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Introduction to Physical Sciences 14e

Part 1

Chapter 1

MEASUREMENT Chapter 1 is important because all quantitative knowledge about our physical environment is based on measurement. Some chapter sections have been reorganized and rewritten for clarity. The 1.2 Section, “Scientific Investigation,” introduces the student to the procedures for scientific investigation. Major terms such as experiment, law, hypothesis, theory and scientific method are introduced. The idea that physical science deals with quantitative knowledge should be stressed. It is not enough to know that a car is going “fast”; it is necessary to know how fast. A good understanding of units is of the utmost importance, particularly with the metric-British use in the United States today. The metric SI is introduced and explained. Both the metric and the British systems are used in the book in the early chapters for familiarity. The instructor may decide to do examples primarily in the metric system, but the student should get some practice in converting between the systems. This provides knowledge of the comparative size of similar units in the different systems and makes the student feel comfortable using what may be unfamiliar metric units. The Highlight, “Is Unit Conversion Important? It Sure Is,” illustrates the importance of unit conversion. The general theme of the chapter and the textbook is the students’ position in his or her physical world. Show the students that they know about their environment and themselves through measurements. Measurements are involved in the answers to such questions as, How old are you? How much do you weigh? How tall are you? What is the normal body temperature? How much money do you have? These and many other technical questions are resolved or answered by measurements and quantitative analyses.

DEMONSTRATIONS Have a meter stick, a yardstick, a timer, one or more kilogram masses, a one-liter beaker or a liter soda container, a one-quart container, and a balance or scales available on the instructor’s desk. Demonstrate the comparative units. The meter stick can be compared to the yardstick to show the difference between them, along with the subunits of inches and centimeters. The liter and quart also can be compared. Pass the kilogram mass around the classroom so that students can get some idea of the amount of mass in one kilogram. Mass and weight may be compared on the balance and scales. When discussing Section 1.6, “Derived Units and Conversion Factors,” have class members guess the length of the instructor’s desk in metric and British units. Then have several students

Introduction to Physical Sciences 14e

independently measure the length with the meter stick and yardstick. Compare the measurements in terms of significant figures and units. Compare the averages of the measurements and estimates. Convert the average metric measurement to British units, and vice versa, to practice conversion factors and to see how the measurements compare. Various metric unit demonstrations are available from commercial sources.

ANSWERS TO MATCHING QUESTIONS a. 19 b. 13 c. 21 d. 14 e. 15 f. 8

g. 10

p. 3 q. 20 r. 16

v. 17 w. 5

s. 22 t. 7 u. 23

h. 2

i. 21

j. 1 k. 9

l. 4

m. 18

n. 6 o. 11

ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. c

2. b

3. d

4. b

5. b

6. b

7. d

8. c 9. d 10. c

11. d

12. c 13. a

ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. biological 7. shorter

2. hypothesis

8. fundamental

3. scientific method 9. time or second

4. sight

5. limitations

6. greater than

10. one-millionth, 10-6 11. liter

12. mass

ANSWERS TO SHORT-ANSWER QUESTIONS 1. An organized body of knowledge about the natural universe by which knowledge is acquired and tested. 2. Physics, chemistry, astronomy, meteorology, and geology. 3. Observations and Measurements. 4. Hypothesis. 5. A law is a concise statement about a fundamental relationship of nature. A theory is a welltested explanation of a broad segment of natural phenomena. 6. That phenomena must be investigated, not speculated. 7. Sight, hearing, touch, taste, and smell. 8. They have limitations and can be deceived. 9. (a) No. (b) Yes. (c) Good luck. (Use a couple of straight edges to determine.) 10. They are the most basic quantities of which we can think. 11. A fixed and reproducible value. 12. A group of standard units and their combinations. 13. km/hour

Introduction to Physical Sciences 14e

14. Yes, officially adopted in 1899. 15.Kilogram, a platinum-iridium cylinder. 16. Mass. Weight varies with gravity. 17. Meter-kilogram-second, International System of Units, centimeter-gram-second. 18. Base 10 easier to use (factors of 10). 19. mega- (M), kilo- (k), centi- (c), milli- (m) 20. Mass of a cubic liter of water. 21. Cubic meter. 22. kg, m, s, and C (electric charge). 23. The compactness of matter. 24. It is given a new name. 25. No. An equation must be equal in magnitude and units. 26. Yes. And it could be confused with “meters” instead of “miles.” 27. To express measured numbers properly. 28. By reading a measurement value from an instrument and rounding according to the general rules. 29. Two.

