MA = =

L

MA = ? L = 8000N E = 2000N

E 8000 2000

=4 D. VR = =

DE

VR = ? DE = 50m DL = 10m

DL 50 10

=5 E. Efficiency = =

MA VR 4 5

× 100%

× 100%

= 80%

104

Simple Machine

Efficiency = ? MA = 4 VR = 5

18

PRESSURE

Introduction Why is it that a tractor can pass through the soft ground without sinking but the car sinks? Why are needles, razor blades and knives sharp? All these use the principle of pressure. Pressure is defined as the force acting on a unit area. The wider the area the lower the pressure. The tires of a tractor are wide than that of a car. Therefore, a tractor applies a lower pressure on the ground. Hence it can pass through a soft ground. Needles, razor blades, and knives with sharp edges create a larger pressure with minimal force applied for easy piercing and cutting. What are the characteristics of pressure in liquid? How do you calculate atmospheric pressure? This unit covers these questions. .

Specific outcomes By the end of this unit, you will be able to: ❖ Pressure: • Define pressure • State the units of pressure

.

•

Work out pressure, force, area calculations using P =

F A

❖ Pressure in liquid: • State factors that affect pressure in liquid • Work out pressure, depth, density calculations using P = hpg • State the characteristics of pressure in liquid ❖ Instruments for measuring pressure: • Describe the barometer, including how it works • State the standard atmospheric pressure • Describe the manometer, including types and how they work ❖ Transmission of pressure in hydraulic system: • Describe how pressure is transmitted in the hydraulic system F F • Work out force, area calculations using A1 = A2 • State uses of hydraulic system ❖ Upthrust and floatation: • Describe the upthrust force • State the Archimedes’ principle • Describe the principle of floatation

1

2

Pressure

105

PRESSURE 18.1 PRESSURE • Pressure is defined as the force acting on a unit area. • SI unit of pressure: Pascal (Pa); 1 Pa = 1 N/m2 (newton per metre squared). • Pressure is directly proportional to force and inversely proportional to the area acting on an object. • Greater force produces higher pressure while less force produces lower pressure. • Greater area produces lower pressure while lower area produces higher pressure. • This is why a needlepoint, razor and knife’s edge are extremely small to create a pressure higher enough to pierce the skin or cut the material easily. • Tractors with wide tyres can pass through the soft ground without sinking because a larger area of tyres reduces the pressure on the ground. • The formula for calculating pressure is given as:

.

P=

𝐅 𝐀

where P = pressure (in Pa or N/m2) F = force (in N) A = area (in m2)

❖ EXAMPLES 1. Calculate the pressure produced by a force of 600N acting on an area of 40m2. 2. Calculate the pressure exerted on the floor when a woman who weighs 620N stands on one foot which has an area of 80cm2. 3. A book has a mass of 600g and a surface area of 250cm2. What pressure is the book exerting on the table? 4. A box puts pressure of 30 Pa on an area of 0.4m2. Find the box of the book on the floor. 5. A balloon exerts a pressure of 45 Pa by a force of 90N. What is the area of the balloon? ❖ SOLUTIONS 1. P = =

F

A 600 40

P=? F = 600 A = 40m2

= 15 Pa 2. P = =

F A 620 0.008

= 77500 Pa = 77.5 kPa 3. F = W W = mg = 0.6 × 10 = 6N P= =

F A 6 0.025

= 240 Pa 106

Pressure

P=? F = 600 A = 0.008m2; 1cm2 = 0.0001m2 80cm2 = 0.008m2

Force acting on the book is weight of the book Weight = mass (m) × acceleration due to gravity (g)

P=? F = 6N A = 0.025m2; 1cm2 = 0.0001m2 250cm2 = 0.025m2

4.

F = PA = 30 × 0.4 = 12N

5.

A= =

F

P = 30 Pa A = 0.4m2 F=? P = 90 Pa F = 45N A=?

P 45 90

= 0.5m2

18.2 PRESSURE IN LIQUIDS 1. DEPTH OF LIQUID • Pressure in a liquid column is different at different depths. • Pressure increases with an increase in depth. • This is because the liquid particles above the liquid column press the particles below, resulting in an increase in pressure below the liquid column.

Liquid comes out with lowest pressure through a hole nearest the surface of the liquid column while with highest pressure through a hole farthest from the surface of the liquid column. This shows that pressure increases with increase in depth. 2. DENSITY OF LIQUID • The density of a liquid affects the pressure of a liquid. • A denser liquid exerts more pressure than a less dense liquid. • This is because particles in a denser liquid are closely packed. As a result, there is more collision of particles in a given area. 3. ACCELERATION DUE TO GRAVITY • The pressure of a liquid is directly proportional to the gravitational strength. • When the gravitational strength is high, the pressure of the liquid increases. • When the gravitational strength is low, the pressure of a liquid decreases. ❖

CALCULATING PRESSURE IN LIQUIDS • The formula for calculating pressure in liquids is given as: P = hpg

where P = pressure (in Pa or N/m2) h = depth (in m) p = density of liquid (kg/m3) g = acceleration due to gravity

Pressure

107

❖

EXAMPLES 1. Calculate the pressure at the bottom of a dam 20m deep. The density of water is 1000kg/m3. Take acceleration due to gravity to be 10N/kg. 2. A sea diver descends 40m from the surface of the sea. Calculate the pressure that water exerts on the diver.

❖

SOLUTIONS 1. P = hpg = 30 × 1000 × 10 = 300000 Pa = 300 kPa 2. P = hpg = 40 × 1000 × 10 = 400000 Pa = 400 kPa

❖

CHARACTERISTICS OF LIQUID PRESSURE 1. PRESSURE AT THE SAME DEPTH IS EQUAL IN ALL DIRECTIONS • Pressure acts in all directions in fluids, that is, liquids and gases. • This is because particles that makeup fluids can move in any direction. • At the same depth, pressure is equal in all directions. • This can be evidenced by making holes around a can bottle of water at the same depth. It can be observed that water comes out at the same speed.

Water comes out with the same speed at the same depth. This shows that water has same pressure at that depth 2. PRESSURE DOES NOT DEPEND ON THE SURFACE AREA, SHAPE OR SIZE OF THE CONTAINER • Pressure in liquids depends on depth, not the container's surface area container, shape or size. • Therefore, in a container where liquid can move freely in various parts, the liquid has the same depth regardless of the container's area, shape, or size.

108

Pressure

3.

LIQUID FINDS ITS OWN LEVEL • Liquid pressure in side B is higher than in side A because the depth of the liquid column in side B is higher. • When the tap is opened, the liquid moves from side B to side A until both sides have the same Level.

18.3 INSTRUMENTS FOR MEASURING PRESSURE 1. BAROMETER

• A barometer is an instrument used to measure atmospheric pressure. • It consists of a glass tube closed at one end, containing mercury, standing upside down and immersed in a container that has mercury. • When the atmospheric pressure exerts pressure on the mercury in the container, it causes mercury to rise inside the glass tube. • The height of the mercury column inside the glass tube is directly proportional to the atmospheric pressure. • When the atmospheric pressure is high, the height of the mercury column increases. • When the atmospheric pressure is low, the height of the mercury column decreases. • The height of the mercury column is used to measure air pressure. • The standard atmospheric pressure is about 760 mmHg. • The higher you ascend, the lower the atmospheric pressure. Therefore, on mountains the atmospheric pressure is lower than 760 mmHg. • Mercury is used in barometer instead of water because it is denser than water. If water is used the glass tube needs to be very tall and might require a large instrument.

Pressure

109

2.

MANOMETER • A manometer is an instrument used to measure gas pressure. • It consists of a U-shaped glass tube containing liquid, usually mercury. • Mercury is usually used because it is a dense liquid. • When the manometer is not connected to a gas supply, the mercury level on both sides is equal. ❖ CLOSED - END MANOMETER

Pgas = hpg • A closed-end manometer has one arm of the tube closed and the other connected to the gas supply containing gas to be measured. • When the gas supply is connected to the manometer, gas exerts pressure on the mercury in the tube and changes the mercury level. • Gas pressure corresponds to the difference in mercury level between the two arms of the tube. ❖ EXAMPLE 1. Calculate the pressure of the gas in the flask connected to the closed-end manometer.

❖ SOLUTION 1. Pgas = hpg = 820 − 50 = 770 mmHg

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Pressure

OPEN-END MANOMETER • An open-end manometer has one arm of the tube open and the other connected to the gas supply containing gas to be measured. • When the gas supply is not connected to the manometer, only the atmospheric pressure exerts pressure. As the arms are open on both ends, the mercury level is the same on both arms. • When the gas supply is connected to the manometer, gas pressure exerts pressure on the mercury in the tube and changes the mercury level. • Gas pressure corresponds to the difference in mercury level between the two arms of the tube. • If the gas pressure from the gas supply is greater than the atmospheric pressure, the mercury level on the left side falls and the right side rises. Therefore, gas pressure (P gas) can be calculated by the following formula:

.

❖

Pgas = hpg + Patm

•

where Pgas = gas pressure from the gas supply hpg = difference in mercury level Patm = atmospheric pressure (760 mmHg)

If the gas pressure from the gas supply is lower than the atmospheric pressure, the mercury level on the left side rises while on the right side falls. Therefore, gas pressure (Pgas) can be calculated by the following Pgas = Patm − hpg

where Pgas = gas pressure from the gas supply hpg = difference in mercury level Patm = atmospheric pressure (760 mmHg)

Pressure

111

❖

EXAMPLES 1. Calculate the pressure of the gas in the flask connected to the open-end manometer. Patm = 760 mmHg. A.

B.

2.

112

Calculate the pressure of gas in the flask connected to the closed-end manometer.

Pressure

2.

Calculate the pressure of gas in the flask connected to the open-end manometer. Patm = 76 cmHg.

❖

SOLUTIONS 1. A Pgas = hpg + Patm = (320 − 20) + 760 = 1060 mmHg B. Pgas = Patm − hpg = 760 − (650 − 75) = 185 mmHg 2. Pgas = hpg = 30 − 15 = 15 cmHg 3. Pgas = hpg + Patm = (86 − 32) + 76 = 130 cmHg

18.4 TRANSMISSION OF PRESSURE IN HYDRAULIC SYSTEMS • A hydraulic system is a machine that transmits force through liquids. • Liquids are used in hydraulic systems because they are almost incompressible, their volume cannot be reduced when compressed. • Therefore, liquids can pass the force applied from one end of the system to the other end. • The simplest hydraulic system consists of a liquid-filled pipe with a smaller piston on one end and a larger piston on the other end. • When force F1 is applied to a smaller piston with are A1, the liquid transmits pressure to the larger piston with A2 producing force F2. • As liquids are incompressible, the pressure on a smaller piston is equal to the pressure on a larger piston. Therefore P1 = P2. • The formula for calculating pressure transmission in hydraulic systems is given as: 𝐅𝟏 𝐀𝟏

=

𝐅𝟐 𝐀𝟐

where F1 = force applied on a smaller piston A1 = area of a smaller piston F2 = force output on a larger piston A2 = area of a larger piston Pressure

113

❖

USES OF HYDRAULIC SYSTEM • Used in car jack • Used in brakes • Used in digging machines

❖

EXAMPLES 1. In a simple hydraulic system, a smaller piston has a cross-section of 0.02m2 and a larger piston of 0.5m2. If a force of 20N is applied to a smaller piston, calculate the force generated in a large piston. 2. The figure below shows a hydraulic system. Piston A has a contact area of 4m 2 and piston B has a contact area of 30m2. Calculate the weight that can be lifted by piston B.

❖

SOLUTIONS 1.

F1 A1

=

F2 = =

F2 A2 F1 × A2 A1 20 × 0.5

F1 = 20N A1 = 0.02m2 F2 = ? A2 = 0.5m2

0.02

= 500N 2. F2 = =

F1 × A2 A1 50 × 30 4

= 375N

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Pressure

F1 = 20N A1 = 0.02m2 F2 = ? A2 = 0.5m2

18.5 UPTHRUST AND FLOATATION ❖ UPTHRUST FORCE • Upthrust force, also known as buoyancy force, is an upward force exerted by the liquid on an object. • When an object is immersed in liquid, it experiences an upward force exerted by the liquid. • Due to the effect of upthrust, an object immersed in liquid appears to be lighter than its actual weight. • The property of liquid to exert an upward force is called buoyancy. • When the weight of an object is greater than the upthrust force, the object sinks into the liquid. • When the weight of an object is lesser than the upthrust force, the object floats on the liquid.

❖ ARCHIMEDES’ PRINCIPLE • Archimedes’ principle states that when an object is partially or completely immersed in a fluid, it experiences an upthrust which is equal to the weight of the fluid displaced by it.

❖ PRINCIPLE OF FLOATATION • When an object floats, it displaces a weight of fluid equal to its weight. • The law of floatation is applied in ships, submarines and hot air balloons. • For a ship to float on seawater it should displace water equal to its weight. • For a submarine to submerge in water, it fills its tank with water so that its weight is greater than the weight of water displaced. To surface, it pumps out water from the tank so that the upthrust is greater than its weight. • For a hot air balloon to rise, the upthrust should be greater than its total weight.

Pressure

115

REVIEW QUESTIONS

.

1. Pressure is defined as A. Force per unit distance B. Force per unit volume C. Force per unit mass D. Force per unit area 2. When the area is constant, increasing force A. Increases pressure B. Decreases pressure C. Pressure remains constant D. None of the above 3. When force is constant, increasing area A. Increases pressure B. Decreases pressure C. Pressure remains constant D. None of the above

4. On the soft ground a narrow wheel is likely to sink than a wide wheel because A. A narrow wheel has a small contact area so it exerts greater pressure on the ground B. A narrow wheel has a large contact area so it exerts greater pressure on the ground C. A narrow wheel has a small contact area so it exerts less pressure on the ground D. A narrow wheel has a large contact area so it exerts less pressure on the ground 5. Regarding pressure in liquids, which factor would generate a higher pressure at the bottom of the liquid column? A. Use a less dense liquid B. Reduce the depth of the liquid C. Increase the depth of the liquid D. All the above 6. Which of the following is a characteristic of pressure in liquids? A. Liquid finds its own level in a container B. At the same depth pressure is equal in all directions C. Pressure does not depend on the size of the container D. All the above 7. Which instrument is used to measure atmospheric pressure? A. Manometer B. Thermometer C. Barometer D. Ammeter 8. Which instrument is used to measure gas pressure? A. Manometer B. Thermometer C. Barometer D. Ammeter 9. Why is mercury often used in pressure measuring instruments instead of water? A. It is attractive B. It is dense C. It is colourless D. It is easy to find

116

Pressure

10. Why liquids are used in hydraulic systems instead of gases? A. Liquids are visible B. Liquids are easy to collect C. Liquids are easily replaced D. Liquids are incompressible

.

11. Which of the following uses hydraulic system? A. Hydraulic brakes B. Digging machines C. Car jacks D. All the above 12. Upthrust is A. The downward force exerted by the fluid on an object B. The upward force exerted by the fluid on an object C. The downward force exerted by an object on the fluid D. The upward force exerted by an object on the fluid 13. A car of mass 1200kg is at rest. Each of its tyres is in contact with 0.5m2 of the ground. What is the total pressure exerted by the car on the ground? A. 24000 Pa B. 6000 Pa C. 2400 Pa D. 600 Pa 14. A 75kg person stands on snowshoes. If the total area of the two snowshoes is 0.4m 2, what is the pressure on the snow? Take g = 10N/kg A. 30 Pa B. 187.5 Pa C. 300 Pa D. 1875 Pa 15. The figure shows the section of the tyre of the bike. The tyre exerts pressure on the ground.

A. Define: i. Pressure, ii. Pascal. B. The mass of the bike and rider is 70kg and the total surface area in contact with the ground is 0.002m2. Calculate the pressure exerted on the ground. C. The temperature of the tyre is increased by heat from the surrounding and friction with the road. Using kinetic energy, explain what happens to the pressure inside the tyre if the size of the tyre does not change. 16. A needle with a point of the area of 0.1mm2 is pushed into the skin with a force of 5N. A. State the relationship between pressure and force. B. Calculate the pressure exerted on the skin by the sharp point of the needle.

Pressure

117

17. The diagram below shows a gas manometer that contains a liquid with a density of 10000kg/m3. The

manometer is connected to the gas and while the other end is open to the atmosphere with a pressure of 100kPa. Calculate the pressure of the gas supply. .

118

Pressure

REVIEW QUESTIONS D A B A C D C A B D D B A D

.

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14.

15. A. i. Pressure is the force acting on a unit area ii. Pascal is the SI unit of pressure B. F = W W = mg = 70 × 10 = 700N P= =

F A 700 0.002

Force acting on the book is weight of the book Weight = mass (m) × acceleration due to gravity (g)

P=? F = W = 700N A = 0.002m2

= 350 000 Pa = 350 kPa C. Pressure is directly proportional to temperature. An increase in temperature increases pressure. This is because as temperature increases particles gain more kinetic energy and more rapidly. This increases the collisions of particles with the walls of the container. As the temperature of the tyre increased, the pressure inside the tyre increased as well. 16. A. Pressure is directly proportional to force. When force increases, pressure increases as well. B. P = =

F A 5 1 × 10−7

= 50 000 000 Pa = 50 000 kPa

P=? F = 5N A = 1 × 10−7m2; 1mm2 = 0.000001m2 0.1mm2 = 0.0000001m2 = 1 × 10−7 m2

17. Pgas = hpg + Patm = (0.2 × 10000 × 10) + 100000 = 120 000 Pa = 120 kPa

Pgas = ? h = 20cm ÷ 100 = 0.2m p = 10000kg/m3 g = 10N/kg Patm = 100 kPa × 1000 = 100000 Pa

Pressure

119

19

THERMAL PHYSICS

Introduction Matter is anything that has mass and occupies space. There are three states of matter; namely, solid, liquid and gas. What happens when matter gains or loses energy? This unit covers this question and many more. It also covers different types of thermometers, how they work and gas laws.

.

Specific outcomes By the unit of this unit, you will be able to: ❖ Kinetic theory of matter: • State the kinetic theory of matter • State the properties of solids, liquids and gases, including the arrangement and movement of particles • Describe the Brownian motion of particles ❖ Change of state of matter ❖ Describe the following change of state: • Melting • Boiling • Evaporation • Condensation • Freezing • Sublimation • Deposition ❖ Differentiate boiling and evaporation ❖ Thermal expansion of matter ❖ Describe the thermal expansion of matter ❖ State the applications of thermal expansion of matter ❖ Kelvin temperature scale: • Describe temperature • State the SI unit of temperature • Convert kelvin from degrees celcius and degrees celcius to kelvin • Differentiate heat and temperature ❖ Thermometer: • State the types of thermometers • Describe the liquid-in-liquid thermometer • State the advantages and disadvantages of mercury • State the advantages and disadvantages of alcohol • State the features of the clinical thermometer • Describe the lower fixed point and upper fixed point • Describe the thermocouple, and explain how it works ❖ Gas laws: • Describe Boyle’s law and work out calculations • Describe Charles’s law and work out calculations • Describe pressure law and work out calculations • Describe the combined gas law and work out calculations

.

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Thermal Physics

THERMAL PHYSICS 19.1 KINETIC THEORY OF MATTER • Matter is anything that has mass and occupies space. • There are three states of matter; solid, liquid and gas. • Kinetic theory of matter states that: 1. Matter is made up of tiny particles, 2. Particles are in constant motion, 3. The degree of movement of particles depends on their temperature.

.

•

The state of matter of a substance depends on the arrangement and movement of particles of that substance. 1. SOLID ❖ Physical properties of solid: • Solids have a fixed shape and volume, • Solids are incompressible, • Solids do not flow, • Solids have high density. ❖ Arrangement and movement of particles: • Particles are tightly packed, • Particles are arranged in a regular pattern, • Particles vibrate at a fixed position. 2. LIQUID ❖ Physical properties of liquid: • Liquids have a fixed volume, • Liquids have no fixed shape, • Liquids are incompressible, • Liquids have high density. ❖ Arrangement and movement of particles: • Particles are loosely packed, • Particles have some freedom to move about, • Particles can move from one point to another. 3. GAS ❖ Physical properties of gas: • Gases no fixed volume or shape, • Gases are compressible, • gases have low density. ❖ Arrangement and movement of particles: • Particles are very far apart, • Particles freely move randomly at high speed.

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121

19.2 BROWNIAN MOTION • Brownian motion provides evidence of the continuous random motion of particles in the air. • A microscope is used to look into a smoke cell that contains smoke particles. • When the light hits the smoke particles, they are observed as bright specks of light. They move in a zigzag path. • The zigzag movement of smoke particles is due to the collision of smoke particles with air particles.

.

19.3 CHANGE OF STATE OF MATTER • A substance can change its state based on the amount of heat it gains or loses. • During the change of state temperature of a substance remain unchanged until it changes the state. This is because all the heat supplied is used to break the force of attraction between particles of the substance. 1. MELTING • Melting is the change of state from solid to liquid. • Particles in solid gain heat energy and increase vibrating, changing the state of a substance from solid to liquid. • The temperature required to change a solid substance to liquid is called the melting point • Substances have different melting points. Water melts at 0oC. 2. BOILING • Boiling is the change of state from liquid to gas. • Particles in a liquid state gain heat energy and move randomly at high speed, therefore changing the state of a substance to a gas. • Particles with high kinetic energy escape from the liquid as gas. • The temperature required to change a liquid substance to gas is called boiling point. • Substances have different boiling points. Water boils at 100oC. • The change of state from liquid to gas below a substance's boiling point is called evaporation. • Evaporation occurs at the surface of a liquid. 3. EVAPORATION • Evaporation is the change of state from liquid to gas below the boiling point of a liquid. • It occurs on a liquid surface. • After evaporation, the substance gets cooler because particles use energy in the substance to excite particles into gas form. • Therefore, evaporation has a cooling effect on substances. 122

Thermal Physics

DIFFERENCES BETWEEN BOILING AND EVAPORATION Boiling Occurs at a fixed temperature It is a quick process Takes place through the liquid Bubbles are formed Source of energy is energy

Evaporation Occurs at any temperature It is a slow process Takes place only at liquid surface No bubbles are formed Energy supplied by surrounding

.

❖

4.

CONDENSATION • Condensation is the change of state from gas to liquid. • Gas particles lose heat energy. This causes gas particles to reduce their movement and change the substance state to liquid. • The temperature required to change gas to liquid is called condensation point. • Substances condense at different temperatures. • Water condenses at 100oC, same as its boiling point.

5.

FREEZING • Freezing is the change of state from liquid to solid. • Particles in liquid substance lose heat energy, reduce their movement and become tightly packed forming a solid state. • The temperature required to change liquid to gas is called freezing point. • Substances freeze at different temperatures. • Water freeze at 0oC, same as its melting point.

6.