ANSWERS TO VISUAL CONNECTION a. meter, b. kilogram, c. second, d. mks, e. foot, f. pound, g. second, h. fps

ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. Scientific laws explain nature, not to regulate society (legal laws), and scientific laws do not compel, just describe. 2. The liter is larger than the quart, and the kilogram is larger than the pound. However, the kilometer itself is shorter than the mile. 3. Intrinsic properties are invariant. Kilogram cylinder and meterstick subject to wear, dirt, and change. 4. A liter, because it is larger than a quart.

Introduction to Physical Sciences 14e

5. (a) No, gold would weigh about 20 times more. (b) Solving for mass from the density equation, this would be 19320 g or 19.32 kg. No playing catch, as weight is about 45 lb. (1 kg = 2.2 lb). 6. 1 m = 3.28 ft. 830 m (3.28 ft/m) = 2.72 x 103 ft; 508 m (3.28 ft/m) = 1.67 x 103 ft. Δ = 1.05 x 103 ft

ANSWERS TO EXERCISES 1. 10,000 cm or 105 cm 2. 2 x l03 Mb 3. 106 mm3 4. 1 m3 = 103 L. 1 m3 = 102 cm x 102 cm x 102 cm = 106 cm3 (1 L/103 cm3) = 103 L = 1000 L. 5. 0.50 L (1 kg/L) = 0.50 kg = 500 g 6. 25 cm  25 cm  35 cm = 22  103 cm3 or mL = 22 L and 12 L (1 kg/L) = 12 kg = 12,000 g 7. (a) 0.55 Ms = 0.55  106 s (b) 2.8 km = 2.8  103 m (c) 12 mg = 12  10–3 g = 1.2  10–6 kg (d) 100 cm = 1.00 m 8. (a) 40 Mb (b) 572.2 mL (c) 540.0 x 102 cm (d) 5.5 kilobucks 9. 6 ft 5 in. = 77 in (2.54 cm/in.) = 196 cm = 1.963 m 10. 6 ft, 7 in. 11. Yes, to two significant figures 12. (a) 55 mi/h (1.61 km/mi) = 89 km/h (90 km/h). (b) 65 mi/h (1.61 km/mi) = 106 km/h (110 km/h) 13. No, 300 L ~ 300 qt (1 gal/4 qt) = 75 gal 14. No. Room would be about 12 m  10 m, which would be about 36 ft.  30 ft. Much too large for a dorm room. 15. See AYK # 6, 1,67 x 103 ft. 16. 890 ft (1 m/3.28 ft) 271 m; 1,900 ft = 579 m 17. cm, km 18. 103 kg (2.2 lb/kg) = 2,200 lb. 103 kg heavier by 200 lb 19.  = m/V = 500 g/63 cm3 = 7.9 g/cm3 (the density of iron) 20. V = m /  = 500 g/ 7.9 g/cm2 = 63 cm3 21. (a) 7,7 (b) 0.0021 (c) 9400 (d) 0.00034

Introduction to Physical Sciences 14e

22. (a) 0.00999 (b) 645 (c) 0.0106 (c) 8430 23. (a) 1.00 (b) 7380 (c) 0.00179 (d) 47.6 24. (a) 3.142 (b) 0.006907 (c) 483.6 (d) 0.02350 25. (3.15 m  1.53 m)/0.78 m = 6.2 m 26. 6.75

(3 sf)