SUBLIMATION AND DEPOSITION • Sublimation is the change of state directly from solid to gas. • In sublimation, heat energy is absorbed by the substance. • Example of sublimation: dry ice changes directly from solid to gas when exposed to air. • Deposition is the change of state from gas to solid. • In deposition, heat energy is lost from the substance. • Example of deposition: water vapour solidifies directly into ice during cold weather.

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123

19.4 THERMAL PROPERTIES OF MATTER • When matter is heated, it expands and when cooled it contracts. • This happens because heat particles in an object vibrate faster and move further apart from each other. Therefore an object expands. • The opposite is also true. For example, when cooled particles of an object vibrate slower and move closer to each other, the object contracts. • The expansion of an object as a result of the increase in heat energy is called thermal expansion.

.

❖ APPLICATION OF THERMAL EXPANSION 1. Railway gaps: Gaps are left in railway lines and bridges to allow for the expansion of metals when heated. 2. Thermometer: Mercury in the thermometer expands due to an increase in heat energy. 3. Bimetallic strip: A bimetallic strip is a strip of two metals with equal length but different expansion properties. The commonest bimetallic consist of two strips of copper and iron or brass and steel. Copper expands more than iron when heated at the same temperature. Therefore, when the bimetallic strip is heated, copper expands more than iron. This causes the bimetallic strip to bend.

i. Thermostat: Thermostats are used in electrical appliances such as electric iron and stove to regulate temperature. They consist of a bimetallic strip. When the appliance is too hot, the bimetallic strip is heated. Copper expands more than iron bending the bimetallic strip and break the circuit. When cooled the bimetallic strip becomes straight and completes the circuit. ii. Fire alarm: Fire alarms contain a bimetallic strip. The heat from the fire bends the bimetallic strip and completes the circuit in the alarm. As a result, the fire alarm rings when the room catches fire. 19.5 KELVIN TEMPERATURE SCALE • Temperature is the measure of the average kinetic energy of particles in an object. • It indicates the hotness or coldness of an object. • The hotter the object the greater kinetic energy of particles there is in that object. • Kelvin (K) is the SI unit of temperature. • However, the temperature is usually measured in degrees celcius (oC). • Kelvin temperature has an absolute zero; the lowest temperature possible, which is 0 K or −273oC. • At absolute zero particle’s energy is at minimum. So, therefore, is it a state of complete lack of thermal energy. • Below absolute zero, the temperature does not exist. • Relationship between kelvin (K) and degrees celcius (oC). Tk = Tc + 273

where Tk = temperature in kelvin (K) Tc = temperature in degrees celcius (oC)

❖ EXAMPLES 1. Convert −5oC to kelvin. 2. Convert 293 K to degrees celcius.

124

Thermal Physics

SOLUTIONS 1. Tk = Tc + 273 Tk = −5 + 273 Tk = 268K

.

❖

2. Tk = Tc + 273 Tc = Tk − 273 Tc = 293 − 273 Tc = 20oC ❖

DIFFERENCES BETWEEN HEAT AND TEMPERATURE Heat It is a form of energy Flows from hot to cold Total amount of internal energy SI unit is joule (J)

Temperature It is a thermal condition of a body (hot or cold) Rise when heated and fall when cooled Average kinetic energy of particles SI unit is kelvin (K)

19.6 THERMOMETER • A thermometer is an instrument used to measure temperature. • There are various types of thermometers, each with a distinctive thermometric substance. • A thermometric substance is a substance used by the thermometer with properties that vary with temperature changes. ❖ TYPES OF THERMOMETERS Thermometer Liquid-in-glass thermometer Thermocouple Resistance thermometer Constant volume gas thermometer

Thermometric properties Volume expansion of liquid Change in electromotive force (emf) Change in electric resistance Pressure changes of a fixed mass of gas

1. LIQUID-IN-GLASS THERMOMETER • A liquid-in-glass thermometer uses the thermal expansion of liquid to measure temperature. • It consists of a bulb containing mercury or alcohol, a stem, temperature scale and a capillary tube arising from the bulb. • When the temperature increases, the liquid in the bulb expands and then rises in the capillary tube in the thermometer. • The temperature is noted by the level of liquid on the scale.

Thermal Physics

125

❖

ADVANTAGES OF MERCURY 1. It can measure high temperatures; measures temperatures from −39oC to 357oC. 2. It is visible. 3. It does not stick to glass. 4. It can measure temperature quickly and precisely. 5. Its expansion is regular.

❖

DISADVANTAGES OF MERCURY 1. It is poisonous, therefore harmful when the tube is broken. 2. It is expensive. 3. Cannot measure temperature below −39oC.

❖

ADVANTAGES OF ALCOHOL 1. It can measure low temperature; measures temperature from −122OC to 78oC. 2. It is not poisonous. 3. It is cheap. 4. Its expansion is regular, but not as mercury.

❖

DISADVANTAGES OF ALCOHOL 1. It sticks to the glass. 2. It is colourless, therefore must be coloured. 3. It evaporates. 4. It cannot measure temperature above 78oC.

2.

CLINICAL THERMOMETER

.

• • •

A clinical thermometer is a mercury-in-thermometer used to measure human body temperature. It is a type of liquid-in-thermometer. Clinical thermometers are calibrated to measure human body temperature accurately.

❖ FEATURES OF A CLINICAL THERMOMETER 1. It is very sensitive: Because it has a relatively larger bulb with a narrow capillary tube. 2. It has a quick response: Because a bulb has a thin wall. 3. It has a constriction: This prevents mercury from flowing back, which can affect the reading. 4. It has a small range scale: Clinical thermometer scale ranges from 35oC to 42oC.

126

Thermal Physics

❖

SCALES OF TEMPERATURE • When making a thermometer, scales are calibrated. • This is done by choosing two known temperatures called fixed points, then dividing them into smaller and equal units of degrees.

.

❖ LOWER FIXED POINT • The lower fixed point is the temperature of pure melting ice at standard pressure. • It is taken at 0oC. • Both ice and water are present. •

3.

UPPER FIXED POINT • The upper fixed point is the temperature of steam above boiling water at standard pressure. • It is taken at 100oC. • Both water and steam are present.

THERMOCOUPLE • A thermocouple is a thermometer used to measure extreme cold and hot substances. • It is capable of measuring temperature from −200oC to 2500oC. • It is very sensitive in measuring temperature. • A thermocouple consists of wires of different materials such as copper and iron joined at both ends called junctions. • When one junction is put in a cold object and another junction is put in a hot object, electromotive force (emf) or voltage is produced and current flows around the circuit. • A galvanometer measures the amount of generated voltage. • The amount of voltage generated varies with temperature.

Thermal Physics

127

19.7 GAS LAWS • Gases consist of particles that are in constant random motion. • These particles are constantly colliding with each other and the walls of the container. • Particles colliding with the walls of the container produce pressure in the container called gas pressure. • The number of collisions gas particles make with the walls of the container determines the size of the gas pressure.

.

1. BOYLE’S LAW • Boyle’s law states that pressure is inversely proportional to the volume at a fixed mass of gas and constant temperature. • In other words, a container with the same mass of gas and kept at a constant temperature; when the volume of the container is decreased, the pressure inside the container increases or vice versa. • The formula for calculating gas pressure involving Boyle’s law is given as: PV = constant

where P = pressure (in N/m2) V = volume (in m3)

• If the initial pressure and volume are P1 and V1, and the final pressure and volume are P2 and V2, then, • The formula for calculating gas pressure involving Boyle’s law is given as: P1V1 = P2V2

where P1 = initial pressure (in N/m2) V1 = initial volume (in m3) P2 = final pressure (in N/m2) V2 = final volume (in m3)

❖ EXAMPLE 1. A sample of air occupies a volume of 0.75 m3 at 1500 N/m2. What volume will it occupy at 2000 N/m2? 2. A gas occupies 1.4 m3 at 2400 N/m2. If the volume is reduced to 0.5 m3, what will be its pressure? ❖ SOLUTIONS 1. P1V1 = P2V2 V2 = V2 =

P1 V1

P2 1500 × 0.75 2000

P1 = 1500 N/m2 V1 = 0.75 m3 P2 = 2000 N/m3 V2 = ?

V2 = 0.56 m3 2. P1V1 = P2V2 P2 = P2 =

P1 V1

V2 2400 × 1.4 0.5

P2 = 6720 m3

128

Thermal Physics

P1 = 1500 N/m2 V1 = 0.75 m3 V2 = 0.5 m3 P2 = ?

CHARLES’S LAW • Charles’s law states that volume is directly proportional to temperature at a fixed mass and constant pressure. • In other words, a container with the same mass of gas at constant pressure, when the volume of gas is increased, the temperature inside the container increases or vice versa. • The formula for calculating gas pressure involving Charles’s law is given as:

.

2.

V T

• •

= Constant

where V = volume (in m3) T = temperature (in K)

If the initial volume and temperature are V1 and T1, and the final volume and temperature are V2 and T2, then, The formula for calculating gas pressure involving Charles’s law is given as: V1 T1

=

V2 T2

where V1 = initial volume (in m3) T1 = initial temperature (in K) V2 = final volume (in m3) T2 = final temperature (in K)

❖ EXAMPLES 1. A balloon had 10 m3 volume of air at 15oC. What would be the volume of air if the temperature was increased to 35oC? 2. A sample occupied 25 m3 of air at 47oC. What would be the temperature if the volume of air was reduced to 20 m3? ❖ SOLUTIONS 1.

V1 T1

=

V2 = V2 =

V2

T2 V1 × T 2 T1 1 × 450 278

V2 = 10.7 m3

2.

V1 T1

=

T2 = T2 =

V2 T2 V2 × T 1 V1 20 × 320 25

V1 = 10 m3 T1 = 15oC + 273 K = 288 V2 = ? T2 = 35oC + 273 = 308 K

V1 = 25 m3 T1 = 47oC + 273 K = 320 K V2 = 20 m3 T2 = ?

T2 = 256 K T2 = 273 − 256 T2 = 17oC

Thermal Physics

129

3.

PRESSURE LAW • Pressure law states that pressure is directly proportional to temperature at a fixed mass of gas and constant volume. • In other words, a container with the same mass of gas and volume, when the gas pressure is increased, increases the temperature inside the container and vice versa. • The formula for calculating gas pressure involving pressure law is given as:

.

P T

• •

= Constant

where P = pressure (in N/m2) T = temperature (in K)

If the initial pressure and temperature are P1 and T1, and the final pressure and temperature are P2 and T2, then, The formula for calculating gas pressure involving pressure law is given as: P1 T1

=

P2 T2

where P1 = initial pressure (in N/m2) T1 = initial temperature (in K) P2 = final pressure (in N/m2) T2 = final temperature (in K)

❖ EXAMPLES 1. A sample of air at 15oC has a pressure of 1000 N/m2. What would be the temperature change if pressure increased to 1500 N/m2? 2. A gas at 102oC has a pressure of 5000 N/m2. If the temperature is reduced to 27oC, what will be the pressure of the gas? ❖ SOLUTIONS 1.

P1

T1

=

T2 = T2 =

P2

T2 P2 × T1 P1 1500 × 288 1000

P1 = 1000 N/m2 T1 = 15oC + 273 K = 288 K P2 = 1500 N/m2 T2 = ?

T2 = 432 K T2 = 432 − 273 T2 = 159oC 2.

P1 T1

=

P2 = P2 =

P2 T2 P1 × T2 T1 5000 × 300 375

P2 = 4000 N/m2

130

Thermal Physics

P1 = 5000 N/m2 T1 = 102oC + 273 K = 375 K P2 = ? T2 = 72oC + 273 K = 300 K

COMBINED GAS LAW • Combined gas law combines Boyle’s law, Charles’s law and pressure law into one law. • The gas laws can be summarised by the following equation (formula):

.

3.

P 1 V1 T1

=

P2 V2 T2

where P1 = initial pressure (in N/m2) V1 = initial volume (in m3) T1 = initial temperature (in K) P2 = final pressure (in N/m2) V2 = final volume (in m3) T2 = final temperature (in K)

❖ EXAMPLE 1. A sample of gas has a volume of 10m3, a pressure of 6000 N/m2 and a temperature of 29oC. What will be the new volume if pressure was changed to 13000 N/m2 and temperature 100oC?

❖ SOLUTIONS P1 V1 P2 V2 1. = T1 T2 V2 = V2 =

P1 V1 T2

T1 P2 6000 × 10 × 373 302 × 13000

V2 = 5.7m3

P1 = 6000 N/m2 V1 = 10m3 T1 = 29oC + 273 K = 302 K P2 = 13000 N/m2 V2 = ? T2 = 100oC + 273 K = 373 K

Thermal Physics

131

REVIEW QUESTIONS 1. According to the kinetic theory of matter, which substances are made of particles? A. Gases only B. Liquid only C. All matter D. All matter except solids

.

2. According to the kinetic theory of matter, particles of matter A. Are in constant motion B. Have different shapes C. Have different colours D. Are always fluid 3. A kind of matter that keeps its shape is A. Solid B. Liquid C. Gas D. Air 4. A liquid has A. Rigid shape B. Fixed volume C. Both rigid shape and fixed volume D. Neither rigid shape nor a fixed volume 5. A gas A. Has a definite volume but no definite shape B. Has a definite shape but no definite volume C. Has fast moving particles D. Has particles that are always close together 6. What happens when a liquid becomes a gas? A. Particles give off energy B. Particles absorb energy C. Particles move closer together D. Particles slow down 7. Which of the following is true about a substance’s melting point? A. It is same as its boiling point B. It is the same as its evaporation point C. It varies depending on temperature D. It is the same as its freezing point 8. An inflated balloon will shrink if placed in a refrigerator. This is because of the gas particles in the balloon move A. Faster and become closer together B. Faster and become further apart C. Slower and become close together D. Slower and become further apart 9. Which term best describes the process by which particles escape from the surface of a liquid below its boiling point? A. Condensation B. Sublimation C. Evaporation D. Melting

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10. According to the principle of thermal expansion, what happens to matter when heated? A. It increases B. It expands C. It decreases D. It contracts

.

11. Which of the following electric devices does not use the principle of thermal expansion? A. Fire alarm B. Thermometer C. Electric iron D. Loudspeaker 12. 0oC in Kevin is equal to A. 0 K B. 273 K C. 100 K D. 150 K 13. What thermometric property does the liquid-in-liquid thermometer use? A. Volume expansion of liquid B. Change in electric resistance C. Change in electromotive force D. Pressure change of a fixed mass of gas 14. Which of the following is not a characteristic of mercury-in-thermometer? A. It sticks to the glass B. It is visible C. It responds quickly to change in temperature D. It expands regularly 15. Which of the following is not a feature of a clinical thermometer? A. It has a constriction B. The scale ranges from 0oC to 100oC C. It has a thin walled bulb D. It has a narrow capillary tube 16. Which feature of the clinical thermometer affects the sensitivity of the thermometer? A. The constriction B. The diameter of the bore C. The length of the glass D. The thickness of the bulb glass 17. Which of the following is not related to the thermocouple? A. It has two metal wires joined at both ends B. It measures temperature based on the produced electromotive force C. It can measure extreme hot temperatures such as 2000oC D. It has mercury 18. Which gas law states that at the same mass and temperature volume is inversely proportional to pressure? A. Boyle’s law B. Pressure law C. Charles’s law D. Combined gas law

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19. The Luangwa bridge in Zambia is an example of a bridge that has one end fixed and the other end is free but on rollers. The use of the rollers is to allow for expansion A. As heavy vehicles cross the bridge B. During hot weather C. When the water levels drop in the river D. When the water level rises in the river

.

20. The figure shows a thermocouple used to measure the temperature of the candle flame.

A. B. C. D. E.

Suggest metal X Name component Y Explain how a thermocouple is used to measure temperature Explain why is it capable of measuring very high temperature A thermocouple is more sensitive than a liquid-in-glass thermometer. Explain what this statement means

21. A. Define the lower fixed point and upper fixed point of s laboratory thermometer. B. Describe how you would check for the accuracy of the fixed points on a mercury-in-glass Thermometer. C. State two differences between heat and temperature D. Draw and label a mercury-in-glass clinical thermometer E. Explain why a clinical thermometer has a: i. Small bulb, ii. Constriction, iii. Bulb made of thin glass, iv. Short range.

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SOLUTIONS C A A B C B D C C B D B A A B B D A B

20. A.

.

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19.

Iron wire

B.

Galvanometer

C.

Thermocouple consists of two wires joined at both ends. When one end is put in a cold object and another end in a hot object, an electromotive force is generated and current passes through the circuit. Galvanometer measures the amount of electromotive generated varies with the amount of temperature.

D.

Thermocouple can measure very high temperature because it converts heat to voltage and use metal wires can withstand high temperature.

E.

Thermocouple is more responsive to small temperature changes than a liquid-in-glass thermometer.

21. A.

Lower fixed point is the temperature of pure melting ice at standard pressure. The upper fixed point is the temperature of steam above pure boiling water at standard pressure.

B. Immerse the thermometer bulb into a container filled with pure melting ice and check if it corresponds with the lower fixed point. Then, place the thermometer bulb into a container filled with pure boiling water so that it is in contact with steam and check if it corresponds with the upper fixed point. If the temperature of pure melting ice corresponds with the lower fixed point and the temperature of steam corresponds with the upper fixed point then the thermometer is accurate. C.

i.

Heat is a form of energy while temperature is a measure of hotness or coldness.

ii. Heat flows from hot object to cold object while temperature rises when heated and falls when cooled.

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135

D.

E.

136

i. ii. iii. iv.

The small bulb contains little mercury therefore absorbs heat in a short period of time. Constriction prevents the back flowing of mercury when it expands. Thin walled glass bulb enables heat to be transferred easily to mercury. Clinical thermometer has a short range of 35oC to 42oC because human temperature rarely goes below 35oC or rises above 42oC.

Thermal Physics

20

TRANSFER OF THERMAL ENERGY

Introduction It is a clear day. Not only can you see the sun, but you feel the warmth from the sun. How is heat transferred from the sun to your body? How about heat energy from the stove, through the pot, to the food being cooked? There are three ways in which heat is transferred from one place to another. These are conduction, convection and radiation. This unit covers each of these and their applications in everyday lives. .

Specific outcomes By the end of this unit, you will be able to: ❖ Define heat transfer ❖ Conduction: • Describe conduction • State applications of conduction • Describe an experiment to compare the thermal conductivity of metals • Describe an experiment to show that water is a poor conductor of heat ❖ Convection: • Describe convection • Describe an experiment to demonstrate convection current in air ❖ Radiation: • Describe radiation • Describe an experiment to show which surface; dull or shiny, is a good absorber of heat • Describe an experiment to show which surface; dull or shiny, is a good emitter of heat ❖ Describe the application of heat transfer in the following: • Vacuum flask • Sea breeze and land breeze • Electric kettle .

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137

TRANSFER OF THERMAL ENERGY 20.1 HEAT TRANSFER • Heat transfer involves the transfer of heat energy between objects. • Heat energy is transferred from an object with a higher temperature to an object with a lower temperature until both objects have the same temperature. • Heat energy is transferred between objects in three ways: 1. Conduction 2. Convection 3. Radiation

.

20.2 CONDUCTION • Conduction is the transfer of heat energy through an object from a hotter part to a colder part without an object moving. • A medium is required for conduction to occur. Medium is simply particles (atoms, molecules or ions) of an object. Without these particles, conduction cannot take place. • Vacuum space does not have particles. Therefore, conduction does not occur in a vacuum. • For heat to be transferred by conduction from one object to another, objects must be in contact. • Heat energy is transferred by vibrating particles colliding and passing their energy to adjacent particles.

• • • •

Metals are good conductors of heat because particles are closely packed passes hence heat energy to adjacent particles easily. The other reason is that they have free electrons that carry heat energy. Liquids and gases are poor conductors of heat because particles are loosely packed; hence it is difficult to pass heat energy to the adjacent far particles. Non-metal solids such as wood and plastics are poor conductors of heat because they do not have free electrons to carry heat energy. Heat energy trend: Metal > non-metal solids > liquid > gas

❖ APPLICATION OF CONDUCTION 1. Metals are used as cooking utensils because they are good conductors of heat. 2. Plastics are used as holding handles because they are poor conductors of heat. 3. Air insulators in roofs trap heat to keep the house cool because air is a bad conductor of heat.

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Transfer of Thermal Energy

❖

EXPERIMENT: TO COMPARE THERMAL CONDUCTIVITY OF METALS ➢ SET UP • Place iron, copper, brass and aluminium rods on a tripod stand so that they are touching on one end. • Stick a paperclip on each end of the metal rod with wax. • Heat the touching ends of the rods with a bunsen burner. • Observe the time taken for the paperclip to fall from each rod.

.

➢ RESULTS • The paperclip from the copper rod falls first, followed by aluminium then brass and finally iron. ➢ CONCLUSION • Copper is the best conductor of heat among aluminium, brass and iron.

❖

EXPERIMENT: TO SHOW THAT WATER IS A POOR CONDUCTOR OF HEAT ➢ SET UP • Put a piece of block in a test tube and hold it with a wire gauge to prevent ice from floating • Add water to the test tube • Heat the test tube just below the water surface until water starts boiling ➢ RESULTS • Ice block at the bottom of the test tube does not melt immediately even if water above is boiling ➢ CONCLUSION • Water is a poor conductor of heat

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139

20.3 CONVECTION • Convection is the transfer of heat energy by the circulation of fluid due to temperature differences within it. • Convection requires a medium to take place. • A vacuum has no particles. Therefore, convection does not occur in a vacuum. • Heat in the fluid, that is, liquid and gas, creates a convection current. As a result, warm fluid moves upward and cold fluid moves downward. • This is because warmer particles move further apart, expanding the fluid and its density reduces. As a result, it becomes lighter and moves upward. Conversely, colder particles are closely packed, so the fluid is dense, therefore moves downward.

.

❖ EXPERIMENT: TO DEMONSTRATE CONVECTION CURRENT IN AIR ➢ SET UP • Fit two glass chimneys at the top of a wooden box with a plane glass window. • Place candle below one of the chimney and lit it. • Introduce smoke into the other chimney by placing a smoldering paper over it • Observe the movement of smoke. ➢ RESULTS • Smoke from the smoldering paper goes down into the wooden box and come out from the other chimney. ➢ CONCLUSION • The smoke movement shows a convention current in the air. ➢ EXPLANATION • The air above the candle is heated and becomes less dense, hence move upward. The cooler air is driven into the wooden box by the other chimney. This creates a convection current, warm air rising and moving out of the wooden box from one chimney while cooler air descending and moving into the wooden box with the other chimney. Air movement is made visible by smoke movement.

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Transfer of Thermal Energy

20.4 RADIATION • Radiation is the transfer of heat energy from one place to another using electromagnetic waves. • Radiation does not require a medium for heat to be transferred. • It transfers heat even in a vacuum. • The heat from the sun reaches us by an electromagnetic wave called infrared rays. • An object that receives infrared rays is called an absorber. • An object that releases infrared rays is called an emitter or radiator. • Black or dull surfaces are good absorbers and good emitters. • White or shiny surfaces are poor absorbers and poor emitters.

.

❖ EXPERIMENT: TO SHOW WHICH SURFACE; DULL (BLACK) OR SHINY (WHITE), IS A GOOD ABSORBER OF HEAT ➢ SET UP • Hold corks with wax on different lids, one dull (or black) and one shiny (or white). • Place the lids near a candle, at the same distance, with corks facing away from the candle. • Observe on which lid; dull or shiny, the cork fall off first. ➢ RESULTS • Wax on the dull lid melts and cork falls off first. ➢ CONCLUSION • Dull (or black) surfaces are good absorbers of heat than shiny (or white) surfaces.

❖

EXPERIMENT: TO SHOW WHICH SURFACE; DULL (BLACK) OR SHINY (WHITE), IS A GOOD EMITTER OF HEAT ❖ SET UP • Place two thermometers near a metal block, one on a dull (or black) surface and the other on a shiny (or white) surface. • Heat the metal block with a Bunsen burner • Observe a thermometer that shows a higher temperature reading. ❖ RESULTS • The thermometer on the dull (or black) surface shows a higher temperature reading than the thermometer on the shiny (or white) surface. ❖ CONCLUSION • Dull (or black) surfaces are good emitters of heat than shiny (or white) surfaces.

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141

20.5 APPLICATIONS OF HEAT TRANSFER 1. VACUUM FLASK • Vacuum flask keeps liquid hot or cold by preventing heat loss or heat gain. • The walls of the vacuum flask have a vacuum space with silvered surfaces. • The vacuum space prevents heat loss by conduction and convection. Conduction and convection do not occur in the vacuum. • Silvered surfaces prevent heat loss by radiation. This is because shiny surfaces are poor absorbers and emitters of heat. • Therefore, heat is not easily transferred from the liquid inside the flask to the surrounding. This keeps the liquid hot for an extended period of time.

.

2. SEA BREEZE AND LAND BREEZE • During the day, lands temperature increases more than the sea. This is because hot air above the land rises and is replaced by cooler sea air. This is called sea breeze. • During the night, sea air is warmer than lands air. This is because warm sea air rises and is replaced by cooler land air. This is called land breeze.

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Transfer of Thermal Energy

ELECTRIC KETTLE • Electric kettle warms water by convection. • Water molecules at the bottom gain heat, vibrate fast and move further apart. This causes warmer water at the bottom to become lighter and moves upward. • Colder water on top has less heat. Hence they have closely packed water particles. The water is dense, therefore, moves downward. • This creates a convection current with warmer water moving upward away from the heater while colder water downward toward the heater.

.

3.

Transfer of Thermal Energy

143

REVIEW QUESTIONS 1. Which method of heat transfer does not need a medium through which to travel? A. Conduction B. Convection C. Radiation D. All the above

.

2. Conduction is A. Transfer of heat by electromagnetic wave B. Transfer of heat by the circulation of fluid C. Transfer of heat through an object from hot to cold part D. All the above 3. Convection is A. Transfer of heat by electromagnetic wave B. Transfer of heat by the circulation of fluid C. Transfer of heat through an object from hot to cold part D. All the above 4. Radiation is A. Transfer of heat by electromagnetic wave B. Transfer of heat by the circulation of fluid C. Transfer of heat through an object from hot to cold part D. All the above 5. Which method transfers heat by the vibration of particles with neighbouring particles? A. Convection B. Conduction C. Radiation D. All the above 6. Why most cooking utensils such as pots and pans are made of metals? A. Metals do not dissolve in water or cooking oil B. Metals are good conductors of heat C. Metals do not break easily D. Metals are attractive 7. Which of the following listed metals is the best conductor of heat? A. Brass B. Iron C. Aluminium D. Copper 8. Which method transfers heat by ascending of warmer particles and descending of colder particles? A. Radiation B. Conduction C. Convection D. All the above 9. The heat from the sun reaches the earth by A. Radiation B. Convection C. Conduction D. None of the above 10. Which of the following statements is correct? A. Shiny surfaces are good absorbers of heat than dull surfaces B. Dull surfaces are good absorbers of heat than shiny surfaces C. Shiny surfaces are poor absorbers of heat as dull surfaces D. Dull surfaces are poor absorbers of heat as dull surfaces 144

Transfer of Thermal Energy

11. Which of the following statements is correct? A. Shiny surfaces are good emitters of heat than dull surfaces B. Dull surfaces are good emitters of heat than shiny surfaces C. Shiny surfaces are poor emitters of heat as dull surfaces D. Dull surfaces are poor emitters of heat as dull surfaces

.

12. The purpose of vacuum space in a vacuum flask is to prevent heat loss by A. Convection only B. Conduction only C. Radiation only D. Convection and conduction 13. The diagram below shows a solar water heating system

A. B. C. D. E.

Explain why the pipes are painted black. State two reasons for the use of copper pipes as opposed to iron pipes. Explain the function of the glass cover over the pipes. Describe and explain how water circulates in the system. Explain why heat rays easily pass through the glass cover to the copper pipes but do not come through the same glass cover that easily.

14. Define the greenhouse effect and explain its effects on the environment.

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145

SOLUTIONS C C B A B B D C A B B D

13. A.

.

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

Pipes are painted black so that they absorb heat energy from the sunlight. Black is a good absorber of heat.

B.

i. Copper is a good conductor of heat. Therefore, it becomes warm easily after exposure to heat. ii. Copper does not rust.

C.

The glass cover allows heat energy to pass through but prevents it from escaping the solar panel.

D.

An electric pump drives cold water from the tank to the solar panel for warming. Warm water is sent back to the storage tank. Circulation of water in the tank relies on convection current. Warm water is light hence is found on the up part of the storage and ready for use. Cold water is dense. Hence it is found at the bottom of the tank. Cooling water in the tank becomes heavy and sinks at the bottom and is pumped to the solar panel to be warmed.

E.

Radiation from the sun passes the glass cover by visible light. Visible light has a lower wavelength. Therefore, it passes the glass easily. When it strikes materials inside the solar panel it turns into infrared, a form of heat energy. Infrared radiations have a higher wavelength so they cannot pass through the glass. Hence, heat energy is trapped inside the solar panel.

14. The greenhouse effect is the trapping of heat from the sun in the earth’s atmosphere by greenhouse gases. Greenhouse gases are carbon dioxide, methane and water vapour. These gases trap heat from escaping the earth’s atmosphere. The greenhouse effect causes global warming and climate change. This change in the earth’s atmosphere disrupts food production causing food shortages. It also has a negative impact on all earth’s inhabitants, in the sea and on land.

146

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21

HEAT CAPACITY AND LATENT HEAT

Introduction When you apply the same amount of heat to water and petrol and measure the temperature change, you couldfind that the temperature of petrol rises faster than that of water. When you continue applying the same heat to both, petrol also boils first. A substance that takes a lot of heat to change its temperature is said to have a high heat capacity. A substance that takes a lot of heat to change its state is said to have high latent heat. Therefore, water has high heat capacity and high latent heat than petrol. How do you calculate the heat capacity and latent heat of a substance? This unit covers these questions. .

Specific outcomes By the end of this unit, you will be able to: ❖ Heat capacity: • Define the following terms: • Heat capacity • Specific heat capacity • Work out heat capacity calculations using Q = C∆θ • Work out heat capacity, mass calculations using Q = mc∆θ ❖ Latent heat: • Define the following terms: • Latent heat • Latent heat of fusion • Specific latent heat of fusion • Latent heat of vaporisation • Specific latent heat of vaporisation • Work out specific latent heat of fusion, mass calculations using Q = mIf • Work out specific latent heat of vaporisation, mass calculations using Q = mIv .

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HEAT CAPACITY AND LATENT HEAT 21.1 HEAT CAPACITY ❖ HEAT CAPACITY • Heat capacity is defined as the amount of heat energy required to raise the temperature of a substance by 1oC or 1K. • The symbol of heat capacity is C. • Units of heat capacity: J/oC or J/K. • Different substances have different heat capacity. • For example, when the same mass of water and petrol are heated, petrol requires less heat while water requires more heat to be raised to the same temperature. Therefore, water has a higher heat capacity than petrol. • The formula for calculating heat capacity is given as:

.

Q = C∆θ

where Q = heat energy (in J) C = heat capacity (in J/oC or J/K) ∆θ = temperature change (in oC or K)

❖ SPECIFIC HEAT CAPACITY • Specific heat capacity is the amount of heat energy required to raise the temperature of 1kg of a substance by 1oC or 1K. • Symbol of specific heat capacity C. • Units of specific heat capacity: J/kgoC or J/kgK. • Different substances have different specific heat capacity. • The specific heat capacity of water is 4200 J/kgoC. This means it takes 4200J of heat to raise the temperature of 1kg of water by 1oC. • Among liquids, water has the highest specific heat capacity. • As 70% of the Earth’s surface is covered by water, its high specific heat capacity plays a very important role in the temperature maintenance of the earth. • When the temperature increases, water absorbs a large amount of heat without a significant increase in its temperature. When the temperature decreases, water releases a large amount of heat without a rapid drop in its temperature. • As 60 − 70% of the human body is water, it plays a significant role in body temperature maintenance. • The formula for calculating specific heat capacity is given as: Q = mc∆θ

where Q = heat energy (in J) m = mass of a substance (in kg) c = specific heat capacity (in J/oC or J/K) ∆θ = temperature change (in oC or K)

❖ EXAMPLES 1. How much heat, in kilojoules, is required to raise the temperature of 2kg of water from 30oC to 40oC. The specific capacity of water is 4200 J/kgoC 2. What is the specific capacity of lead if it takes 96J to raise the temperature of a 75g block by 10oC? 3. What is the specific heat capacity of silicon if it takes 192J to raise the temperature of 45g of silicon by 6oC?

148

Latent Heat and Heat Capacity

❖

SOLUTIONS 1. Q = mcΔθ = 2 × 4200 × (40 − 30) = 84 000 J = 84kJ

2. c = =

Q mΔθ 96 75 × 10

= 0.13 J/goC

3. c = =

Q mΔθ 192 45 × 6

= 0.71 J/goC

Q=? m = 2kg c = 4200 J/kgoC Δθ = 40 − 30 = 10oC

Q = 96J m = 75g c=? Δθ = 10oC

Q = 192J m = 45g c=? Δθ = 6oC

21.2 LATENT HEAT • Latent heat is the amount of heat energy absorbed or released when a substance changes its state without a change in temperature. • Liquid water has loosely packed particles but has limited freedom of movement. For water to change its state from liquid to water vapour, its particles gain heat energy and move farther apart • During the change of state, all the heat energy supplied to the substance is used to overcome the force of attraction that holds particles together in a substance. Therefore, the substance’s temperature does not change until it changes its state. • For example, ice at 0oC has closely packed particles that have limited freedom of movement. For ice to change its state from solid to liquid, its particles gain heat energy and move farther apart. During the change of state, the temperature does not change until all the ice changes to liquid water • Latent heat, therefore, is used to overcome the force of attraction holding particles together in a substance. ❖ TYPES OF LATENT HEAT 1. LATENT HEAT OF FUSION • Latent heat of fusion is the amount of heat energy absorbed when a substance changes from solid to liquid without a change in temperature. • Specific latent heat of fusion (If ) is the amount of heat energy absorbed when 1kg of a substance changes from solid to liquid without change in temperature. • Specific latent heat of fusion of ice is 334000 J/kg. This means it requires 334000J of heat to change 1kg of ice at 0oC to liquid water at 0oC. • The formula for calculating specific latent heat of fusion is given as: Q = mIf

where Q = heat energy (in J) m = mass of a substance (in kg) If = specific latent heat of fusion (in J/kg) Latent Heat and Heat Capacity

149

2.

LATENT HEAT OF VAPORISATION • Latent heat of vaporisation (Iv ) is the amount of heat absorbed when a substance changes from liquid to gas without a change in temperature. • Specific latent heat of vaporisation (Iv ) is the amount of heat absorbed when 1kg of a substance changes from liquid to gas without change in temperature. • Specific latent heat of vaporisation of water is 2260000 J/kg. This means that 2260000J of heat is required to change 1kg of liquid at 100oC to gas at 100oC. • The formula for calculating specific latent heat of vaporisation is given as: Q = mIv

where Q = heat energy (in J) m = mass of a substance (in kg) Iv = specific latent heat of vaporisation (in J/kg)

❖ EXAMPLES 1. How much heat is needed to change 3kg of ice at 0oC to water at 0oC? If = 334000 J/kg. 2. How much heat is needed to change 3kg of water at 100oC to water vapour at 100oC? Iv of water is 2260000 J/kg. 3. Calculate the amount of heat is needed to melt 40g of ice at 0oC. Take Iv of ice to be 334J/g. ❖ SOLUTIONS 1. Q = mIf = 3 × 334000 = 1002000J = 1002kJ

150

Q=? m = 3 kg If = 334000 J/kg

2. Q = mIv = 3 × 2260000 = 6780000J = 6780kJ

Q=? m = 3 kg Iv = 2 260 000 J/kg

3. Q = mIf = 40 × 334 = 13360J

Q=? m = 40 kg If = 334 J/kg

Latent Heat and Heat Capacity

REVIEW QUESTIONS 1. The amount of heat energy required to raise the temperature of a substance by 1oC is called A. Latent heat B. Specific heat capacity C. Heat capacity D. Latent heat of fusion

.

2. The amount of heat energy required to raise the temperature of 1kg of a substance by 1oC is called A. Latent heat B. Specific heat capacity C. Heat capacity D. Latent heat of fusion 3. The amount of heat energy absorbed or released when a substance changes its state without a change in temperature is called A. Latent heat B. Heat capacity C. Latent heat of fusion D. Latent heat of vaporisation 4. The amount of heat energy absorbed when a substance changes from solid to liquid without a change in temperature is called A. Latent heat B. Heat capacity C. Latent heat of fusion D. Latent heat of vaporisation 5. The amount of heat absorbed when a substance changes from liquid to gas without a change in temperature is called A. Latent heat B. Heat capacity C. Latent heat of fusion D. Latent heat of vaporisation 6. The specific heat capacity of water is A. 1000 J/kgoC B. 334000 J/kg C. 4200 J/kgoC D. 2260000 J/kg 7. The specific latent heat of fusion of ice is A. 1000 J/kgoC B. 334000 J/kg C. 4200 J/kgoC D. 2260000 J/kg 8. The specific latent heat of vaporisation of water is A. 1000 J/kgoC B. 334000 J/kg C. 4200 J/kgoC D. 2260000 J/kg 9. What is meant by the statement “the specific heat capacity of water is 4200 J/kgoC? 10. What is meant by the statement “the specific latent heat of fusion of ice is 334000 J/kg? 11. What is meant by the statement “the specific latent heat of vaporisation of water is 2260000 J/kg? 12. What is the specific capacity of aluminium if it takes 3600J to raise the temperature of a 2kg aluminium block by 2oC?

Latent Heat and Heat Capacity

151

13. In an experiment to determine the specific heat capacity of substance P, 50g of a substance was heated and the temperature rose from 20oC to 65oC in 20 minutes using 9450J energy. A. What is meant by specific heat capacity? B. Calculate the specific heat capacity of substance P. C. What would substance P likely be?

.

14. The following was collected and recorded during an experiment to determine the specific heat capacity of water. ➢ Heater rating: 42W, 12V ➢ Mass of water: 300g ➢ Initial water temperature: 20oC ➢ Final water temperature: 40oC ➢ Time taken to heat water: 10 minutes A. Calculate the specific heat capacity of water from the data collected. B. To the 300g of water at 40oC, 200g of cold water at 20oC was added. Calculate the final temperature of water.

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Latent Heat and Heat Capacity

SOLUTIONS C B A C D C B D

.

1. 2. 3. 4. 5. 6. 7. 8.

9. 4200J of heat is required to raise 1kg of water by 1oC 10. 334000J of heat is used to change 1kg of ice at 0oC to water at 0oC 11. 2260000J of heat is used to change 1kg of water at 100oC to gas at 100oC 12. c = =

Q mΔθ 3600 2×2

= 900 J/kgoC

Q = 3600J m = 2kg c=? Δθ = 2oC

13. A. Specific heat capacity is the amount of heat energy required to raise 1kg of a substance by 1oC. B. c = =

Q mΔθ 9450 50 × 45

= 4.2 J/goC

Q = 9450J m = 50g c=? Δθ = 60oC − 20oC = 45 oC

C. Substance P is water. The specific heat capacity of water in joules per gram degree Celsius is 4.2 J/goC. 14. A. Energy = power × time = 42 × 600 = 25200J

c= =

Q mΔθ 25200 300 × 20

= 4.2 J/goC

P = 42W E=? t = 10 min × 60 = 600sec Q = 25200J m = 300g c=? Δθ = 40oC − 20oC = 20 oC

B. Energyloss = Energygain mcΔθloss = mcΔθgain 300 × 4.2 × (40 − T) = 200 × 4.2 × (T − 20) 50400 − 1260T = 840T − 16800 2100T = 67200 T = 32oC Final temperature of water is 32oC

• • • • •

Cold water will warm up, the warm water will cool down Energy lost by warm equals energy gain by cold water Cold and warm water will wind up at the same temperature T Cold water warms up from 20 oC to T, therefore Δθ = T − 20 oC Warm water cools down from 40oC to T, therefore Δθ = 40 oC − T Latent Heat and Heat Capacity

153

22

WAVE MOTION

Introduction What can you see when you throw a stone in the water body such as dam, river or pond? Ripples form on the surface of water. Water ripples are examples of wave motions. A wave motion transfers energy from one point to another. There are two types of wave motions: transverse waves and longitudinal waves. .

Specific outcomes This unit covers the properties of these waves. By the end of this unit, you will be able to: ❖ Define a wave ❖ Describe a transverse wave, and give examples of transverse wave ❖ Describe a longitudinal wave, and give examples of longitudinal wave ❖ Define the following wave properties: • Crest • Trough • Amplitude • Wavelength • Frequency • Period • Speed ❖ Differentiate transverse wave and longitudinal wave ❖ Calculate the wavelength, frequency, period and speed of the wave .

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Wave Motion

WAVE MOTION 22.1 WAVE • Wave is defined as a disturbance travelling in a medium that transports energy from one location to another. • As a wave is travelling, energy is transported through the medium without the medium moving. • There are two forms of wave motions: 1. Transverse wave, 2. Longitudinal wave.

.

22.2 TRANSVERSE WAVE • A transverse wave is a wave in which the direction of disturbance is at a right angle to the direction of travel of the wave. • Examples of transverse waves: ripples on the surface of water, electromagnetic waves, vibrations in a guitar string.

The direction of disturbance of particles is at right angle to the direction of the wave

22.3 LONGITUDINAL WAVE • A longitudinal wave is a wave in which the direction of disturbance is parallel to the direction of travel of the wave. • Examples of longitudinal waves: sound waves, waves on spring.

The direction of disturbance of particles is parallel to the direction of the wave

Wave Motion

155

22.4 WAVE PROPERTIES

. 1. CREST • Crest is the highest point the particle of the medium rises to. 2. TROUGH • Trough is the lowest point the particle of the medium sinks to. 3. AMPLITUDE • Amplitude is the maximum displacement of the particle in a medium from its rest position. • It is also defined as the distance from the resting position to crest or trough. • Amplitude indicates the amount of energy of the wave. • High amplitude means high energy wave and low amplitude means low energy wave. • The symbol of amplitude is A. • SI unit of amplitude: metre (m). 4. WAVELENGTH • Wavelength is the distance between two successive crests or troughs. • The symbol of wavelength is λ. • SI unit of wavelength is a metre (m). 5. FREQUENCY • Frequency is the number of complete waves generated per second. • Frequency indicates how fast particles of the medium vibrate when the wave is passing. • It is inversely proportional to wavelength. So the higher the frequency, the lower the wavelength or vice versa. • The symbol of frequency is f. • SI unit of frequency is Hertz (Hz). 6. PERIOD • Period is the time taken for one complete vibration of a particle. • It is inversely proportional to frequency. Thus, the higher the period, the lower the frequency or vice versa. • The symbol of period of the wave is T. • SI unit of period of the wave is seconds (s). • The formula the calculating period (T) or frequency of the wave (f). T=

156

Wave Motion

1 f

where T = period (in s) f = frequency (in Hz)

SPEED • Speed is the distance travelled by a wave per unit time. • It is directly proportional to wavelength and inversely proportional to frequency. As the wavelength of a wave increases, speed also increases. However, as the frequency increases, speed decreases • The symbol of speed of the wave is V. • SI unit of speed of the wave is metres per second (m/s). • The formula for calculating speed, frequency or wavelength of the wave is given as:

.

7.

V = fλ

where V = speed (m/s) f = frequency (in Hz) λ = wavelength (in m)

❖ EXAMPLES 1. The speed on a rope is 12 m/s and its wavelength is 60cm. What is the frequency? 2. A travelling wave has a wavelength of 0.4m, amplitude of 0.15m and frequency of 8 Hz. A. What is the period of the wave? B. What is the speed of the wave? ❖ SOLUTIONS 1. V = fλ f= f=

V λ 12

V = 12 m/s λ = 60cm ÷ 100m = 0.6m f=?

0.6

f = 20 Hz

2. T =

1 f 1

T=

8

T = 0.13s V = fλ V = 8 × 0.4 V = 3.2 m/s

V=? λ = 0.4m f = 8 Hz

Wave Motion

157

REVIEW QUESTIONS 1. Which of the following is true about transverse waves? A. The particles vibrate along the same direction as the wave motion B. The particles vibrate in a perpendicular direction with respect to the wave motion C. The particles move along the same direction as the wave motion D. The particles move against the direction of the wave motion

.

2. Which of the following is true about the longitudinal wave? A. The particles vibrate along the same direction as the wave motion B. The particles vibrate in a perpendicular direction with respect to the wave motion C. The particles move along the same direction as the wave motion D. The particles move against the direction of the wave motion 3. A wave created by shaking a rope up and down is called A. Sound wave B. Longitudinal wave C. Transverse wave D. None of the above 4. Sound wave is an example of A. Longitudinal wave B. Transverse wave C. All the above D. None of the above 5. The distance between the midpoint to the crest or trough is called A. Wavelength B. Frequency C. Period D. Amplitude 6. The distance from one top of the crest to the top of the crest of the next is called A. Wavelength B. Frequency C. Period D. Amplitude 7. The number of cycles per one unit of time is called A. Wavelength B. Frequency C. Period D. Amplitude 8. The time taken for a wave to make one complete cycle is called A. Wavelength B. Frequency C. Period D. Speed 9. Distance travelled by a wave per unit of time is called A. Wavelength B. Frequency C. Period D. Speed 10. Hertz is a unit of A. Wavelength B. Frequency C. Period D. Speed 158

Wave Motion

11. A wave produces 20 complete cycles in 5 seconds. What is its frequency? A. 20 Hz B. 5 Hz C. 100 Hz D. 4 Hz

.