Introduction to Physical Sciences 14e

Chapter 2

MOTION This chapter covers the basics of the description of motion. The concepts of position, speed, velocity, and acceleration are defined and physically interpreted, with applications to falling objects, circular motion, and projectiles. A distinction is made between average values and instantaneous values. Scalar and vector quantities are also discussed. Also, an interesting Highlight on Galileo and the Leaning Tower of Pisa discusses the status of the tower. Problem solving is difficult for most students. The authors have found it successful to assign a take-home quiz on several questions and exercises at the end of the chapter that is handed in at the beginning of class. (It may save time and be instructive to have students exchange and grade papers as you go over the quiz.) This may be followed by an in-class quiz on one of the take-home exercise, for which the numerical values have been changed. The procedure provides students with practice and helps them gain confidence.

DEMONSTRATIONS A linear air track may be used to demonstrate both velocity and acceleration. If an air track is not available, a 2-in.  6-in.  12-ft wooden plank may be substituted. It will be necessary to have a V groove cut into one edge of the plank to hold a steel ball of about 1-in. diameter. The ball will roll fairly freely in the V groove. Also, various free-fall demonstrations are commercially available. (General references to teaching aids are given in the Teaching Aids section.)

ANSWERS TO MATCHING QUESTIONS a. 16

b. 13 c. 1 d. 6

o. 17

p. 11

q. 4

e. 14

f. 2

g. 3

h. 12 i. 5

j. 15 k. 18

l. 8

m. 10 n. 7

r. 9

ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. d

2. c

3. d

4. d

5. d

7. c

8. d

9. d

10. d

11. c

6. b 12. c

Introduction to Physical Sciences 14e

ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. position 2. scalar

3. vector

4. distance

5. speed

6. constant or uniform

7. time, t2 8. gravity

9. m/s2 10. centripetal (center-seeking)

11. 4

12. acceleration

ANSWERS TO SHORT-ANSWER QUESTIONS 1. Mechanics. 2. An origin or reference point. 3. Length per time (length/time). 4. A scalar has magnitude, and a vector has magnitude and direction. 5. Distance is the actual path length and is a scalar. Displacement is the directed, straight-line distance between two points and is a vector. Distance is associated with speed, and displacement is associated with velocity. 6. They both give averages of different quantities. 7. (a) They are equal. (b) The average speed has a finite value, but the average velocity is zero because the displacement is zero. 8. Either the magnitude or direction of the velocity, or both. An example of both is a child going down a wavy slide at a playground. 9. Yes, both (a) and (b) can affect speed and therefore velocity. 10. No. If the velocity and acceleration are both in the negative direction, the object will speed up. 11. Initial speed is zero. Initial acceleration of 9.8 m/s2, which is constant. 12. The object would remain suspended. 13. Yes, in uniform circular motion, velocity changing direction, centripetal acceleration. 14. Center-seeking. Necessary for circular motion. 15. Yes, we are in rotational or circular motion in space. 16. Inwardly toward the Earth's axis of rotation for (a) and (b). 17. g and vx 18. Greater range on the Moon, gravity less (slower vertical motion). 19. Initial velocity, projection angle, and air resistance. 20. No, it will always fall below a horizontal line because of the downward acceleration due to gravity.

Introduction to Physical Sciences 14e

21. Both have the same vertical acceleration. 22. Less than 45o because air resistance reduces the velocity, particularly in the horizontal direction. ANSWERS TO VISUAL CONNECTION a. speed, b. uniform velocity, c. acceleration (change in velocity magnitude), d. acceleration (change in velocity magnitude and direction) ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. More instantaneous. Think of having your speed measured by a radar. This is an instantaneous measurement, and you get a ticket if you exceed the speed limit. 2. (a) The orbital (tangential) acceleration is small and not detected. (b) The apparent motion of the Sun, Moon, and stars. 3. (a) toward the center of the Earth, (b) toward the axis, (c) zero 4. Yes, neglecting air resistance. 5. d  ½ gt 2 , so t  2d / g 