12. The period of the wave is 10 seconds. What is its frequency? A. 10 Hz B. 1 Hz C. 0.1 Hz D. 100 Hz 13. A pendulum swings one complete up and down in 3 seconds. What is its period? A. 3 sec B. 0.3 sec C. 1 sec D. 30 sec 14. The frequency of a certain wave is 0.5 Hz. What is the period? A. 2 sec B. 0.5 sec C. 5 sec D. 10 sec 15. A certain wave has a frequency of 5 Hz and a wavelength of 20 m. What is the speed of the wave? A. 20 m/s B. 100 m/s C. 5 m/s D. 0.25 m/s 16. A leaf in a pool oscillates up and down three complete cycles in 1 second as a wave passes. The wave’s wavelength is 2 metres. What is the wave’s speed? A. 3 m/s B. 1 m/s C. 6 m/s D. 2 m/s 17. A wave of frequency 12 000 Hz travels 1200m in 5 seconds. What is the wavelength of the wave? A. 5m B. 0.1m C. 2m D. 0.02m 18. There are two types of waves. A. State the two types of waves B. State the differences between the two types of waves 19. Draw a labelled diagram showing a longitudinal wave covering two complete wavelengths?

Wave Motion

159

SOLUTIONS B A C A D A B C D B B C A A B C D

18. A. B.

.

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17.

Transverse wave and longitudinal wave Transverse wave Particles vibrate perpendicular to the direction of wave motion Consists of crests and troughs Examples; water waves, light wave

19.

160

Wave Motion

Longitudinal wave Particles vibrate along the direction of wave Consists of regions of compression and rarefaction Examples; sound wave, waves on spring

23

ELECTROMAGNETIC SPECTRUM

Introduction Likely, you have talked on a cellphone, listened to the radio, turned on a light, or used a microwave. If so, you have experienced electromagnetic radiations. Electromagnetic radiations are self-propagating waves that transfer energy from one place to another. These radiations have both electrical and magnetic properties, hence the name. The range of all electromagnetic radiations is known as the electromagnetic spectrum. All waves of the electromagnetic spectrum travel at the same speed but have different wavelengths and frequencies. When placed in the order of increasing frequency, they are gamma rays, x-ray, ultraviolet, visible light, infrared, microwave, and radio. Most of the electromagnetic radiations on earth originate from the sun.

.

Specific outcomes This unit covers the waves of the electromagnetic spectrum. By the end of this unit, you will be able to: ❖ Arrange the components of the electromagnetic spectrum based on increasing wavelength and frequency ❖ State the properties of the electromagnetic spectrum ❖ Describe the following components of the electromagnetic spectrum: • Gamma ray • X-ray • Ultraviolet • Visible light • Infrared • Microwave • Radio

.

Electromagnetic Spectrum

161

ELECTROMAGNETIC SPECTRUM 23.1 The electromagnetic spectrum is divided into seven regions based on wavelength and frequency.

. Electromagnetic spectrum components memory aid: based on increasing frequency Rich Radio Men Microwave In Infrared Vegas Visible light Use Ultraviolet Xpensive X-ray Gadgets Gamma ray

23.2 PROPERTIES OF ELECTROMAGNETIC SPECTRUM 1. All travel at a speed of 3 × 108 m/s. 2. They can travel in a vacuum. 3. They are all transverse waves. 4. They carry energy from one place to another. 5. They obey V = fλ equation. 6. They can be absorbed or emitted by an object.

23.3 COMPONENTS OF ELECTROMAGNETIC SPECTRUM 1. GAMMA RAY • Gamma rays are high energy and penetrating rays. • Earth’s atmosphere absorbs any gamma ray that reaches the planet. • They can destroy living cells. • They are medically used to kill cancer cells in cancer treatment.

162

Electromagnetic Spectrum

2.

X − RAY • X-ray are high energy rays. • They are able to penetrate hard objects. • They are used in imaging technology such as X-rays in hospitals, airport luggage screening.

3.

ULTRAVIOLET • Ultraviolet are energetic enough to cause problems to life. • They can cause sunburn and skin cancer. • The ozone layer absorbs most ultraviolet rays from the sun.

4.

VISIBLE LIGHT • Our eyes are sensitive to visible light. • This is a wave that makes seeing possible. • Visible light consists of seven colours of different wavelengths: red, orange, yellow, green, blue, indigo and violet.

5.

INFRARED RAYS • Infrared rays are responsible for transferring heat energy. • The warmth we feel from the sun is infrared. • They are used in remote control.

6.

MICROWAVE • Microwaves have a lower wavelength. • They can penetrate objects and deposit heat. • They are used in the microwave oven. • They are also used in satellite communications.

7.

RADIO WAVE • Radio waves have the lowest wavelength in the electromagnetic spectrum. • They are used in telecommunications; TV, radio, cell phones.

.

Electromagnetic Spectrum

163

REVIEW QUESTION 1. Which of the following is not an electromagnetic wave? A. Light wave B. Radio wave C. Sound wave D. Infrared wave

.

2. Which of the following is not a property of the electromagnetic spectrum? A. They can travel in a vacuum B. They travel at 3 × 108 m/s C. They transfer energy from one place to another D. They are longitudinal waves 3. Electromagnetic wave used in television is A. Gamma rays B. Radio wave C. Infrared waves D. X – rays 4. What type of wave is absorbed by the ozone layer to protect us from the sun? A. Radio wave B. Visible light C. Microwave D. Ultraviolet 5. Which electromagnetic wave has the longest wavelength and shortest frequency? A. Gamma rays B. Radio C. Visible light D. Infrared 6. Which electromagnetic wave has the longest frequency and shortest wavelength? A. Gamma rays B. Radio C. Visible light D. Infrared 7. What type of wave is used for airport security scanning? A. Gamma ray B. Ultraviolet C. Infrared D. X – ray 8. What type of wave is used for medical images? A. Gamma ray B. Ultraviolet C. Infrared D. X – ray 9. What type of wave is used medically to destroy cancer cells? A. Gamma ray B. Ultraviolet C. Infrared D. X – ray 10. What type of wave is used for cooking? A. Visible light B. Microwave C. Radio D. Infrared 164

Electromagnetic Spectrum

11. What is the most important source of electromagnetic energy for the earth? A. Electricity B. Moon C. Sun D. Water

.

12. The figure shows components in the electromagnetic spectrum in order of decreasing frequency. Decreasing frequency Gamma

X-rays

X

Light

Infrared

Microwaves

Y

Two components X and Y have not been maned A. Name radiation X and Y B. Define frequency of a wave C. State the speed of the wave in the figure above D. State one property, other than the speed, that all electromagnetic waves have in common 13. The electromagnetic spectrum accommodates all the electromagnetic waves. The figure below represents the electromagnetic spectrum. 1

X-rays

2

3

Infrared

4

TV and Radio waves

A. Name the radiations 1, 2, 3, and 4, B. State the two properties common only to electromagnetic waves and not any other transverse waves. C. What is the source of the television waves?

Electromagnetic Spectrum

165

SOLUTIONS C D B D B A D D A B C

.

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

12. A. X = Ultraviolet Y = Radio wave B. Frequency of a wave is the number of complete waves generated per second. C. 3 × 108 m/s D. They can travel in a vacuum. 13. A. 1 = Gamma ray 2 = Ultraviolet 3 = Visible light 4 = Infrared B. They all travel at the speed of 3 × 108 m/s They can travel in a vacuum. C. Television and radio stations.

166

Electromagnetic Spectrum

24

LIGHT

Introduction What is enabling you to see the printed words of this book, colours on this page, the colour of your pen, pencil or anything surrounding you? It is light! For you to see any object, light from the object enters your eye. The object can be a source of light such as a light bulb, candlelight, light from the phone’s screen or the object can reflect light from other sources such as sun, moon or streetlight.

.

What is light? It is an electromagnetic wave. Therefore, it has the properties of other electromagnetic waves. How does light travel? Can light bend? Can it be split into components? This unit covers these and many more questions.

Specific outcomes By the end of this unit, you will be able to: ❖ State the properties of light ❖ Describe the formation of shadow ❖ Reflection of light: • Define reflection • Draw the reflected ray, normal, incident ray and identify the angle of reflection and angle of incidence • State the law of reflection • Differentiate regular reflection and diffuse reflection • State the characteristics of the virtual image • Describe how to locate the virtual image in a plane mirror ❖ Refraction: • Define refraction • Draw the refracted ray, normal, incident ray and identify the angle of refraction and angle of incidence • Explain why light bends • Describe how light bends when it travels • Describe an experiment to demonstrate that light bends when it passes between different medium • State Snell’s law ❖ Critical angle: • Describe the critical angle and total internal reflection • State the applications of optical fibres ❖ Lenses: • Differentiate convex lens and concave lens • Differentiate real image and virtual image • Diagram images on a ray diagram when • Object is beyond 2F • Object at 2F • Object is between 2F and F • Object at F • Object is between F and C ❖ Dispersion of light: • Describe the dispersion of light • State the colours of the spectrum in order

.

Light

167

LIGHT 24.1 RECTILINEAR PROPAGATION OF LIGHT • Light, also called visible light, is a component of the electromagnet spectrum. • It is a transverse wave. Thus, it has properties of transverse waves. • The light wave travels at a speed of 3 × 108 m/s. • Like other electromagnetic waves, the light wave can travel in a vacuum. • The light wave travels rectilinearly. Rectilinear means travelling in a straight line. In simple words, light travels in a straight line. • A ray is a path along which light energy travels. • A beam is a group of rays of light.

.

24.2 SHADOW FORMATION AND OPAQUE • Opaque objects are objects which do not allow light to pass through. • As light travels in a straight line, opaque objects cast a shadow behind them. • When the moon, an opaque object, comes between the sun and earth, the moon’s shadow falls on the earth. This phenomenon is observed as an eclipse. • The illustration below shows a metal ball casting a shadow on the whiteboard.

•

There are two ways in which light can change direction: 1. Reflection 2. Refraction

24.3 REFLECTION OF LIGHT • Reflection is the bouncing back of the light. • Reflection helps us to see objects. • Objects that produce light such as the sun, light bulb and candle are called luminous objects. • Objects that do not produce light are called non-luminous objects. Most objects are non-luminous • Without light, we cannot see non-luminous objects. • To see a non-luminous object, light strikes an object and reflects into our eyes. ❖ REFLECTION TERMS • INCIDENT RAY: The ray of light that strikes the surface. • REFLECTED RAY: The ray of light that leaves the surface. • POINT OF INCIDENCE: The point where a ray of light strikes the surface. • NORMAL: A perpendicular line drawn from the point of incidence. • ANGLE OF INCIDENCE: The angle between the incident ray and normal. • ANGLE OF REFLECTION: The angle between the reflected ray and normal.

168

Light

❖

LAW OF REFLECTION • The incident ray, reflected ray and the normal all lie in the same plane. • The angle of incidence is equal to the angle of reflection. ∠i=∠r

where ∠ i = angle of incidence ∠ r = angle of reflection

❖

REGULAR AND DIFFUSE REFLECTION • If a parallel beam of light strikes the plane minor and is reflected as a parallel beam, the reflection is called regular reflection. This reflection is produced by smooth surfaces such as a mirror. • If a parallel beam of light strikes the surface and is reflected in many directions, the reflection is called diffuse reflection. • Diffuse reflection occurs on surfaces that are not smooth. These surfaces produce a distorted image • Most objects show a diffuse reflection of light.

❖

PLANE MIRROR • When we see into a mirror, we see an image behind the mirror. This image is called a virtual image. • Rays from an object are reflected at the mirror and appear to our eyes as though coming from an object behind the mirror. ❖ CHARACTERISTICS OF A VIRTUAL IMAGE 1. The virtual image is upright. 2. The virtual image is laterally inverted. Thus, the right part of an object is on the left side of the virtual image and vice versa. 3. The virtual image in the mirror has the same size as an object. 4. The distance from the mirror to the virtual image is equal to the distance from the mirror to an object. Light

169

❖

HOW TO LOCATE A VIRTUAL IMAGE IN A PLANE MIRROR 1. Draw an incident ray starting from an object. 2. Draw a normal line where the incident ray strikes the mirror. 3. Use a protractor to draw a reflected ray such as ∠ i = ∠ r (angle of incident equal to the angle of reflection). 4. Repeat steps 1, 2 and 3 with another incident ray. 5. Extend the reflected rays behind the mirror until they intersect. 6. The point of intersection is the location of the virtual image. 7. Remember to use solid lines for the incident and reflected rays and dashed lines to locate the image.

24.4 REFRACTION • Refraction is the bending of light as it passes from one medium to another. ❖ REFRACTION TERMS • INCIDENT RAY: The ray that strikes the boundary of the medium. • REFRACTED RAY: The ray that bends into another medium. • NORMAL: A perpendicular line drawn from the point of incidence. • ANGLE OF INCIDENCE: The angle between the incident ray and normal. • ANGLE OF REFRACTION: The angle between the refracted ray and normal.

❖ WHY LIGHT BEND? • Refraction of light is due to the different speeds of light as it travels through different mediums of different densities. 170

Light

1.

FROM LESS DENSE MEDIUM TO DENSE MEDIUM • Light bends toward the normal when it travels from a less dense medium to a dense medium. • For example, when light travels from the air (a less dense medium) to water (a dense medium), it bends toward the normal.

2.

FROM DENSE TO LESS DENSE • Light bends away from the normal when it travels from a dense to a less dense medium. • For example, when light travels from water (a dense medium) to the air (a less dense medium), it bends away from the normal.

3.

PARALLEL TO NORMAL LINE • When light travels through the normal, it is not refracted.

Light

171

❖

EXPERIMENT: TO DEMONSTRATE THAT LIGHT BENDS WHEN IT PASSES BETWEEN DIFFERENT MEDIUM

.

❖ SET UP • Place a glass block on the plain paper. • Trace the outline of the glass block in pencil. • Put two pins on one side of the glass and label them as A and B. • Draw a straight line connecting A and B to the surface of the glass block. Label the intersect as N. • Look through the glass on the opposite side of pin A and B until you can see these two pins aligned. • Put two pins in line with pins A and B and label them as C and D. • Draw a straight line connecting C and D to the surface of the glass block. Label the intersect as point M. • Draw a straight line connecting points N and M. • Construct normal at point N and point M.

❖ RESULTS • The line ABN bends toward the normal at N, and line MCD bends away from the normal at M. ❖ CONCLUSION • Light bends when it travels between different mediums. • Light bends toward the normal when it travels from a less dense medium (air) to a dense medium (glass block). • Light bends away from the normal when it travels from a dense medium (glass block) to a less dense medium (air).

❖

LAW OF REFRACTION (SNELL’S LAW) • The incident ray, refracted ray and the normal all lie in the same plane. • The ratio of the sine angle of incidence to the sine angle of refraction is constant.

sine i sine r

= constant (n)

where sine i = sine angle of incidence sine r = sine angle of refraction constant = refractive index

❖ EXAMPLES 1. Light passes from air to the unknown material at an angle of incidence of 43o and an angle of refraction of 28o. What is the refractive index of the unknown material? 2. Light passes from air into a crown glass with a refractive index of 1.52, at an angle of 65o. What is the angle of refraction? 3. How deep is the swimming pool when the apparent depth is 15m and the refractive index of water is 1.33 172

Light

SOLUTIONS sine i 1. n = sine r sine 43 n= sine 28 n = 1.45

2. n =

sine i

sine r sine i sine r = n sine 65 sine r = 1.25 sine r = 0.725 r = 46.5o

3. refractive index =

sine 43 = 0.682 sine 28 = 0.469 n=?

.

❖

n = 1.25 sine 65 = 0.906 sine r = 0.725 r = sine−1(0.725)

real depth

apparent depth real depth = refractive index × apparent depth real depth = 1.33 × 10 real depth = 19.95 real depth = 20m

refractive index = 1.33 apparent depth = 15m real depth = ?

24.5 CRITICAL ANGLE • When an angle of incidence from a dense medium to a less dense medium reaches a certain critical value, the refracted ray lies along the boundary at 90o to normal. This angle of incidence is called the critical angle. • It is the largest angle at which refraction can still occur. • Any angle of incidence greater than the critical angle produces a reflected ray instead of a refracted ray. • This reflection of light is called total internal reflection.

Light

173

❖

CRITICAL ANGLE AND OPTICAL FIBRES • Optical fibre is a thin rod of glass. • It is designed with an angle of incidence greater than the critical angle for total internal rotation of light. • Light entering at one end undergoes repeated total internal rotation until it comes out of the other end.

.

❖ APPLICATIONS OF OPTICAL FIBRES 1. Endoscope: An instrument used in hospitals to view patients inside. 2. Internet: Optical fibre cables transmit a large amount of data at a very high is speed. 3. Sparkles of diamond: Diamond internally reflection of light is observed as beautiful sparkles.

24.6 LENSES • A lens is an optical instrument used to focus or adjust the size of an object. • Lenses are used in cameras, microscopes, magnifying glass, and our own eyes have lenses. ❖ TYPES OF LENSES 1. CONVERGING (CONVEX) LENS • Converging or convex lens has a thick centre and a thin rim. • It converges (bends inward) rays of light. • Instruments that use converging lens include a microscope, magnifying glass, camera, telescope, projector. The lens in the human eye is a converging lens. 2. DIVERGING (CONCAVE) LENS • A diverging or concave lens has a thin centre and a thick rim. • It diverges (bend outward) rays of light. • Instruments that use diverging lens include eyeglasses, laser, peephole, camera.

❖ LENS TERMS • OPTICAL CENTRE (C): The centre of the lens. • PRINCIPAL AXIS: The line passing through the optical centre (C) at a right angle to the lens. • FOCAL POINT (F): The point where rays of light meet. • FOCAL LENGTH: The distance between the optical centre (C) and the focal point.

174

Light

❖

RAY DIAGRAM • There are three actions the incident ray takes when it passes through the lens. 1. Rays parallel to the principal axis are converged through the focal point (F). 2. Rays passing through the focal point (F) are refracted parallel to the principal axis. 3. Rays passing through the optical centre (C) continue in a straight line.

❖

REAL AND VIRTUAL IMAGES

.

Real image Formed by converging rays Image is inverted Formed by a convex lens Can be projected or seen on a screen

❖

Virtual image Formed by diverging rays Image is upright Formed by a concave lens Cannot be projected or seen on screen

VARIOUS IMAGES ON RAY DIAGRAM 1. OBJECT BEYOND 2F • Image is formed between F and 2F. • Image is real. • Image is inverted (upside down). • Image is smaller than the object. • Examples: camera, the human eye.

Light

175

OBJECT AT 2F • Image is formed at 2F. • Image is real. • Image is inverted (upside down). • Image has the same size as an object. • Example: photocopier.

3.

OBJECT BETWEEN 2F AND F • Image is formed beyond 2F. • Image is real. • Image is inverted (upside down). • Image is larger than an object. • Example: microscope, projector.

4.

OBJECT AT F • Image is formed at infinity. • Rays travel parallel to each other.

176

Light

.

2.

5.

OBJECT BETWEEN F AND C • Image is formed at F and 2F. • Image is virtual. • Image is erect (upright). • Image is larger than an object. • Example: magnifying glass.

24.7 DISPERSION OF LIGHT • A prism is a transparent object with smooth and polished surfaces which refract light. • When white light passes through the prism, a band of colours called spectrum is formed. This effect is known as the dispersion of white light. • White light is a mixture of many colours and each travels at a different speed. • A prism bends each colour according to its speed. • Red travels the fastest hence located on top of the spectrum, whilst violet travels the slowest hence located at the bottom of the spectrum. • Red has the longest wavelength and shortest frequency. Violet has the shortest wavelength and longest frequency. • Colours in the spectrum are red, orange, yellow, green, blue, indigo and violet (ROY G BIV).

Light

177

REVIEW QUESTIONS

.

1. Which of the following statements is not true? A. Light is a transverse wave B. Light travels at 3.0 × 108 m/s C. Light travels in a straight line D. All are true

2. Which of the following statements is true? A. Transparent objects cast a shadow because light cannot pass through them B. Transparent objects cast a shadow because light can pass through them C. Opaque objects cast a shadow because light cannot pass through them D. Opaque objects cast a shadow because light can pass through them 3. The bouncing back of light as it hits an object is called A. Reflection B. Diffraction C. Refraction D. Dispersion 4. The law of reflection states that A. The angle of refraction is equal to the angle of reflection B. The angle of incidence is equal to the angle of reflection C. The angle of incidence is equal to the angle of refraction D. The angle of incidence is equal to the angle of dispersion 5. When a parallel beam of light hits the surface and is reflected as a parallel beam of light. This reflection is known as A. Regular reflection B. Regular refraction C. Diffuse reflection D. Diffuse refraction 6. Which of the following is not a characteristic of a virtual image on a plane mirror? A. The image is laterally inverted B. The image is erect C. The image is larger than the object D. The image is formed behind the mirror 7. A plane mirror forms an image that is A. Real and inverted B. Real and upright C. Virtual and inverted D. Virtual and upright 8. The law of refraction states that A. The ratio of the sine angle of incidence to the sine angle of refraction is always constant B. The ratio of the sine angle of reflection to the sine angle of refraction is always constant C. The angle of incidence is equal to the angle of refraction D. The angle of refraction is equal to the angle of reflection 9. The principle on which optic fibres is based is A. Reflection B. Refraction C. Dispersion D. Total internal reflection

178

Light

10. A lens that is thicker at the centre and the edges is A. Convex lens B. Diverging lens C. Concave lens D. None of the above

.

11. A lens that is thinner at the centre than the edges is A. Convex lens B. Converging lens C. Concave lens D. None of the above 12. A convex lens has a focal length f. An object is placed beyond 2f. The image formed is A. Upright B. Smaller than the object C. Located at 2f D. Virtual 13. A convex lens has a focal length f. An object is placed at 2f. The image formed A. Has the same size as an object B. Is located between f and 2f C. Is upright D. Is virtual 14. A convex lens has a focal length f. An object is placed between 2f and f. The image formed is A. Larger than the object B. Inverted C. Real D. All the above 15. A convex lens has a focal length f. An object is placed between f and the lens. The image formed is A. Real B. Upright C. Smaller than the object D. All the above 16. The splitting of white light into its component colours is called A. Dispersion B. Diffraction C. Refraction D. Reflection

Light

179

17. The virtual image may be produced by a plane mirror or a convex lens. Figures show a plane mirror and an object X. Figure 2 shows a convex lens and an object Y.

A. B. C. D.

180

Draw rays of light to locate the position of the images on both figures. What is meant by virtual image? State three characteristics of the image formed in figure 1 and figure 2. Mention one difference between the two images.

Light

SOLUTIONS D C A B A C D A D A C B A D B A

.

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16.

17. A.

B. C.

Virtual image is an image that cannot be projected on the screen Characteristics of the image in figure 1 i. it is erect ii. it has the same size as an object iii. it is laterally inverted Characteristics of the image in figure 2 i. it is erect ii. it is larger than the object iii. it is virtual D. Image in figure 1 has the same size as an object while the image in figure 2 is larger than the object.

Light

181

25

SOUND WAVE

Introduction How does sound travel from your cellphone, computer, TV, audio system or any sound source to your ear? Or first of all, what is a sound wave? It is a longitudinal wave. It travels in a medium; air, liquid or solid, by vibrating the medium particles in the direction of the wave. As sound travels, air particles are pushed together forming a slightly higher pressure area in the air. Behind, it leaves air particles that are further apart, a slightly lower pressure area. What are the slightly higher and lower pressure areas called? What are the other properties of a sound wave? What is the speed of sound in the air? How can it be calculated? This unit covers these questions.

.

Specific outcomes By the end of this unit, you will be able to: ❖ State the properties of sound wave ❖ Describe an experiment to show that sound wave needs a medium to travel ❖ Describe an experiment to determine the speed of sound in air by direct method ❖ Describe an experiment to determine the speed of sound in air by echo method ❖ Explain what is meant by pitch of sound ❖ Explain what is meant by loudness of sound ❖ Define echo ❖ Define ultrasound ❖ State applications of ultrasound ❖ Work out speed, wavelength, frequency of sound calculations using V = fλ

.

182

Sound Wave

SOUND WAVE 25.1 SOUND WAVE • A sound wave is a longitudinal wave made of vibrations in the air. • When something makes a sound, it vibrates air particles which transmit sound through the air until it reaches our ears.

.

❖ PROPERTIES OF SOUND WAVE 1. A sound wave is a longitudinal wave. 2. Particles do not move across the medium but only vibrate to and from. Sound wave transmits only the vibrations. 3. It has regions of compression; slightly higher-pressure areas. 4. It has regions of rarefaction; slightly lower-pressure areas. 5. Sound waves do not travel in a vacuum. 6. The speed of sound wave increases with an increase in temperature. 7. The sound wave travels faster in a denser medium. It is fastest in solids, then liquids and least in gases. air = 340 m/s water = about 1500 m/s glass = about 5000 m/s steel = 5940 m/s

❖ EXPERIMENT: TO SHOW THAT SOUND NEEDS A MEDIUM TO TRAVEL ➢ SET UP i. Suspend an electric bell in a bell jar. ii. Connect the bell jar to a vacuum pump. iii. Turn on the bell and listen if sound can be heard. iv. Then remove air from the bell jar using a. vacuum pump and listen if sound can be heard. ➢ RESULTS • Before air is removed, the sound of the bell is heard. • After the air is removed, sound can no longer be heard. ➢ CONCLUSION • Sound needs a medium to travel.

Sound Wave

183

❖

EXPERIMENT: TO DETERMINE THE SPEED OF SOUND IN AIR 1. DIRECT METHOD

.

i. You should have two persons, person A with a starting pistol and person B with a stopwatch, ii. Position person A and person B a distance apart in an open field, such as 400m apart in a playing ground, iii. When person A fires a pistol, person B should start the stopwatch immediately he sees smoke from the pistol and stops the stopwatch when he hears the sound, iv. Calculate the speed of sound from the distance and time taken to hear the sound. Speed =

2.

distance between person A and person B time taken to hear sound

ECHO METHOD

i.

Have two persons, person A and person B with a stopwatch positioned a distance from the wall, such as 100m from the playing ground wall ii. When person A makes a clap, person B should record the time taken to hear the echo iii. Calculate the speed of sound Speed =

distance between two persons and wall × 2 time taken to hear echo

Note: For accuracy time, repeat claps and find the average time. For example, record time taken for 30 claps. Time for each clap =

184

Sound Wave

time taken for 30 claps 30 claps

❖

EXAMPLES 1. In an experiment to measure the speed of sound by direct method, it took 1.47s for the pistol sound to be heard by the person with a stopwatch positioned 500m apart. Calculate the speed of sound. 2. In an experiment to measure the speed of sound by echo method, when a student stands 80m away from a wall 25 claps are heard over 11.75s. Calculate the speed of sound.

❖

SOLUTIONS distance between persons 1. Speed = time taken to hear sound 500 Speed = 1.47 Speed = 340 m/s

.

2. Time for each clap = Time for each clap =

time taken for 25 claps 25 claps 11.75

25 Time for each clap = 0.47s

Speed = Speed =

distance between persons and wall × 2 time taken to echo 80 × 2

0.47 Speed = 340 m/s

25.2 PITCH OF SOUND • Pitch of sound describes how high or low the sound is. • It depends on frequency. High frequency produces high pitched sound while low frequency produces a low pitched sound. • A bird’s sound is high pitched than a lion’s sound. Likewise, a woman’s voice is high pitched than a man’s voice. • The human ear can hear sound with frequency ranging from 20Hz to 20000Hz (20Hz − 20kHz). A sound below or above this range cannot be heard.

25.3 LOUDNESS OF SOUND • Loudness of sound depends on the amplitude of a sound wave. • The larger the amplitude, the more energy the wave has and the louder the sound. • Loudness of sound is measured in decibels (dB). • Sound above 80dB becomes noise to the human ear. • The table on the next page shows the loudness of sound from various sources

Sound Wave

185

Sound source Breathing Whisper Refrigerator Conversation Car Helicopter Police siren Fireworks

Decibel (dB) 10 30 40 60 70 100 120 140

25.4 ECHO • Echo is produced when a sound wave is reflected from the surface. • It is heard separate from the original sound if the sound source is closer to the observer than the reflecting surface. • To reduce the effects of echoes, building walls can be roughened up with padding and floors covered with rugs or carpets. This scatters the incident ray of sound.

25.5 ULTRASOUND • This is the sound with a frequency of more than 20kHz. • The human ear cannot hear ultrasound. ❖ APPLICATION OF ULTRASOUND 1. Prenatal scan to check the development of the baby in the womb. 2. Used in ships to find the depth of the seabed. 3. Used to check for cracks in small pipes that are too small for naked eyes to see.

186

Sound Wave

REVIEW QUESTIONS 1. Which of the following is true about the sound wave? A. It is a transverse wave B. It can travel in a vacuum C. It is part of the electromagnetic spectrum D. It travels slower than light wave

.

2. In a sound wave, molecules A. Vibrate perpendicular to the movement of the wave B. Vibrate parallel to the movement of the wave C. Move in the direction of the wave D. Vibrate in all directions 3. Compressions are formed where air pressure is A. Lower B. Higher C. Zero D. All the above 4. Rarefactions are formed where air pressure is A. Lower B. Higher C. Zero D. All the above 5. The speed of sound in the air is A. 210 m/s B. 1000 m/s C. 340 m/s D. 3 × 108 m/s 6. The pitch of the sound depends on the sound wave’s A. Wavelength B. Amplitude C. Frequency D. Speed 7. The loudness of the sound depends on the sound wave’s A. Wavelength B. Amplitude C. Frequency D. Speed 8. The range of human hearing is between A. 10Hz to 100Hz B. 20Hz to 2000Hz C. 10 to 10kHz D. 20Hz to 20kHz 9. A sound wave with a frequency above 20000Hz is called? A. Ultrasound B. Echo C. Decibel D. Pitch 10. Sound is transmitted best in A. Vacuum B. Gases C. Liquids D. Solids Sound Wave

187

11. Regarding sound and temperature A. Speed of sound decreases with an increase in temperature B. Speed of sound increases with an increase in temperature C. Speed of sound is independent of temperature D. Speed of sound is equal to the temperature

.

12. Which of the following does not produce a sound wave? A. Hammer hitting a block a rubber B. Gunfire in a room with no echoes C. Bell ringing underwater D. Explosion in outer space 13. Which of the following uses ultrasound waves? A. Prenatal scanning B. Optical fibres cables C. Telephones D. Killing cancerous cells 14. Explain why the speed of sound is fastest in solids than liquid and gas? 15. Explain why the speed of sound increases with an increase in temperature? 16. A short pulse of sound waves produces an echo from a wall 40m away. The echo arrives back at the source of the sound 0.23s after the pulse was produced. A. Calculate the speed of sound. B. If the wavelength of the sound wave is 20mm, what is the frequency of the sound?

188

Sound Wave

SOLUTIONS D B B A C C B D A D B D A

.

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13.

14. Sound wave is transmitted by particles in a medium. The denser the medium, the faster sound travels. Solids have a denser medium than liquid and gases. Solids have a denser medium as particles are closely packed. As solids have closely packed particles, they transmit sound waves quickly. 15. Increasing the temperature of a medium increases the vibration of particles. Since particles vibrate faster, a sound wave is transmitted quickly.

16. A.

Speed = Speed =

distance between sound source and wall × 2 40 × 2

time taken to hear echo

0.23 Speed = 347.8 Speed = 348 m/s B.

V = fλ f= f=

V λ 348 0.02

V = 348 m/s λ = 20mm ÷ 10cm ÷ 100m = 0.02m f=?

f = 17400Hz

Sound Wave

189

26

MAGNETISM

Introduction Speakers, transformers and electric motors depend on magnetism to work. What is magnetism? It is a force of attraction or repulsion exerted by magnets on other objects. What are the properties of magnets? What materials are attracted by a magnet? This unit covers these questions.

.

Specific outcomes By the end of this unit, you will be able to: ❖ State the properties of a magnet ❖ Differentiate a permanent magnet and a temporary magnet ❖ Describe the methods of magnetisation by: • Stroking method • Electrical method ❖ State the methods of demagnetisation ❖ Magnetic fields: • Describe the movement of magnetic field lines • Describe how to draw magnetic field lines ❖ Describe the movement of the magnetic field due to a current-carrying wire ❖ Describe the movement of the magnetic field due to current in a solenoid ❖ State the applications of magnetism

.

190

Magnetism

MAGNETISM 26.1 PROPERTIES OF A MAGNET • Magnets have a pair of poles at both ends. These two opposite poles are: • North pole • South pole • Like poles repel each other, while unlike poles attract each other: • North pole and north pole repel. • South pole and south pole repel. • North pole and south pole attract each other.

.

• • • • • •

The Force of attraction of a magnet is greater at its poles than at the middle. Magnets attract only certain materials such as iron, steel, nickel and cobalt. Materials that are attracted by a magnet are called magnetic materials. Materials that are not attracted by a magnet are called non-magnetic materials. Most substances are non-magnetic materials. Aluminium, copper, brass and lead are examples of metals that are non-magnetic. When a magnet is broken, each half becomes a new magnet with both north pole and south pole.

A bar magnet is broken down into two smaller pieces. Each piece becomes a new magnet, each with north pole and south pole

26.2 PERMANENT AND INDUCED MAGNET ❖ PERMANENT MAGNET • These are magnets that retain their magnetism. • They create their magnetic field. • They can repel another magnet. The ability to repel another magnet is a test for a permanent or temporary magnet. Permanent can repel another magnet while a temporary magnet can repel another magnet. • They are mainly made of steel. ❖ INDUCED (TEMPORARY) MAGNET • These are magnets that only become magnetic when placed in a magnetic field; that is, near a permanent magnet. • Once removed from the magnetic field, they lose their magnetism. • An induced or temporary magnet cannot repel another magnet. • The process by which a magnetic material becomes temporarily magnetised when placed near or in contact with a magnet is called induced magnetism. Magnetism

191

26.3 MAGNETISATION • The magnetic material can be made into a magnet, temporarily or permanently, by magnetisation. • For example, if an iron nail is placed in contact with a magnet, it becomes magnetised and can attract other nails. • The attracted end of a magnetic material acquires the opposite pole of the magnet.

• • • •

When iron and steel are placed in contact with a permanent magnet, iron easily becomes more magnetised than steel. When a permanent is removed, iron loses its magnetism easily but steel retains the magnetism. Iron is a soft magnetic material because it easily becomes magnetised but loses its magnetic when disconnected from the permanent. Steel is a hard magnetic material because it is hard to become magnetised but becomes a permanent magnet when magnetised.

❖ METHODS OF MAGNETISATION 1. STROKING METHOD • Stroke the unmagnetised steel bar with a permanent magnet repeatedly in the same direction. • The steel eventually gets magnetised and becomes a permanent magnet. • The diagram below depicts how magnetisation by stroking method is performed

2. ELECTRICAL METHOD • Place a steel bar in a solenoid coil. • A solenoid is a long coil of wire wrapped many turns. • If the direct current passes through the solenoid, magnetic fields are produced around it, magnetising the steel bar. • The magnet produced by the solenoid is called an electromagnet. • The strength of the electromagnet depends on the current, the number of turns of the solenoid coil and the material of the magnetised bar. Iron is easily and strongly magnetised. • The poles of the magnet depend on the direction of the current.

192

Magnetism

26.4 METHODS OF DEMAGNETISATION 1. HEATING: If a magnet is heated above 80oC it quickly loses its magnetic properties. 2. REVERSE FIELD WITH ALTERNATING CURRENT: Passing a magnet through an alternating current with reversed magnetic field demagnetises a magnet. 3. HAMMERING: Drilling and hitting a magnet hard enough repeatedly causes a magnet to demagnetise.

26.5 MAGNETIC FIELD • A magnetic field is a region around a magnet where the force of magnetism acts. • The fields can be observed by sprinkling iron filings around a magnet or moving a compass around a magnet. • The magnetic field lines move from the north pole to the south pole. • The magnetic field lines are closer to each other at the poles and far apart at the centre of the magnet. • The magnet field lines do not intersect each other. ❖ HOW TO DRAW MAGNETIC FIELD LINES 1. Place a magnet bar on a plain paper. 2. Place a compass near the edge of the magnet. 3. Draw a dot over the arrowhead of the compass. 4. Move the compass so that the tail of the compass needle is over the dot and draw another dot over the compass needle. 5. Continue with the process until the other pole of the magnet is reached. 6. Remove the compass and draw a line connecting the dots with arrows indicating the direction the compass pointed. 7. Choose another place near the compass and repeat steps 2 to 6.

❖ MAGNETIC FIELDS BETWEEN LIKE AND UNLIKE POLES

Magnetism

193

26.6 MAGNETIC FIELD DUE TO A CURRENT-CARRYING WIRE • A wire carrying an electric current produces magnetic field lines around the wire. • If a magnetic compass is placed near the wire, the needle is deflected, indicating magnetic field lines. • The right-hand grip rule determines the direction of the magnetic. • The right-hand grip rule states that if a current-carrying wire is held in the right hand such that the thumb points in the direction of the current, then the fingers wrapped around the wire show the direction of the magnetic field.

.

26.7 MAGNETIC FIELD DUE TO CURRENT IN A SOLENOID • A solenoid is a long coil of wire wrapped many turns. • When current passes through the solenoid, magnetic fields are produced similar to the magnetic field produced by the bar magnet. • One end of the solenoid behaves like a north pole and the other end behaves like a south pole. • The pole produced on each side of the solenoid depends upon the direction of the current. • The magnet field inside the solenoid is almost uniform and parallel to the axis of the solenoid. • The strength of the magnetic field depends on the strength of the current and the number of turns of the solenoid coil.

26.7 APPLICATIONS OF MAGNETISM 1. Used in transformers. 2. Used in MRI (magnetic resonance imaging), a device in hospital for imaging. 3. Used in speakers. 4. Used in electric motors. 5. Used in electric generators.

194

Magnetism

REVIEW QUESTIONS

.

1. Magnetic poles that are unlike A. Attract each other B. Repel each other C. Cancel each other D. All the above 2. Magnetic poles that are alike A. Attract each other B. Repel each other C. Cancel each other D. All the above

3. What happens when a bar magnet is broken into two halves? A. The bar magnet is demagnetised B. The magnetic poles are separated C. The magnetic field of each piece becomes stronger D. Two new magnets are formed each with north and south pole 4. Which of the following materials is a magnetic material? A. Copper B. Rubber C. Nickel D. Water 5. Which of the following will prove that a metal bar is a permanent magnet? A. If it attracts another magnet B. If it conducts electricity C. If it repels another magnet D. If it attracts magnetic materials 6. What happens to an induced magnet when it leaves the magnetic field? A. It becomes a permanent magnet B. It loses its magnetism C. It creates its own magnetic fields D. It becomes a non-magnetic material 7. In which direction are the arrows on a magnetic field line? A. From north to south B. From south to north C. From north to north D. From south to south 8. Which of the following can increase the strength of a magnet produced by electrical method? A. Increase the current B. Use an iron core C. Increase the number of turns of the solenoid D. All the above 9. Which statement describes an example of induced magnetism? A. A bar magnet swinging freely comes to rest pointing north and south B. A bar magnet attracts a piece of soft iron C. A bar magnet loses its magnetism after it is repeatedly dropped D. Two north poles repel each other while north and south attract each other 10. The region around a magnet where the magnetic force is exerted is called A. North pole B. South pole C. Magnetic field D. Magnetic material Magnetism

195

.

11. Magnetism can be considered as a A. Physical property B. Chemical property C. Both D. None

12. The figure below shows a solenoid made from the wire wound around a plastic cylinder. A current in the solenoid produces a magnetic field.

Draw the pattern of the magnetic field lines inside and outside the cylinder.

196

Magnetism

SOLUTIONS A B D C C B A D B C A

.

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

Magnetism

197

27

ELECTROMAGNETIC INDUCTION

Introduction What happens when you pass a wire between two magnets? Electricity is produced in a wire! This was discovered by Michael Faraday and opened a way of generating electricity using magnets, a process called electromagnetic induction. This unit covers electromagnetic induction. It also covers how generators and transformers work.

.

Specific outcomes By the end of this unit, you will be able to: ❖ Electromagnetic induction: • Define electromagnetic induction • Find the direction of current, force and magnetic fields using Fleming’s left-hand rule • Find the direction of current flow using Lenz’s law • State the factors that increase the induced electromotive force according to Faraday’s law ❖ Simple generator: • Define a generator • Label the parts of an AC generator, and state the function of each component • Label the parts of a DC generator, and state the function of each component • Differentiate an AC generator and a DC generator ❖ Transformer: • Define a transformer • State the components of a transformer • Explain how a transformer works • Differentiate a set-up transformer and a set-down transformer

.

• •

198

Work out voltage, number of turn in coil calculations using Work out voltage, current calculations using

Electromagnetic Induction

Vp Vs

=

Is Ip

Vs

Vp

=

Ns

Np

ELECTROMAGNETIC INDUCTION 27.1 ELECTROMAGNETIC INDUCTION • Electromagnetic induction is the production of electromotive force due to the cutting of magnetic field lines. • Voltage is induced in an electrical conductor such as copper wire whenever it cuts magnetic field lines; that is, it moves across them, but not when it moves along them or at rest.

.

❖ FLEMING’S LEFT-HAND RULE • If a current-carrying wire is placed in a magnetic field, it will experience a magnetic force. • The direction of force can be found using Fleming’s left-hand rule: • Thumb: Direction of force. • First finger: direction of magnetic fields. • Second finger: direction of current.

Electromagnetic Induction

199

❖

LENZ’S LAW • Lenz’s law states that an electric current flows in a direction such that the current opposes the change that induced it. • If a magnet moves toward the solenoid, the solenoid produces a magnetic field that tends to repel it. • If the north of the magnet moves toward the solenoid, it produces a current so that its fields have a north pole to repel the incoming magnet. • If the south of the magnet approaches the solenoid, it produces a current in that its fields have a south pole to repel the incoming magnet. • If the magnet moves away from a solenoid, it produces a magnetic field that tends to attract it • Current flows according to the magnetic field produced.

❖

FARADAY’S LAW • Faraday’s law states that the size of induced electromotive force is directly proportional to the rate at which the conductor cuts magnetic field lines. • The electromotive force increases with an increase of: 1. Speed of the magnet. 2. Number of turns of the coil. 3. Strength of the magnet.

.

200

Electromagnetic Induction

27.2 SIMPLE AC AND DC GENERATOR • A generator is a device that converts mechanical energy into electric current by electromagnetic induction. • There are two types of generators: 1. Alternating current (AC) generator, 2. Direct current (DC) generator. • Both involve rotating a coil of wire in a fixed magnetic field, cutting through the lines of the field which in turn generate an electric current in the coil.

.

1. SIMPLE AC GENERATOR • An AC generator is a device that generates an alternating current by electromagnetic induction. • An alternating current is a current which changes its direction of current flow periodically. • The voltage level also reverses along with the current. ❖ COMPONENTS OF AN AC GENERATOR 1. PERMANENT MAGNETS: They produce a magnetic field. 2. COIL: As the coil spins, it cuts through the magnetic field inducing current in the coil. The direction changes after every turn. 3. SLIP RINGS: Rotate the coil, provide electric contact with brushes and transmit electricity from the coil to the output line. 4. CARBON BRUSH: They are in contact with slip rings. They conduct generated electricity from slip rings to the output line.

Electromagnetic Induction

201

2.

DC GENERATOR • A DC generator is a device that generates direct current using electromagnetic induction. • It contains similar components as the AC generator. • The only structural difference is that the DC generator contains split commutators instead of slip rings. • Split commutators enable the current to flow only in one direction.

.

27.3 TRANSFORMER • A transformer is a device that changes the voltage of an alternating current. • It works according to the principle of electromagnetic induction. ❖ COMPONENTS OF A TRANSFORMER 1. PRIMARY COIL: Coil of copper wire connected to the alternating current source. 2. SECONDARY COIL: Coil of copper wire connected to the load. 3. IRON CORE: Where coils are wounded.

202

Electromagnetic Induction

❖

HOW TRANSFORMERS WORK • If an electric current is supplied to the primary coil, it produces a changing magnetic field in the iron core. • The magnetic field in the secondary core induces an electric current in the secondary coil by electromagnetic induction.

❖

TYPES OF TRANSFORMERS 1. STEP-UP TRANSFORMER • A step-up transformer increases the voltage of the primary source. • It has a greater number of coil turns in the secondary coil than in the primary coil.

2. STEP DOWN TRANSFORMER • A step-down transformer decreases the voltage of the supply. • It has a greater number of coil turns in the primary coil than in the secondary coil.

Electromagnetic Induction

203

❖

OUTPUT VOLTAGE CALCULATIONS • The output voltage depends on the number of turns of the primary coil and secondary coil • The formula for calculations involving voltage and number of turns

.