2(11 m)  1.5 s Balloon lands in front of prof. Student gets 9.8 m/s2

an “F” grade. 6. (a) updraft, slow down, reach terminal velocity later. (b) downdraft, speed up, terminal velocity sooner. 7. Escaping air stabilizes chute – prevents rocking. 8. Streamlines. Prevents air blocking. ANSWERS TO EXERCISES 1. 7 m 2. 5 m south of east 3. v = d/t = 100 m/12 s = 8.3 m/s 4. 1.6 m/s 5. t = d/v = 7.86  1010 m/ 3.00  108 m/s = 2.62  l02 s. Speed of light (constant). 6.. t = d/v = 750 mi/(55.0 mi/h) = 13.6 h 7. (a) d = v t = (52 mi/h)(1.5 h) = 78 mi (b) v = d/t = 22 mi/0.50 h = 44 mi/h (c) v = d/t = 100 mi/2.0 h = 50 mi/h 7. v = d/t = 7.86  1010 m/ 2.62  l02 s = 3.00  108 m/s. Speed of light (constant).

Introduction to Physical Sciences 14e

8. (a) d/150 s. (b) d/192 s., (c) d/342 s. Omission. d inadvertently left out. Assuming 100 m, (a) 100 m/150 s = 0.667 m/s. (b) 100 m/192 s = 0.521 m/s. (c) 200 m/ 342 s = 0.585 m/s. 9. (a) v = d/t = 300 km/2.0 h = 150 km/h, east. (b) Same, since constant. 10. (a) v = d/t = 750 m/20.0 s = 37.5 m/s, north. (b) Zero, since displacement is zero. 11. a = (vf – vo )/t = (12 m/s – 0)/6.0 s = 2.0 m/s2 12. (a) a = (vf – vo )/t = (0 – 8.3 m/s)/1200 s = –6.9  10–3 m/s2 (b) v = d/t = (5.0 

m)/(1.2 

s) = 4.2 m/s (Needs to start slowing in plenty of time.)

13. (a) a = (vf – vo )/t = (8.0 m/s – 0)/10 s = 0.08 m/s2 in direction of motion. (b) a = (12 m/s – 0)/15 s = 0.80 m/s2 in direction of motion. 14. (a) (a) 44 ft/s/5.0 s = 8.8 ft/s2, in the direction of motion. (b) 11 ft/s2, (c) -7.3 ft/s2 (b) a = (88 ft /s – 44 ft /s)/4.0 s = 11 ft /s2 in direction of motion. (c) (66 ft /s – 88 ft /s)/3.0 s = –7.3 ft /s2 opposite direction of motion. (d) a = (66 ft /s – 0)/12 s = 5.5 ft /s2 in direction of motion. 15. No, d = ½ gt 2 = ½ (9.8 m/s2) (4.0)2 = 78 m in 4.0 s. 16. v = vo + gt = 0 + (9.8 m/s2)(3.5 s) = 34 m/s 17. d = ½ gt2, t = sq.root [2(2.71 m)/9.80 m/s2] =7.4 s 18. d = ½ gt2. t as in 17. 4.3 s – 2.5 s = 1.8 s. 19. (a) ac = v2/r = (10 m/s)2/ 70 m = 1.4 m/s2 toward center. (b) ac /g = (1.4 m/s2 )/(9.8 m/s2 ) = 0.14 or 14%‚ yes. 20. 90.0 km/h = 25.0 m/s. ac = v2/r = (25.0 m/s)2/500 m = 1.25 m/s2. 21. 0.55 s. Vertical distance is the same. 22. 45o – 37o = 8o, so 45o + 8o = 57o.