Vs Vp

• • • •

=

Ns

where Vs = voltage in secondary coil (V) Vp = voltage in primary coil (V) Ns = number of turns in secondary coil (turns) Np = number of turns in primary coil (turns)

Np

For an ideal transformer, the power of the primary coil (Pp) is equal to the power of the secondary coil (Ps): Pp = Ps Power is equal to current times voltage. Therefore the Pp = Ps can be written as: IpVp = IsVs IpVp = IsVs is further rearranged into the following formula below. The formula for calculations involving voltage and current is given as: Vp Vs

=

where Vp = voltage in primary coil (V) Vs = voltage in secondary coil (V) IP = current in primary coil (A) Is = current in secondary coil (A)

Is Ip

❖ EXAMPLE 1. A set up transformer has a primary coil with 250 turns and a secondary coil with 500 turns. The primary coil has a voltage of 90V. A. What is the voltage in the secondary circuit? B. If the current in the secondary circuit is 2A, how much current flows through the primary circuit? ❖ SOLUTION 1. A.

Vs

Vp

=

Vs = Vs =

Ns Np Ns × Vp Np 500 × 90

Vp = 90V Vs = ? Np = 250 turns Ns = 500 turns

250

Vs = 180V

B.

Vp Vs

=

Ip = Ip =

Is Ip I s × Vs Vp 2 × 180 90

Ip = 4A

204

Electromagnetic Induction

Vp = 90V Vs = 180V Ip = ? Is = 2A

REVIEW QUESTIONS 1. The process of inducing electromotive force by changing the magnetic field in the coil is called? A. Static electricity B. Electromagnetic induction C. Magnetisation D. All the above

.

2. If a wire is placed between magnets electromotive force is induced in a wire by A. Moving a wire up and down B. Rotating the wire C. Moving a wire sideways D. All the above 3. A magnet is moved in and out of a solenoid. Which of the following would increase the voltage induced? A. Reduce the number of coils of solenoid B. Place the magnet stationary C. Use a weak magnet D. Increase the speed of magnet in and out 4. What law states that the magnetic field of the induced current is in a direction to produce a field that opposes the change causing it? A. Fleming’s left-hand rule B. Faraday’s law C. Lenz’s law D. Ohm’s law 5. An electric current that changes direction at regular intervals is A. Alternating current B. Direct current C. Both A and B D. None of the above 6. A device that changes mechanical energy into electrical energy by electromagnetic induction is A. Motor B. Generator C. Transformer D. None of the above 7. A device that increases or decreases the voltage of alternating current is A. Motor B. Generator C. Transformer D. None of the above 8. Which statement is true concerning how the transformer works? A. Electricity is transferred from the primary coil to the secondary through an iron core B. Electricity flows from the primary coil to the secondary coil C. Electricity is induced in a rotating coil D. A changing magnetic field is transferred from the primary coil to the secondary coil 9. A step-up transformer A. Increases voltage B. Decreases voltage C. Maintains voltage D. All the above

Electromagnetic Induction

205

10. A step-down transformer A. Increases voltage B. Decreases voltage C. Maintains voltage D. All the above 11. Why is alternating current (AC) commonly used in power distribution? A. Because DC wastes electricity B. Because AC was established first C. Because DC does not last long D. Because AC is easy to step up and step down 12. The voltage in the primary coil that has 100 turns is 50V. What is the voltage in the secondary coil if it has 40 turns? A. 40V B. 20V C. 80V D. 10V

.

13. The figure shows a DC motor.

A. B. C. D. E. F.

Name parts labelled L and M. What can be used to identify the direction of current? Which terminal N or O is positive? Describe how an electric motor works. State two ways in which the coil in the electric motor can be made to rotate faster. State two changes that can be made to the construction of the DC motor to make it run as an AC generator.

14. The figure below shows a transformer.

A. B. C. D.

206

How can you tell from the diagram that this is a step-up transformer? Calculate the output voltage of the transformer. If the current in the primary circuit is 12A, how much current flows through the secondary circuit? Suggest a suitable material that could have been used to make the core of the transformer.

Electromagnetic Induction

SOLUTIONS B A D C A B C D A B D B A.

.

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13.

L = split commutator M = carbon brush B. Fleming’s left-hand rule C. Terminal O D. The current-carrying coil produces a magnetic field in wires. This induced magnetic field interact with the magnetic field from the magnets producing a force that rotate the coil. E. i. Increase the current ii. Use a stronger magnet F. i. Replace split commutators with slip rings ii. Add a shaft to rotate the coil

14. A. B.

It has a greater number of turns in the secondary coil Vs Vp

=

Vs = Vs =

Ns Np Ns × Vp Np 250 × 100

Vp = 100V Vs = ? Np = 50 turns Ns = 250 turns

50

Vs = 500V

C.

Vp Vs

=

Is = Is =

Is Ip I p × Vp Vs 100 × 12

Vp = 100V Vs = 500V Ip = 12A Is = ?

500

Is = 2.4A D.

Iron

Electromagnetic Induction

207

28

STATIC ELECTRICITY

Introduction Likely you have seen a lightning flash before. If so, you saw the effects of static electricity. All objects around us consist of atoms. Most of the atoms are neutral because they contain the same number of positive and negative charges. However, when one type of charge becomes more than the other, the object becomes electrically charged. Charges are static because they do not move or flow. However, these charges can be transferred from the electrically charged object to a neutral object or an object with an opposite charge. Thus, what is observed as a lighting is a transfer of charges from the clouds to the ground or other clouds. .

How do objects become electrically charged? What are the applications of static electric electricity? This unit answers these questions.

Specific outcomes By the end of this unit, you will be able to: ❖ Static charges: • Describe the formation of static electricity • State properties of static charges • Describe the process of static induction ❖ Describe an experiment to demonstrate static electricity ❖ Describe how static electricity is applied in the following: • Spraying paint • Photocopier • Inkjet printer • Electrostatic precipitator • State dangers of static electricity .

208

Static Electricity

STATIC ELECTRICITY 28.1 STATIC CHARGES • All matter is made up of atoms. • An atom consists of positively charged protons and neutral neutrons found in the nucleus and negatively charged electrons in the outer shells. • The number of positively charged protons and negatively charged electrons is equal. Therefore, the net charge of an atom is zero. • If an atom loses an electron, it becomes positively charged. • If an atom gains an electron, it becomes negatively charged. • The build-up of charged particles on the surface of an object is called static electricity. • These charges are static because they are at rest on the surface of an object instead of flowing. • Static charges are formed on the surface of objects by rubbing objects with each other. • When objects are rubbed (friction), some electrons move from one object to another. • The object that loses electrons becomes negatively charged while the object that gains electrons become negatively charged. • Charged particles tend to attract other charged or neutral atoms. • Like charges repel each other while unlike charges attract each other. • The diagram below depicts the interaction of charges. Positive charge and positive charge repel each other. Negative charge and negative charge repel each other. Positive charge and negative charge attract each other

.

• • •

If one object is charged while another is neutral, the charged object induces the opposite charge in the neutral object. When a neutral object becomes charged without direct contact with the charged object, the charging is called induction of charges. When a neutral object become charged by direct contact with the charged object, the charging is called conduction of charges.

Static Electricity

209

. ❖

EXPERIMENT: TO DEMONSTRATE STATIC ELECTRICITY ❖ SET UP • Get a pen and a piece of paper. • Cut the paper into smaller pieces. • Place the pen near the pieces of paper and observe if they will be attracted to the pen. • Rub the pen repeatedly through your head’s hair for about 5 seconds. • Then place the pen near the pieces of paper again and observe if papers will be attracted to the pen. ❖ RESULTS • Before rubbing the pen through the hair, it did not attract pieces of paper. • After rubbing the pen through the hair, it attracted pieces of paper. ❖ EXPLANATION • Before rubbing the pen through the hair, both surfaces of the pen and pieces of paper were neutral. • After rubbing the pen through the hair, the pen attracted pieces of paper because it became charged.

28.2 APPLICATIONS OF STATIC ELECTRICITY 1. SPRAY PAINTING • Spray painting is used to coat metal surfaces with paint particles using a spray gun. • The spray gun contains an electrostatic charging unit that has negatively charged ink. • The object to be painted is induced with positive charges. • Negatively charged ink from the gun is attracted by the positively charged object and sticks to its surface.

210

Static Electricity

2.

PHOTOCOPIER • The photocopier uses electrostatic charges to produce a copy. • The original image is placed onto a sheet of glass and an image is projected onto a positively charged drum. • The drum produces a positively charged image on which the black powder or toner which is negatively charged is attracted to producing a copy of the original image. • The copy of the original image is finally transferred to a plain paper.

3.

INKJET PRINTER • Inkjet printer consists of an inkjet nozzle, electrostatic charging unit and a deflecting plate. • The inkjet release ink droplets which are electrostatically charged by the charging unit. • The deflector deflects the charged ink droplet based on the charge of the ink to the paper.

4.

ELECTROSTATIC PRECIPITATOR • An electrostatic precipitator is a filtration device that removes dust and smoke from exhaust fumes of industries and power stations. • It consists of thin negatively charged wires and flat positively charged metal plates. • As exhaust fumes pass through the negatively charged wires, smoke particles and dust acquire negative charges. • The positively charged metal plates trap these negatively charged particles.

.

28.3 DANGERS OF STATIC ELECTRICITY 1. LIGHTNING • The rubbing of clouds and air causes clouds to be charged. • Lightning is the discharge of electrons occurring between clouds and the earth. • Lightning is very dangerous. They cause fire, can kill wildlife and even humans. • Lightning can be prevented by earthing. This involves placing a lightning conductor from the building to the earth's ground. This transfers the discharge of charges from the clouds immediately to the earth. 2. FIRE HAZARD • The flowing of flammable liquids in pipes and engines can build up static electricity. • This can create a spark that causes fire.

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211

REVIEW QUESTIONS 1. The build-up of charges on the surface of an object is called A. Static electricity B. Current electricity C. Pressure D. Magnetism

.

2. What happens when an object loses electrons? A. It becomes neutral B. It becomes positively charged C. It becomes negatively D. None of the above 3. What happens when an object gains electrons? A. It becomes neutral B. It becomes positively charged C. It becomes negatively D. None of the above 4. What happens when a positive charge and brought closer to a positive charge? A. Charges will attract each other B. Charges will repel each other C. Nothing happens D. None of the above 5. What happens when a positive charge and brought closer to a negative charge? A. Charges will attract each other B. Charges will repel each other C. Nothing happens D. None of the above 6. Why do objects become charged by rubbing? A. Objects share electrons B. One object shares a piece with another object C. One object transfers electrons to another object D. All the above 7. When a neutral object rearranges charges without direct contact with a charged object the charging is known as A. Friction B. Conduction C. Induction D. All the above 8. When a neutral object become charged by direct contact with a charged object the charging is known as A. Friction B. Conduction C. Induction D. All the above

212

Static Electricity

SOLUTIONS A B C B A C C B

.

1. 2. 3. 4. 5. 6. 7. 8.

Static Electricity

213

29

CURRENT ELECTRICITY

Introduction Imagine how life would be if there were to be a worldwide blackout for a year. Food might be scarce as most food requires electricity to produce, preserve and cook. Industries might shut down and hospitals might stop functioning properly. All of us would feel the impact of the blackout. This is because our daily lives are highly dependent on electricity.

.

What is electricity? It is a form of energy that results from the presence or flow of electrical charges. The previous unit, static electricity, covered the build-up of non-moving electrical charges on an object. How about if electrical charges were able to flow in an object such as a copper wire? The form of electricity formed in that case is current electricity, a type of electricity we use to power our cellphones, computers, TV, light the bulb, power industry machines, hospitals, etc.

Specific outcomes This unit covers current electricity. By the end of this unit, you will be able to: ❖ Current: • Define current • How current is transmitted in a conductor • State the SI unit of current and charge • Instrument used to measure, and how to measure current

.

•

Work out current, charge, time calculations using I =

❖ Electromotive force: • Define electromotive force • State the units of electromotive force •

Calculate electromotive force using V =

Q t

E Q

❖ Potential difference: • Define potential difference • State the unit of potential difference • Instrument used to measure, and how to measure potential difference •

Calculate potential difference using V =

E

Q

❖ Resistance: • Define resistance, resistor and rheostat • State the unit of resistance • State Ohm’s law •

Work out resistance, voltage, current calculations using R =

• Describe the factors that affect resistance ❖ Circuit diagram: • Differentiate series circuit and parallel circuit • Calculate current, voltage and resistance in a series circuit • Calculate current, voltage and resistance in a parallel circuit • State advantages and disadvantages of a series circuit • State advantages and disadvantages of a parallel circuit

214

Current Electricity

V I

CURRENT ELECTRICITY 29.1 CURRENT • Current is defined as the rate of flow of electrical charges. • Electricity is carried in an electric conductor such as a copper wire by electrons. • Therefore, for current to flow in a conductor, electrons flow in a wire transmitting current. • SI unit of current is Ampere (A). • SI unit of charge is Coulomb (C). • Instrument used to measure current is Ammeter. • The ammeter is connected in series with the other components to measure the current flowing in a circuit. • The formula for calculating current with charge and time is given as:

.

I=

Q t

where I = current (in A) Q = charge (in C) t = time (in s)

❖ EXAMPLES 1. What is the current if a charge of 10 coulombs flows in 5 seconds? 2. If a charge of 9000 coulomb flows for 30 minutes, what is the current? 3. A current of 2.5A flows through a bulb for 5 minutes. Calculate the amount of charge that flows through the bulb in this time. 4. How long would it take for 50 coulombs of charge to flow if the current is 2A? ❖ SOLUTIONS 1. I = I=

Q

t 10 5

I=? Q = 10C t = 5s

I = 2A 2. I = I=

Q t 9000 1800

I = 5A

3. Q = It Q = 2.5 × 300 Q = 750C

4. t = t=

Q I 50 2

I=? Q = 9000 t = 30min × 60s = 1800s

I = 2.5A Q=? t = 5min × 60s = 300s I = 2A Q = 50C t=?

t = 25s

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215

29.2 ELECTROMOTIVE FORCE (E.M.F) • Electromotive force is defined as the energy supplied per unit charge of the power source. • It indicates how much chemical energy is converted to electrical energy by the battery when one charge passes through it. • The symbol of electromotive force is V. • Units of electromotive force is Volt (V), 1V = 1 J/C. • The formula for calculating electromotive force with energy and charge through the cell is given as: V=

E Q

where V = electromotive force (in V) E = energy supplied by the cell (in J) Q = charge flow through the cell (in C)

❖ EXAMPLE 1. What is the electromotive force of the cell when 2000J of energy flows through a circuit with a charge of 400C? ❖ SOLUTION 1. V = V=

E

Q 2000 400

V=? E = 2000J Q = 400C

V = 5V

29.3 POTENTIAL DIFFERENCE • Potential difference is defined as the difference in potential energy between two points in a circuit. • It indicates the amount of energy converted to other forms by a component per unit charge passing through that component. • For instance, if the circuit has a battery and a bulb, the potential difference indicates how much electric energy is converted to light by a bulb when one charge passes through it. For example, if a battery supplied 2V before the bulb the circuit has 2V, while after the bulb the circuit has 0V. Therefore, the potential difference is 2V. • The symbol of potential difference is V. • Unit of potential difference is Volt (V). • The instrument used to measure potential difference: voltmeter. (Don’t confuse with voltameter, an electrolytic cell). • A voltmeter is connected in parallel in the circuit to measure the potential difference between two points. • The formula for calculating potential difference with energy and charge through the component is: V=

216

E Q

Current Electricity

where V = potential difference (in V) E = energy converted to other forms (in J) Q = charge flow through the component (in C)

❖

EXAMPLES 1. 650J is required to transfer 100 coulombs of charge between a component. What is the potential difference? 2. How much energy is needed to transfer 10 coulomb charges through a potential difference of 4 volts? 3. It requires 20J to transfer charges between a component with 5V. How much charge is transferred?

❖

SOLUTIONS

.

1. V = V=

E

V=? E = 650J Q = 100C

Q 650 100

V = 6.5V 2. E = VQ E = 4 × 10 E = 40J

3. Q = Q=

E

V = 4V E=? Q = 10C

V = 5V E = 20J Q=?

V 20 5

Q = 4C

29.4 RESISTANCE • Resistance is the ratio of the applied voltage to the current through the component. • Resistance opposes the flow of current. • A resistor is a device that restricts the flow of electric current. • A rheostat is an adjustable resistor used to control current. It can increase or decrease resistance. • Unit of resistance: Ohms (Ω). • Ohm’s law states that the current flowing in a circuit is directly proportional to the voltage and inversely proportional to the resistance. • The formula for calculating resistance with voltage (potential difference) and current. R=

❖

V I

where R = resistance ( in Ω) V = potential difference (in V) I = current (in A)

FACTORS AFFECTING RESISTANCE 1. LENGTH OF THE WIRE • The longer the wire, the more the resistance. • This is because there are more collisions between electrons carrying electric current and atoms of the wire in a long wire than a shorter wire.

Current Electricity

217

2.

CROSS-SECTIONAL AREA OF THE WIRE • The wider the wire, the lower the resistance. • As the cross-sectional area of a wire increases, the number of atoms of the wire that oppose flow current decreases. • As a result, the resistance decreases.

3.

MATERIAL OF THE CONDUCTOR • Materials like copper, aluminium and silver are good conductors of electric current. • Silver is the best conductor, but it is cost. Therefore copper is mainly used in household circuits. • Hard rubber is among the least conductors, therefore mainly used as an electric insulator.

4.

TEMPERATURE • Increasing the conductor’s temperature increases resistance. • As the conductor’s temperature increases atoms of the conductor vibrate more vigorously, increasing collisions between electrons carrying electric current and atoms of the conductor. • Result resistance increases.

❖

EXAMPLES 1. How many ohms of resistance must be present in a circuit that has 120V and a current of 10A? 2. A circuit contains two 1.5 batteries and a bulb with a resistance of 3Ω. Calculate the current. 3. A light bulb has a resistance of 4Ω and a current of 2A. What I the voltage across the bulb?

❖

SOLUTIONS 1. R = R=

V

I 120 10

V = 120V I = 10A R=?

R = 12Ω 2. I = I=

V R 3

V = 3V (1.5v × 2 batteries) R = 3Ω I=?

3 I = 1A

3. V = IR V=2×4 V = 8V

218

Current Electricity

I = 2A R = 4Ω V=?

29.5 CIRCUIT DIAGRAM • A circuit diagram is a graphical representation of how components are connected. • Each component is represented by a symbol. • Below are symbols commonly used in a circuit diagram.

❖ TYPES OF CIRCUITS 1. SERIES CIRCUIT

• • • •

In a series circuit, all components are arranged in a chain. All components have the same current: I = I1 = I2 Electromotive force (emf) is the sum of potential difference across all components: emf = V1 + V2 Total resistance is equal to the sum of each resistor in the circuit: RT = R1 + R2

❖ EXAMPLE

From the diagram, calculate: A. Total resistance, B. Current through the cell, C. Potential difference at each resistor.

Current Electricity

219

SOLUTION A. RT = R1 + R2 RT = 2 + 4 RT = 6Ω

B.

I= I=

R1 = 2Ω R2 = 4Ω RT = ?

V

.

❖

V = 12V R = 6Ω I=?

R 12 6

I = 2A C.

2.

Potential difference at 2Ω V = IR V=2×2 V = 4V

V=? R = 2Ω I = 2A

Potential difference at 4Ω V = IR V=2×4 V = 8V

V=? R = 4Ω I = 2A

PARALLEL CIRCUIT

• • • •

In a parallel circuit, the current has more than one pathway The total current is equal to the sum of the current in each pathway IT = I1 + I2 All components have the same potential difference as the voltage supply (emf) emf = V1 = V2 The reciprocal of total resistance is equal to the sum of the reciprocal of each resistance 1

RT

220

=

1

R1

Current Electricity

+

1

R2

❖

EXAMPLE

. From the diagram above, calculate A. V1 and V2 B. I1, I2 and total resistance (IT) C. Total resistance

❖

SOLUTION A. emf = V1 = V2 = 12V V1 = 12V V2 = 12V B.

V1

I1 = I1 =

V1 = 12V R1 = 2Ω I1 = ?

R1 12 2

I1 = 6A I2 = I2 =

V2

V2 = 12V R2 = 6A I2 = ?

R2 12 4

I2 = 3A IT = I1 + I2 IT = 6 + 3 IT = 9A C.

1 RT 1 RT 1 RT

= =

1

+

IT = ? I1 = 6A I2 = 3A 1

R1 R2 1 1 2

+

4

RT = ? R1 = 2Ω R2 = 4Ω

= 0.75

RT = 1.33Ω

Current Electricity

221

OVERVIEW OF SERIES AND PARALLEL CIRCUITS Current

Voltage Total resistance

❖

Parallel circuit • Shared amongst pathways • Increases as more components are added • Same across all components •

1 RT

=

1 R1

+

1 R2

+…

ADVANTAGES AND DISADVANTAGES OF SERIES AND PARALLEL CIRCUITS Circuits Series

Advantages • All components receive same current • All components in circuit are controlled by one switch • Requires few wires

Disadvantages • If one component is disconnected or burnt, other components stop receiving current • Adding more components increase resistance

Parallel

•

•

•

222

Series circuit • Same everywhere • Does not increase as more components are added • Shared between components • RT = R1 + R2 + …

.

❖

Current Electricity

If one component is disconnected or burnt, other components still receive current Each component receives voltage supplied by the source

• •

Adding more components requires more current Battery runs out quickly Requires a lot of wires

REVIEW QUESTIONS

.

1. The rate of flow of electric charges is called A. Voltage B. Resistance C. Current D. Potential difference 2. What is the symbol for the current? A. C B. A C. V D. Ω

3. The type of current that flows in only one direction is called A. Alternating current B. Direct current C. Potential difference D. Voltage 4. The type of current which flows in one direction then in the opposite direction is known as A. Alternating current B. Direct current C. Potential difference D. Voltage 5. Electric charge is measured in A. Ampere B. Volt C. Ohms D. Coulomb 6. How is current affected if the time interval decreases while the amount of charge remains the same? A. The current increases B. The current decreases C. The current stays the same D. The current becomes zero 7. The instrument used to measure current is called? A. Rheostat B. Voltmeter C. Voltameter D. Ammeter 8. If a charge of 27000 coulomb flows for 2 hours, what is the current? A. 2.5 A B. 4 A C. 3.75 A D. 1.8 A 9. The energy supplied per unit charge of the power source is called A. Current B. Potential difference C. Resistance D. Electromotive force 10. The unit for electromotive force is called A. Coulomb B. Volt C. Ampere D. Ohms Current Electricity

223

11. What is the electromotive force of the cell when 6.8kJ of energy flows through a circuit with a charge of 850C? A. 8 V B. 6 V C. 12 V D. 10 V

.

12. The difference in potential energy between two points of a circuit is called A. Resistance B. Current C. Potential difference D. Electromotive force 13. The instrument used to measure potential difference is called? A. Ammeter B. Voltmeter C. Voltameter D. Rheostat 14. 2.1kJ is required to transfer 420 coulombs of charge between a component. What is the potential difference? A. 4 V B. 6.5 V C. 2 V D. 5 V 15. The resistance to the flow of electrical current is known as A. Current B. Resistance C. Electromotive force D. Potential difference 16. A circuit element that restricts the flow of electric current is known as A. Circuit breaker B. Closed circuit C. Resistor D. Power 17. The combination of two or more cells in series is referred to as? A. Power B. Generator C. Battery D. Motor 18. What energy conversion occurs in a battery? A. Chemical energy to electrical energy B. Electrical energy to chemical C. Chemical energy to potential energy D. Potential energy to kinetic energy 19. What factors affect materials resistance? A. Length of the material B. Type of the material C. Cross-section of the material D. All the above

224

Current Electricity

20. The diagram showing electric circuits using circuit symbols is known as A. Series circuit B. Parallel circuit C. Circuit diagram D. Closed circuit

.

21. How does the thickness of a wire affect the resistance through the wire? A. The thicker the wire the higher the resistance B. The thicker the wire the lower the resistance C. The thinner the wire the lower the resistance D. Resistance is not affected by the thickness of the wire 22. A circuit that has two or more components in one electrical pathway is known as A. Open circuit B. Series circuit C. Closed circuit D. Parallel circuit 23. A circuit that has two or more electrical pathways for components to have electricity is known as A. Circuit diagram B. Closed circuit C. Series circuit D. Parallel circuit 24. The direction of flow of electric current is from A. Positive to negative B. Negative to positive C. Positive to positive D. Negative to negative 25. What conditions must be met for charges to flow through a circuit? A. The circuit must be closed B. Potential difference must exist in the circuit C. Must have a source of charge D. All the above 26. The figure shows an electrical circuit containing two resistors

A. i

On the diagram above, draw a voltmeter to measure the potential difference across the 6Ω resistor ii. When the switch is closed, calculate the current through the ammeter A

B. i. What is the potential difference across the 6Ω? ii. If the resistor 6Ω is replaced with the 4Ω resistor in parallel, find the current in the circuit.

Current Electricity

225

27. A. State ohm’s law and define resistance. B. Derive an expression for determining total resistance when two resistors in series are connected to a pair of resistors in parallel in the same circuit. C. The figure below shows a circuit designed for effective resistance.

.

Neglecting the battery resistance, calculate the ammeter reading when the: i. switch is open, ii. switch is closed. 28. The figures below show two electrical circuits. Two circuits are using the identical batteries

A. State the circuit with parallel connection and that with series connection of resistors. B. The two resistors P and Q are light bulbs. State two disadvantages of connecting the bulbs as shown in figure 1. C. In figure 1, the ammeter reads 1.5A when the switch S0 is closed. Calculate the voltmeter reading in the circuit. D. When switches S1 and S2 in figure 2 are closed, what is the combined resistance of the circuit? 29. In the experiment to verify Ohm’s law for a conductor made from constantan wire, the following results were obtained at 25oC. Voltage (volts) Current (amperes)

0 0

1.5 0.2

3.0 0.4

4.5 0.6

6.0 0.8

7.5 1.0

9.0 1.2

A. Draw a circuit diagram for the experimental setup used to obtain the results shown in the table Above. B. i. Plot a graph of voltage against current. ii. Calculate the gradient of the graph. In what units is the gradient of the graph? iii. Does a constantan conductor obey Ohm’s law. Explain your answer. C. What is the effect on the resistance of a conductor when the: i. length is increased ii. temperature is increased 226

Current Electricity

SOLUTIONS C A B A A A D C D B A C B D B C C A D C B B D A D

.

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25.

26. A. i.

ii.

RT = R1 + R2 =6+2 = 8Ω I= =

V R 16 8

= 2A

Current Electricity

227

26. B. i. V = IR =2×6 = 8V 1

1

=

RT 1 RT 1 RT

= =

RT = I=

.

ii.

1

+

R1 R2 1 1 4 3

4 4 3

+

2

Ω

V R

= 16 ÷ = 16 × = 12A

4 3 3 4

27. A. Ohm’s law states that the current flowing in a circuit is directly proportional to the voltage and inversely proportional to the resistance. Resistance is the opposition to the current flow in a circuit. B. RT = R1 + R2 +

R3 R4 R3 + R4

C. i. RT = R1 + R2 =3+2 = 5Ω I= =

V R 20 5

= 4A ii. RT = R1 + =3+

R2 R3 R2 + R3 2×2 2+2

=3+1 = 4Ω I= =

V R 20 4

= 5A

228

Current Electricity

28. A. Parallel circuit; figure 2 Series circuit; figure 1 B. Two disadvantages of figure 1 (series) circuit i. If one component is disconnected or burnt, other components stop receiving current and therefore stops working. ii. Adding more components increases resistance. C. RT = R1 + R2 =3+6 = 9Ω

.

V = IR = 1.5 × 9 = 13.5V D.

1 RT 1 RT 1 RT

= = =

1

+

1

R1 R2 1 1 3 1

+

6

2

RT = 2Ω

29. A.

Current Electricity

229

29. i. B.

. ii.

Gradient =

3 − 1.5 0.4 − 0.2

= 3.75Ω Units of gradient: ohms (Ω) iii. Yes. Constantan wire obey Ohm’s law because a straight slope on the graph shows that voltage is directly proportional to current. C.

230

i. Increasing the length of the conductor increases resistance. ii. Increasing the temperature of the conductor increases resistance.

Current Electricity

30

PRACTICAL ELECTRICITY

Introduction Most countries have energy providing companies that generate, transmit, distribute and supply electricity in the country. For instance, in Zambia, there is ZESCO. How do these companies charge electric power consumption? How can electricity be used safely? This unit covers these questions. .

Specific outcomes By the end of this unit, you will be able to: ❖ Electric power: • Define electric power • State the unit of power • Work out power, voltage, current calculations using P = VI ❖ Calculate the cost of electricity using; cost = power × time × cost per unit ❖ Safe use of electricity: • Label the parts of a 3-pin plug • State the colour and function of the live wire, neutral wire and earth wire • Explain how the following provide safety: • Fuse • Double insulation • Circuit breaker ❖ State the hazards of electricity .

Practical Electricity

231

PRACTICAL ELECTRICITY 30.1 ELECTRIC POWER • Electric power is defined as the rate of consuming electric energy in a circuit. • Unit of electric power is Watt (W). • Appliances consume electric energy differently. • The formula for calculating electric power:

.

P = VI

where P = power (in W) V = potential difference (in V) I = current (in A)

❖ EXAMPLES 1. Brian builds a circuit with a bulb, an ammeter and a 12V battery. If the ammeter reads 2.5A, what is the power of the bulb? 2. If an appliance is rated 12A and 50V, what is its power? 3. If the bulb uses 100W when current 4A, what is voltage? 4. Calculate the current if an appliance uses 20W and voltage is 5V.

❖ SOLUTIONS 1. P = VI P = 12 × 3 P = 30W

2. P = VI P = 50 × 12 P = 600W

3. V = V=

P I 100 4

P=? V = 12V I = 3A

P=? V = 50V I = 12A P = 100W I = 4A V=?

V = 25V

4. I = I=

P V 20 5

I = 4A

232

Practical Electricity

P = 20W V = 5V I=?

30.2 COST OF ELECTRICITY • Electric energy supplying companies charge power consumption in units. • One unit is equivalent to one kilowatt per hour (1 unit = 1 kWh). • To calculate the cost of electricity, first find the unit used then multiply by the cost per unit. • The energy supplying company provides the cost of each unit. • The formula for calculating the cost of electricity is given as:

.

Unit used (energy consumed) = power (in kilowatt) × time (in hours) E = Pt

Total cost = units used × cost per unit Total cost = power (in kW) × time (in hrs) × cost per unit

❖ EXAMPLES 1. If 1 unit costs K1, how much does it cost to leave a 100W bulb on for 30 days? 2. If a TV uses 250W and 1 unit costs K1, how much does it cost to watch the TV for 4 hours? ❖ SOLUTIONS 1. Total cost = power × time × cost per unit Total cost = 0.1 × 720 × K1 Total cost = K72

2. Total cost = power × time × cost per unit Total cost = 0.25 × 4 × 1 Total cost = K1

Total cost = ? Power = 100W ÷ 1000W = 0.1kW Time = 30 days × 24 hrs = 720 hrs Total cost = ? Power = 250W ÷ 1000W = 0.25kW Time = 4 hrs

30.3 SAFE USE OF ELECTRICITY • Electrical appliances come with a plug. • To connect the appliance to the electricity, the plug is inserted into the socket. • The plug consists of pins connected to electric cables from the appliance: • 2-pin plug: has a live and neutral wire, • 3-pin plug: has live, neutral and earth wire. The figure below shows a 3-pin plug and the arrangement of electric wires.

Practical Electricity

233

1.

LIVE WIRE • A brown cable. • It takes electricity to the appliance. • It has high voltage. • It can cause electric shock.

2.

NEUTRAL WIRE • A blue cable. • It takes electricity away from the appliance. • It has zero voltage.

3.

EARTH WIRE • A green or yellow cable. • It is connected to the metal casting of an appliance. • It protects from electric shock if the live wire is connected to the metal cast of an appliance.

❖

FUSE • An electrical safety device that stops current from flowing to the appliance when over-current. • It is connected to the live wire. • If too much current flows to the appliance, the fuse melts and disconnects the circuit.

❖

DOUBLE INSULATION • Some appliance consists of double insulation. • The first layer insulates the electric cables. • The second layer of plastic casting prevents the possibility of the live wire touching the appliance casing. • Appliances with double insulation do not require earth wire.

❖

CIRCUIT BREAKER • A device that interrupts over-current. • It automatically or manually disconnects the circuit when overloaded and prevents electrical damage.

30.4 HAZARDS OF ELECTRICITY • Electricity can cause electric shock and fire hazards. • Danger can arise from: 1. Damaged insulation 2. Overheating 3. Wet condition

234

Practical Electricity

REVIEW QUESTIONS 1. The rate of consuming electric energy in a circuit called? A. Work B. Force C. Power D. Energy

.

2. The basic unit of electric power is A. Volt B. Ampere C. Coulomb D. Watt 3. How many watts are in one kilowatt? A. 10 000 B. 1000 C. 100 D. 10 4. A small heater operates at 12V, 2A. how much can energy will it use when it runs for 5 minutes? A. 30J B. 120J C. 1800J D. 7200J 5. Which unit measures the rate at which charges flow through an element of an electric kettle? A. Ampere B. Kilowatt hour C. Volt D. Watt 6. A small radio is rated 160W, 240V. What value of a fuse would be suitable for the plug of this radio? A. 2A B. 5A C. 10A D. 15A 7. What would be the best fuse to be used where an element rated 240V, 1000W is being used correctly? A. 3A B. 5A C. 10A D. 13A 8. Which of the following will cost the most if operated from the mains supply? A. 5000W electrical cooker for 1 minute B. 1000W electric fire used for 10 minutes C. 500 W electric iron used for 1 hour D. 100W lamp use for 1 day 9. If the cost of 1 unit (1kWh) of electricity is K 0.50, what is the cost of running a 2kW electric fire for 6 hours? A. K 4.00 B. K 4.80 C. K 6.00 D. K 7.20

Practical Electricity

235

10. An immersion water heater is marked 240V, 2kW and it is operated for 180 minutes. If the cost of 1 unit (kWh) of electricity is K 5.00, what is the cost of running this water heater? A. K 6.00 B. K 10.00 C. K 15.00 D. K 30.00

.

11. The standard colour for the live wire is A. Brown B. Green C. Blue D. Yellow 12. The function of the live wire is to A. Transmit electric current from the appliance B. Protects from electric shock C. Transmit electric current to the appliance D. All the above 13. The standard colour for the neutral wire is A. Yellow B. Brown C. Red D. Blue 14. The function of the neutral wire is to A. Transmit electric current from the appliance B. Protects from electric shock C. Transmit electric current to the appliance D. All the above 15. Which of the following is the standard colour for the earth wire? A. Yellow B. Brown C. Red D. Blue 16. Which electric cable can cause electric shock? A. Earth B. Neutral C. Live D. All the above 17. When using a 3-core wiring (live, neutral and earth leads), where should the fuse be fitted? A. Only in the live wire B. Only in the neutral wire C. Only in the earth wire D. In either the live or neutral wire 18. The purpose of the fuse is A. Increase current flow B. Reduce current flow C. Increase voltage D. Break current flow when over current

236

Practical Electricity

19. What happens when a 250V, 2500W water heater is connected to the main supply using a plug fitted with a 5A fuse? A. The fuse in the plug melts B. The heater burns out C. The heater runs at half power D. The fuse in the plug works normally

.

20. The covering of the electrical cables with a material that does not conduct electricity is known as A. Plating B. Galvanisation C. Casting D. Insulation 21. An electric stove has two plates, each with a rating of 240V, 3000W. One plate is switched on for 5 hours. For the same period of time a heater rated 230V, 2300W and a geyser with an element rated 240V, 2500W are switched on. A. Calculate the electrical energy in kWh used by the stove, heater and geyser in these 5 hours. B. If a kWh of electricity cost K 1.2, what is the cost of using the stove, heater and geyser for 5 hours? C. What advice would you give regarding the switch on of many electrical appliances at the same time? 22. A fuse in a three-pin plug for a one plate electric cooker has blown off. A grade 8 girl changed the plug and connected the new one as shown in the figure below.

A. What was wrong with the connections made by the girl? B. Will the cooker connected to this plug work? Explain your answer. C. Draw a three-pin plug showing correct connections of the wires.

Practical Electricity

237

23. The figure shows a household circuit diagram showing a fridge, stove and geyser of power rating 0.8kW, 0.9kW and 0.6kW respectively.

. If the appliances are left ‘ON’ for 8 hours daily for seven days, find: A. The total electrical power consumed in one week, B. The electricity bill for one week if one electrical unit costs $ 2.

238

Practical Electricity

SOLUTIONS C D B D A A B D C D A C D A A C A D A D

.

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

21. A. Energy consumed by stove E = Pt =3×5 = 15kWh Energy consumed by heater E = Pt = 2.3 × 5 = 11kWh Energy consumed by geyser E = Pt = 2.5 × 5 = 12.5kWh Total energy consumed = 15 + 11 + 12.5 = 38.5kWh B.

Cost = power × time × cost per unit = 38.5 × 1.2 = K 46.2

C.

Using many appliances at the same time increases the rate of consuming electricity.

Practical Electricity

239

22. A. The earth wire is connected where the live wire should be connected. B. The cooker will not work. The cooker cannot receive electric current the live wire is not connected to the power source.

.

C.

23. A. Total power = 0.8 + 0.9 + 0.6 = 2.3kW B.

240

Cost = power × time × cost per unit = 2.3kWh × 168 hr × $ 2 = $ 772.8

Practical Electricity

Cost = ? Power = 2.3kW Time = 1 wk × 7 days × 24 hrs = 168 hrs Cost per unit = $ 2

31

BASIC ELECTRONICS

Introduction Do you use a smartphone, calculator, computer or TV? If so, you use a device that applies electronics. Electronics is a branch of physics that deals with the emission, flow and control of electrons. Do you know that when you heat a metal at a very high temperature it emits electrons? That is called thermionic emission! In this unit, basic electronics is introduced by thermionic emission. The unit also covers the cathode ray oscilloscope (CRO).

.

Specific outcomes By the end of this unit, you will be able to: ❖ Describe how thermionic emission occur ❖ State the characteristics of cathode rays ❖ Cathode ray oscilloscope (CRO): • Label the parts of the CRO, and state the function of each part • State the uses of the CRO • Calculate the voltage input • Calculate period and frequency

.

Basic Electronics

241

BASIC ELECTRONICS 31.1 THERMIONIC EMISSION

. • • • • • • • • •

The figure above shows a vacuum glass with a filament connected to the 6V power supply, a positively charged plate (anode) and a negatively charged plate (cathode) connected to the 400V power supply. When the filament heats the cathode, the ammeter deflects, indicating the current flow in the circuit even though there is a vacuum gap between the cathode and anode. From the above setup, it can be conducted that when the cathode is heated it emits electrons that flow to the anode and transmit the current between the gap. The process by which electrons are emitted from the surface of a heated metal is called thermionic emission. As the cathode is heated by a filament, the kinetic energy of free electrons of the cathode increases. With sufficient energy, the force of attraction between these electrons and protons within the nuclei of the cathode is overcome. As a result, they are released into the surface. Free electrons which are emitted are called thermions. Since electrons have a negative charge, they are repelled by the cathode and attracted to the anode The flow of beams of electrons is called cathode rays. For the metal to emit electrons, it has to be heated at a very high temperature. The minimum amount of temperature needed to start the emission of electrons is called threshold temperature. Different metals have different threshold temperatures for thermionic emission.

31.2 CHARACTERISTICS OF CATHODE RAYS • Cathode rays are beams of electrons. • They travel in a straight line from the cathode (a negatively charged plate) to anode (a positively charged plate). • The flow of electrons is opposite to the direction of conventional current. This is because electric current moves from the positive to the negative terminals, while cathode rays move from the negative to the positive plate. • Cathode rays are deflected by the magnetic fields.

242

•

Cathode rays are deflected by electric fields.

Basic Electronics

31.3 CATHODE RAY OSCILLOSCOPE (CRO)

•

A cathode ray oscilloscope consists of three main parts: an electron gun, defecting system and a fluorescent screen. ❖ ELECTRON GUN 1. Filament: It is used to heat the cathode. 2. Cathode: It emits electrons by thermionic emission. 3. Control grid: It controls the brightness of the spot on the screen by controlling the number of electrons passing through it. 4. Anode: They focus and accelerate the beam of electrons. ❖ DEFLECTING SYSTEM 1. Y-plates: They deflect electron beams up or down. 2. X-plates: They deflect electron beam left or right.

❖ FLUORESCENT SCREEN • A fluorescent screen is a glass surface coated with fluorescent material such as zinc sulphide or phosphor. • It converts the kinetic energy of electrons to heat and light energy. • A bright light is produced on the screen when the electron beam hits it. • It displays the waveform.

Basic Electronics

243

❖

USES OF CRO 1. MEASURING VOLTAGE • CRO is used to measure both ac and dc voltage. • It acts as a voltmeter. • It visually displays the peak voltage as a waveform. • When the potential difference (p.d) is connected to the Y-plate, the vertical deflection (Y-gain) on the screen is directly proportional to the p.d. • For example, if the Y-gain setting is 2V/cm, it means 1 cm of height on the screen shows 2V input. Then 2cm corresponds to 4V, 5cm corresponds to 10V and so on. ❖ EXAMPLES 1. Find the Y-input of the CRO. Both are set to 4V/cm. A.

B.

❖ SOLUTIONS 1. A. 1cm → 4V 3cm → x x=3×4 = 12V B. 1cm → 4V 5cm → x x=5×4 = 20V 244

Basic Electronics

2.

MEASURING SHORT TIME INTERVAL • CRO can be used to measure short time intervals. • To measure time intervals, CRO needs to have a calibrated time base. • For example, a time base of 2ms/cm means the light spot takes 2ms to sweep through 1cm on the x-axis of the screen. Then, 3cm corresponds to 6ms, 5cm corresponds to 10ms and so on. • The period (T) of the input can be calculated by multiplying the time base with the distance covered in one wave. • Frequency (f) can be calculated by using the equation below: f=

1

T

❖ EXAMPLES 1. The diagram below shows the screen of CRO with a time base set to 4sm/cm. If one square is 1cm, find” A. Period of the input ac signal, B. Frequency of the input ac signal.

2. A CRO has a time base set to 2s/cm. If one wave has a length of 10cm, calculate: A. Period, B. Frequency. ❖ SOLUTIONS 1. A. T = 4ms/cm × 10cm = 40ms = 0.04s B. f = =

1 T 1

1s = 1000ms

f = ? Hz T = 0.04s

0.04

= 25Hz 2. A. T = 2s/cm × 10cm = 2s B. f = =

1 T 1

f = ? Hz T = 2s

2

= 0.5Hz Basic Electronics

245

REVIEW QUESTIONS

.

1. Thermionic emission is the loss of A. Heat by hot objects B. Electrons by heated metal surfaces C. Elections by protons D. Heat by electrons

2. The flow of the beam of electrons is referred to as A. X-ray B. Gamma ray C. Light ray D. Cathode ray 3. The diagram below shows a beam of electrons entering a magnetic field.

In which direction are electrons being deflected? A. Into the page B. Out of the page C. Toward the bottom of the page D. Toward the top of the page 4. In a cathode ray oscilloscope (CRO), what does the deflection system consists of? A. Y-plates B. X-plates C. X-plates and Y-plates D. Cathode, grid and anodes 5. Which component of the cathode ray oscilloscope focus and accelerate the beam of electrons? A. Anode B. Cathode C. Control grid D. Fluorescent screen 6. Which component of the cathode ray oscilloscope controls the brightness of the spot? A. Element B. Control grid C. Y-plates D. X-plates 7. Which component of the cathode ray oscilloscope deflects the beam of electrons vertically? A. Element B. Fluorescent screen C. Y-plates D. X-plates 8. Which component of the cathode ray oscilloscope deflects the beam of electrons horizontally? A. Cathode B. Electron gun C. Y-plates D. X-plates

246

Basic Electronics

9. Which of the following can be used to produce a series of voltage pulses? A. Voltmeter B. Bistable circuits C. Astable circuits D. Cathode ray oscillography

.

10. The diagram below shows a sine wave for an a.c input on the screen of a cathode ray oscilloscope. The gain control is set at 0.50V/cm and the time base at 10ms/cm.

What is the peak voltage and frequency of the a.c signal? Peak voltage Frequency A. 2.5V 10Hz B. 2.5V 12Hz C. 5V 10Hz D. 5V 12Hz 11. The figure shows an experiment to demonstrate the emission of electrons.

A. Explain how the electrons are emitted from the cathode B. What term is used to describe the process in question A. above C. Explain why the electrons move toward the anode

Basic Electronics

247

12. The figure below shows a cross section of the cathode ray oscilloscope.

. A. State one property of cathode rays. B. Explain why: i. electrons are emitted from the cathode, ii. electrons accelerate after they leave the cathode, iii. a vacuum is needed in the tube. C. A charge of 1.8 × 10−3C passes through the cathode ray oscilloscope per second when the voltage between the anode and the cathode is 2000V. Calculate the energy carried by the cathode ray beam in 8 seconds. D. When an alternating potential difference of very low frequency is applied across the deflecting plates in the figure above, the spot of light on the screen is seen to move. Describe and explain the movement of the spot. E. State one use of the cathode ray oscilloscope. 13. Describe an experiment to show that cathode rays are not electromagnetic radiations. 14. The figure below shows the waveform obtained on a cathode ray oscilloscope. The gain control is set at 2.5V/cm and the time base at 15ms/cm.

Calculate the: A. Peak voltage, B. Period of the wave, C. Frequency of the wave.

248

Basic Electronics

SOLUTIONS B D C C A B C D D A

.

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

11. A. As the cathode is heated by a filament, the kinetic energy of free electrons of the cathode increases. With sufficient energy, the force of attraction between these electrons and protons within the nuclei of the cathode are overcome. As a result, they are released into the surface. B. Thermionic emission. C. Electrons are negatively charged. Therefore they are attracted by the positively charged anode. 12. A. Cathode rays travel in a straight line from the cathode to the anode. B. i.

As electrons are heated, they gain kinetic energy and move from the surface of the cathode.

ii. Electrons are accelerated from the cathode to the anode due to the potential difference between the cathode and anode. iii. Vacuum prevents electrons in the beam from scattering due to collision with other gas Molecules. C. Charge of electrons = current × time = 1.8 × 10−3C/s × 8s = 0.0144 coulomb KE = charge of electrons × voltage = 0.0144 × 2000 = 28.8J D. Alternative current connected to the deflecting plates produces a spot in form of the waveform which is displayed on the fluorescent screen. Vertical movements are caused by the deflection of beams of electrons by the Y-plates. They are proportional to the voltage input. Horizontal movements are caused by the deflection of beams of electrons by the X-plates. They are proportional to the time E. CRO is used to measure voltage. 13. Cathode rays are beams of electrons that are negatively charged. Therefore, they are deflected by magnetic fields and electric fields. Electromagnetic radiations have no charge and are not deflected by magnetic fields or electric fields. To demonstrate that cathode rays are not electromagnetic radiation, cathode rays are passed through magnetic fields or electric fields. The deflection of these rays indicates that they are not electromagnetic radiation.

Basic Electronics

249

14. A. Voltage = 2.5V/cm × 2cm = 5V B. T = 15ms/cm × 4cm = 60ms = 0.06s C. f = =

1 T 1 0.06

= 16.7Hz

250

Basic Electronics

1s = 1000ms

f = ? Hz T = 0.06s

32

ATOMIC STRUCTURE

Introduction Everything you are seeing; this book, pen, pencil, cellphone, computer, table, chair, a class block and so on consists of tiny particles called atoms. These atoms are the basic unit of matter. They vary in their mass. What is the composition of atoms? How do you calculate their mass? This unit gives an introduction to atomic structure and covers these questions.

.

Specific outcomes By the end of this unit, you will be able to: ❖ Differentiate electron, proton and neutron on charge and mass ❖ Define atomic number ❖ Define mass number ❖ Write the chemical symbol of the given element ❖ Write the electronic configuration of the given element ❖ Define an isotope, and give examples of isotopes ❖ State uses of isotopes

.

Atomic Structure

251

ATOMIC STRUCTURE 32.2 SUBATOMIC PARTICLES • Atom is the smallest unit of matter with the properties of an element. • An atom contains three particles that are responsible for the mass of the atom and charge. i. Proton ii. Neutron iii. Electron • Protons and neutrons are found in the nucleus of an atom located at the centre. Thus, the nucleus of an atom is called nucleon. • Electrons orbit the nucleus in electron shells. • Protons are positively charged, electrons are negatively charged and neutrons are neutral. Therefore, an atom has a net charge of zero. Therefore, it is said to be neutral. • The mass of proton and electron is one atomic mass unit (amu). However, the mass of an electron is too small; therefore, it does not contribute to the mass of an atom. • The structure and characteristics of an atom are shown below:

.

5.2

ATOMIC MASS • The mass of an element is calculated from the number of protons and neutrons. • The mass of electrons is too small. Therefore it is neglected. • Atomic number (also called proton number) is the number of protons in an atom. It distinguishes one element from another. Atomic mass is used to put elements in the periodic table in a logical order. • Mass number (also called nucleon number) is the total number of protons and neutrons of an element. This is the mass of an element.

5.3

CHEMICAL SYMBOL • Each element is represented by a unique chemical symbol, • The atomic number is denoted by Z. • The mass number is denoted by A.

•

252

Mass number

A

Atomic number

Z

X

Chemical symbol of an element

For example, oxygen (with 8 protons and 8 neutrons) is represented by

Atomic Structure

.

5.4

ELECTRONIC CONFIGURATION • Electrons are placed in orbits called shells. • The first shell can contain up to a maximum of 2 electrons. This is known as the duplet or duet rule. • The second and third shells can contain up to a maximum of 8 electrons. This is known as the octet rule. • Electrons that are found in the outermost shell are called valence electrons. • Valence electrons determine the chemical properties of an atom. • Valence electrons are also responsible for the chemical bond formation of an atom with other atoms. • For example, chlorine has 17 electrons. • Electronic configuration; of chlorine, for example, is written as 2,8,7, with each comma separating one shell from another.

5.5

ISOTOPE • Isotopes are atoms of the same element with different mass number. • They have the same number of protons and electrons but different numbers of neutrons. • For example, carbon-12, carbon-13 and carbon-14 isotopes, all have 6 protons and electrons but 6, 7 and 8 neutrons respectively. • Hydrogen-1 (protium), hydrogen-2 (deuterium) and hydrogen-3 (tritium) isotopes, all have 1 proton and electron but with 0, 1 and 2 neutrons respectively. ❖ PROPERTIES OF ISOTOPES • Isotopes of the same element have the same chemical properties but with slightly different physical properties. • Isotopes can be radioactive. They emit radiations in the form of alpha, beta or gamma radiation. Gamma radiation can penetrate and destroy body cells. Medically, gamma rays are used to treat cancer by killing cancer cells.

Atomic Structure

253

REVIEW QUESTIONS 1. The three subatomic particles that make up an atom are A. Proton, electron, nucleus B. Proton, neutron, nucleus C. Proton, neutron, electron D. Neutron, electron, nucleus

.

2. Which subatomic particle has no charge? A. Proton B. Electron C. Neutron D. Nucleus 3. Which subatomic particle has a negative charge? A. Proton B. Electron C. Neutron D. Nucleus 4. Which subatomic particle has a positive charge? A. Proton B. Electron C. Neutron D. Nucleus 5. Nucleon refers to A. Protons and neutrons B. Protons and electrons C. Electrons and neutrons D. Electrons only 6. Electrons present in the outermost shell are called A. Orbit electrons B. Octet electrons C. Duplet electrons D. Valence electrons 7. Which of the following statements accurately describes the masses of subatomic particles? A. Mass of 1 proton equals the mass of 1 electron B. Mass of 1 neutron equals the mass of 1840 proton C. Mass of 1 neutron equals the mass of 1 proton D. Mass of 1 electron equals the mass of 1840 neutron 8. Which of the following is used to identify an element? A. Number of electrons B. Number of nucleons C. Number of nuclei D. Number of protons 9. The atomic number is the same as A. Neutron number B. Nuclei number B. Proton number C. Electron number 10. The isotope deuterium of hydrogen has A. One proton and no neutron B. One proton and one neutron C. One neutron and two protons D. One neutron and no proton 254

Atomic Structure

11. How many nucleons are in one neutral atom of radium isotope A. 88 B. 140 C. 176 D. 228

228 88Ra?

.

12. A. Define an isotope. B.

16 8O

and 188O are both atoms of oxygen: i. Draw a diagram of an atom of one of these isotopes, ii. Give one use of radioactive isotopes.

Atomic Structure

255

SOLUTIONS C C B A A D C D B B D

.

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

12. A. Isotopes are atoms of the same element with different mass number. B. i.

ii. Carbon-14 is used to detect the age of an organism or an artifact. (for more uses of radioactive isotopes see section 33.6 of this book).

256

Atomic Structure

33

RADIOACTIVITY

Introduction An atom consists of three types of particles; protons, neutrons and electrons. Protons and neutrons are found in the nucleus of an atom located at the centre. Electrons orbit the nucleus in electron shells. When the number of protons and neutrons are equal, an atom is said to be stable. Most atoms are stable. However, when the number of protons and neutrons is not equal, an atom is said to be unstable. To be stable, these unstable atoms emit protons, neutrons or electrons in the form of energy called radiations. There are three types of radiations: alpha (α) particles, beta (β) particles, and gamma rays. The process by which these radiations are emitted is called radioactivity.

.

Specific outcomes This unit covers the properties of these radiations and many more. By the end of this unit, you will be able to: ❖ State the characteristics of: • Alpha (α) particles • Beta (β) particles • Gamma rays ❖ Radioactive decay: • Define radioactive decay • Describe alpha decay • Describe beta decay • Describe gamma radiation ❖ Half-life: • Define half-life • Plot the half-life graph of the given substance • Calculate the half-life of the given substance ❖ State the uses of radioactivity ❖ State the dangers of radiations ❖ State the safety precautions for radiations

.

Radioactivity

257

RADIOACTIVITY 33.1 RADIOACTIVITY • Radioactivity is the emission of radiation caused by a spontaneous disintegration of the atomic nuclei. • Radioactivity is a random process. • There are three radiations emitted: 1. Alpha (α) particles 2. Beta (β) particles 3. Gamma rays • Radioactive elements are elements that emit radiations. • Elements such as Uranium and radium are radioactive.

.

33.2 CHARACTERISTICS OF RADIATIONS α-particle What it is 2 proton + 2 neutrons Helium nucleus Charge Positive +2 Ionising power Strong Penetrating power Very weak Absorbed by A sheet of paper Range in air Few mm Deflection in Yes electric field Deflection in Yes magnetic field

β-particle Electron Negative −1 Weak weak 5mm aluminium Few cm Yes

γ-ray Electromagnetic wave Photon Neutral (zero) Very weak Strong 5cm lead Infinite no

Yes

no

33.3 RADIOACTIVE DECAY • Radioactive decay occurs when an unstable atom emits radiations until a stable atom is reached. • An atom in a stable state has an equal number of protons and neutrons. • However, when the number of protons and neutrons are not equal the atom becomes unstable. • For the unstable atom to become stable and balance its nucleus it emits radiations. The emission of these radiations is what is referred to as radioactive decay. • There are three types of radioactive decay: 1. Alpha (α) decay 2. Beta (β) decay 3. Gamma radiation ❖ TYPES OF RADIOACTIVE DECAY 1. ALPHA (α) DECAY • Alpha decay involves the emission of α-particles from the nucleus of a radioactive element. • For example, when Radium (Ra) undergoes α-decay, it emits 2 protons and 2 neutrons. As a result, the mass number of Ra decreases from 226 to 222. The atomic number also decreases from 88 to 86. • Therefore, after α-decay Radium (Ra) changes to Radon (Rn), a stable element with mass number 222 and atomic number 86.

258

Radioactivity

2.

BETA (β) DECAY • Beta decay involves the emission of β-particle from the nucleus of a radioactive element. • For example, when Radium (Ra) undergoes β-decay, it emits one electron. Therefore, the mass number does not change but the atomic number increases from 88 to 89 because one neutron changes to a proton. Therefore, after β-decay Radium (Ra) changes to Actinium (Ac).

3.

GAMMA (γ) RADIATION • This type of decay involves a nucleus change from a higher energy state to a lower energy state through the emission of electromagnetic radiation known as photon. • It often takes place after α-decay or β-decay has occurred. • The atomic number and mass number do not change. Therefore, the parent element is the same as the daughter element. • Gamma radiation is the most dangerous radiation of the three types of radiations.

Radioactivity

259

33.4 HALF-LIFE • The half-life of a radioactive element is the time taken for the unstable nuclei of a radioactive element to decay by half • For example, the half-life of iodine-131 is 8 days. If there is 80g of iodine-131, in 8 days 40g decays (remaining with 40g), in 16 days 20g of the 40g remaining decays (remaining with 20g), in 24 days 10g of the remaining 20g decays (remaining with 10g), in 32 days the remaining will be 5g • Half-life can be represented graphically as shown below:

.

33.6 HALF-LIFE CALCULATIONS • Half-life problems can be solved by using the following formula • The formula for calculating half-life: 𝒕

𝟏 𝒉

A = Ao (𝟐)

where A = amount of substance remaining Ao = initial amount of substance t = time elapsed h = half-life of the substance

❖ EXAMPLES 1. The half-life of potassium-42 is 12.4 hours. If the initial mass of potassium-42 is 750 grams, what mass is left after 62 hours? 2. Sodium-24 has a half-life of 15 hours. How much sodium-24 remains in the 18g sample after 60 hours? 3. Polonium-214 has a relatively short half-life of 164 seconds. How many seconds would it take for 8.0g of this isotope to decay to 0.25g? 4. If 100 grams of carbon-14 decays until only 25 grams of carbon is left after 11 460 years, what is the half-life of carbon-14?

260

Radioactivity

SOLUTIONS 𝑡

1 ℎ 1. A = Ao ( ) 2

A = ? grams Ao = 750 grams t = 62 hr h = 12.4 hr

62

1 4 = 750( ) 2

.

❖

= 23.4g

𝑡

1 ℎ 2. A = Ao ( ) 2

A = ? grams Ao = 18 grams t = 60 hr h = 15 hr

60

1 15 = 18( ) 2

= 1.13g

𝑡

1 ℎ 3. A = Ao ( ) 2 𝑡

1 164 0.25 = 8( ) 2

A = 0.25 grams Ao = 8 grams t=? h = 164 sec

𝑡

1 164 8 (2) = 0.25 𝑡

1 164 (2) = 0.03125 𝑡 1 164 𝑡 164 𝑡 164

log ( =

) = log (0.03125)

2 log (0.03125) 1

log (2)

=5

t = 5 × 164 t = 820 sec

Radioactivity

261

𝑡

25 = 100( 1

11460 ℎ

1

11460 ℎ

(2) (2)

11460 ℎ 11460 ℎ 11460 ℎ

1

)

2

=

25 100

= 0.25

log ( =

11460 ℎ

A = 25 grams Ao = 100 grams t = 11 460 years h = 30 years

.

4.

1 ℎ A = Ao ( ) 2

1

) = log (0.25)

2 log (0.25) 1

log (2)

=2

2h = 11460 h = 5730 years 33.6 USES OF RADIOACTIVITY 1. Medical treatment: Gamma rays are used medically for cancer treatment. They kill cancer cells. 2. Sterilisation: Gamma rays kill bacteria. Therefore they are used to sterilise pieces of medical equipment and preserve food. 3. Thickness measurement: Beta particles are used to monitor the thickness of papers, plastics or metal sheets during manufacturing. 4. Tracers: Small amounts of a radioisotope can be injected into the human body and traced to detect blockage in organs or plants to monitor nutrient intake or pipes to detect leaks. 5. Archaeology: Carbon-14 is used to detect the age of an organism or an artifact. 6. Checking welds: Gamma camera is used to identify cracks in welds. 7. Smoke detector. 33.7 DANGERS OF RADIATIONS • Radiation can cause the following to humans: 1. Radiation burns 2. Damage living cells 3. Genetic changes to living cells 4. Cancer 5. Death 33.7 SAFETY PRECAUTIONS FOR RADIATIONS 1. Avoid unnecessary exposure. 2. Avoid direct contact with the source. 3. Store radioactive sources in places out of reach of the public. 4. Store radioactive sources in a shielded container. 5. Label radioactive sources clearly. 6. Wear protective clothing when handling radioactive materials.

262

Radioactivity

REVIEW QUESTIONS 1. Geiger-Marsden’s experiment on the structure of an atom indicates that most of the space in an atom is empty, except at the centre where there’s some little mass which A. Carries no charge B. Attracts alpha particles C. Carries a negative charge D. Carries a positive charge

.

2. The result of the α-particle scattering experiment gave evidence for which of the following A. Nuclear fusion B. Radioactive decay C. The existence of isotopes D. The nuclear atom 3. The ejection of an α-particle from a nucleus results in A. An increase in the atomic number by one B. An increase in atomic number by two C. A decrease in atomic number by one D. A decrease in atomic number by two 4. The ejection of an β-particle from a nucleus results in A. An increase in the atomic number by one B. An increase in atomic number by two C. A decrease in atomic number by one D. The atomic number is unchanged 5. The ejection of an γ-particle from a nucleus results in A. An increase in the atomic number by one B. An increase in atomic number by two C. A decrease in atomic number by one D. Atomic number is unchanged 6. Uranium-238 ( 230 92U) emits one beta particle to form a new daughter element whose symbol is 𝑁𝑝 . Which of the following is the correct decay equation? 234 0 A. 238 92U → 90𝑁𝑝 + −1𝑒 B. C. D.

238 92U → 238 92U → 238 92U →

236 91𝑁𝑝 238 92𝑁𝑝 238 93𝑁𝑝

+ −10𝑒 + −10𝑒 + −10𝑒

7. A nucleus is represented by 230 91Z. It emits one alpha particle and then one beta particle. What is the resulting nucleus X? A. 230 88Z B. C. D.

226 89Z 226 90Z 230 89Z

8. The half-life of isotope X is four days and its initial mass is 32g. What mass of the isotope will remain after twelve days? A. 4g B. 8g C. 12g D. 16g

Radioactivity

263

9. A radioactive substance has a half-life of 15 minutes. If the original mass is 10kg, what mass remains undecayed after 1 hour? A. 625g B. 740g C. 820g D. 960g

.

10. A radioactive substance has a mass of 4.8kg and its half-life is 20 minutes. What mass remains undecayed after one hour? A. 1200g B. 800g C. 600g D. 300g 11. The half-life of a radioisotope is 2400 years. The activity of a sample is 720 counts/s. How long will it take for the activity to fall to 90 counts/s? A. 300 years B. 2400 years C. 7200 years D. 19200 years 12. A nucleus consists of 90 protons and 144 neutrons. After emitting two beta particles followed by an alpha particle, this nucleus has A. 86 protons and 140 neutrons B. 86 protons and 142 neutrons C. 90 protons and 142 neutrons D. 90 protons and 140 neutrons 13. Nuclear radiation with the strongest ionisation power is A. β-particle B. α-particle C. γ-particle D. All have equal penetrating ability 14. People working with radioactive materials use photographic film badges covered in paper. The badge is used to monitor the level of their radiation exposure. Which radiation is detected? A. Alpha particles only B. Beta particles only C. Gamma rays and beta particles D. Gamma rays 15. Why are radioactive sources stored in boxes lined with lead? Lead A. Absorbs radiation and stops it from escaping into the room B. Absorbs the radioactive source and make it safe to handle C. Completely stops the source from decaying and so it lasts longer D. Slows down the rate at which the radioactive source decays 16. A sheet of paper will stop A. α-particle B. β-particle C. γ-particle D. all the above 17. The most penetrating of the three common types of nuclear radiation is the A. α-particle B. β-particle C. γ-particle D. All have equal penetrating ability 264

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18. The diagram below shows three different types of radiation X, Y and Z.

. Which order correctly identifies X, Y and Z A B C D

X γ-rays α-particles β-particles β-particles

Y α-particles β-particles γ-rays α-particles

Z β-particles γ-rays α-particles γ-rays

19. Which of the following would a scientist most likely use to calculate the age of a piece of a bone? A. γ -rays B. Carbon-14 C. Hydrogen-3 D. Uranium-235 20. A. Name three main types of nuclear radiation and state the nature of each. B. State any two properties of each of the three nuclear radiations. 21. A. Explain the term radioactive decay. B. Explain what background radiation is and give two sources of background radiation. C. State the nature of beta particles. 22. Uranium 238 92U decays to form a nucleus of thorium by emission of an alpha particle. Thorium has the symbol 234 90Th. A. What is the meaning of nucleon number? B. State the proton number and nucleon number of an alpha particle/ 234 C. Write a decay equation to show how Uranium 238 92U decays to 90Th after emitting an alpha particle. 234 D. Thorium-234 ( 90Th) decays to an isotope protoactinium (Pa) by beta decay. Write the decay equation to show this process. 23. Strontium-90 emits a β-particle and has a half-life of 30 years. A. Explain the term ‘half life’. B. What happens to the atomic number of an atom when it emits a β-particle? C. What happens to the mass of an atom when it emits a β-particle? D. A sample of strontium-90 gives a reading of 80 counts per second on a radiation detector (Rate meter). Calculate how long it will take for the count to drop to 20 counts per second. 24. In an experiment to determine the half-life of radon-220 ( 220 86Rn), the following results were obtained as shown in the table below after allowing for the background count. Time/s Count rate/s−1

0 30

10 26

20 23

30 21

40 18

50 16

60 14

72 12

A. What is the origin of the background count? B. How is the background count determined? C. By plotting the graph of count rate (vertically) against the time (horizontally), determine the half-life of 220 86Rn. Show clearly on your graph how you obtain your answer/ Radioactivity

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SOLUTIONS D D D A D D C A A C C D B C A A C D B

.

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19.

20. A. α-particle. It is a helium nucleus with two protons and two electrons β-particle. It is a negatively charged electron γ-ray. It is an electromagnetic wave B α -particle It has a high ionising power It has a weak penetrating power. It can be absorbed by a sheet of paper β-particle It has a weak ionising power It has a weak penetrating power. It can be absorbed by a 5mm piece of aluminium γ-ray It has a very weak ionising power It has a very strong penetrating power. It can be absorbed by a 5cm piece of lead 21. A. Radioactive decay is when an unstable atom emits radiation until it reaches a stable atom. B. Background radiation is the radiation present in the environment that is not due to the delicate release of radiation sources. Sources of background radiation: i. Natural sources Cosmic rays; radiation from space Rocks and building materials; gives off radon gas ii. Artificial sources Medical sources such as x-ray Food and drinks Nuclear power and weapon test C. β-particle is a negatively charged electron

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22. A. Nucleon number is the total number of protons and electrons of an element B. Proton number of α-particle: 2 Nucleon number of α-particle: 4

.

4 → 234 90Th + 2He

C.

238 92U

D.

234 90Th

0 → 234 91Pa + −1e

23. A. Half-life is the time taken for the unstable nuclei of a radioactive element to decay by half B. The atomic number increases by one C. The mass number remains the same 𝑡

1 ℎ D. A = Ao ( ) 2 𝑡

1 30 20 = 80( ) 2

A = 20 counts per second Ao = 80 counts per second t=? h = 30 years

𝑡

1 30 20 (2) = 80 𝑡

1 30 (2) = 0.25 𝑡 30 𝑡 30 𝑡 30

log (0.5) = log (0.25) =

log (0.25) log (0.5)

=2

t = 2 × 30 t = 60 years

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24. A. The ground radiations B. Radiations are detected by the Geiger-Muller (GM) tube, also called the Geiger counter. To determine the background count, first remove all sources of radiations and zero the GM tube. Switch on the Geiger counter and start the stopwatch. Determine the count for 20 minutes. Then, divide the count by 20 to find the count per minute

.

C.

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