Introduction to Physical Sciences 14e

Chapter 3

FORCE AND MOTION This chapter is one of the most important in the textbook because it deals with Newton’s laws of motion and gravitation, as well as the concepts of linear and angular momentum. Also, involving a force, buoyancy and Archimedes’ principle is discussed in this chapter. The material naturally follows that of Chapter 2. With the foundations of kinematics established, the agents that produce motion are considered. This branch of mechanics is known as dynamics. Sufficient time should be spent on this material to be sure that students have a firm understanding of these concepts. Force and net force are discussed in an initial chapter section because of the importance of understanding these concepts. It is suggested that students be required to make complete statements of Newton’s laws and to give examples. When stating Newton’s second law of motion, stress that the force is the unbalanced force acting on the total mass, and that the mass is the total mass being accelerated. Also that the acceleration is in the direction of the unbalanced, or net, force. Acceleration (or deceleration) is evidence of the action of an unbalanced force. The Highlight on automobile airbags includes side airbags and “depowering” features. Also new to the thirteenth edition is the Highlight: Surface Tension, Water Striders, and Soap Bubbles.

DEMONSTRATIONS A linear air track can be used to illustrate Newton’s laws of motion and the concept of linear momentum. An Atwood machine provides an excellent demonstration for illustrating Newton’s second law of motion. Best results can be obtained if the student is led through the demonstration (see the Laboratory Guide) by questions rather than having the instructor merely perform the experiment. An apple may be brought to class to illustrate the idea of one newton of weight. Be sure the apple weighs about 3.6 ounces. Also, a spring balance calibrated in newtons can be displayed supporting a 1-kg mass. Free-fall can be demonstrated with a feather and a coin in a glass tube from which the air can be evacuated. Let your students handle the glass tube for the best results. Newton’s third law of motion can be demonstrated by using a toy rocket that holds water under pressure-equal and opposite forces are demonstrated as the rocket accelerates along a string. Releasing a blown balloon also illustrates the law.

Introduction to Physical Sciences 14e

The law of conservation of angular momentum is demonstrated dramatically using a turntable or rotating stool and two masses (for example, 1 kg each) held in the hands. While rotating, the masses are brought closer to the body (reduced moment of inertia), and the rate of rotation increases. When beginning the demonstration, point out that you can't get started by yourself. You must have an external force or torque, which can be supplied by a student. Students often will ask to try the demonstration. Permission should be granted with caution. The rotation can make a person quite dizzy. (General references to teaching aids are given in the Teaching Aids section.)

ANSWERS TO MATCHING QUESTIONS a. 14 b. 5 c. 11 d. 16 i. 19

j. 10 k. 1

e. 2

l. 12

f. 9

m. 7

g. 15

h. 6

n. 4 o. 17

p. 13 q. 8

r. 18 s. 3

ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. d

2. d

3. c

4. d

5. a

6. d

7. d

8. d

9. c

10. a 11. c

12. a

13. b

14. d 15. c

ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. capable

2. vector

7. kg  m/s

8. static, kinetic or sliding 9. different

3. could

4. force

5. mass 6. inversely 10. universal

11. greater

12. more 13.

net or unbalanced 14. torque

ANSWERS TO SHORT-ANSWER QUESTIONS 1. No. The force may be balanced and the net force zero. 2. They are the same. 3. (a) goes forward (b) goes backward 4. Dishes and glasses remain in place because of inertia. 5. Inertia of hammer head continues motion and tightens on handle. 6. A quick jerk gives inertia for tearing. Larger rolls have more inertia and resistance to motion. 7. If a is zero in F = ma, then F = 0, and an object is at rest or moving with a constant velocity (first law). 8. Yes if the sum of the forces is zero. 9. Yes, by the first law of motion. 10. Zero, since velocity is constant.

Introduction to Physical Sciences 14e

11. Less force to keep moving that to initiate motion. 12. (a) Ten times the force, but also ten times the inertia, or mass; therefore, it falls at the same rate. (b) Similar, except that the acceleration of the rocks would be less—that is, g/6. 13. No, the pad is there only to hold the rocket. The expanding combustion gases exert a force on the rocket, and the rocket exerts a force on the gases. Consider firing a rocket in space-there is nothing to “push against” there. 14. The two forces of the force pair act on different objects. 15. There are equal and opposite forces for all forces, and the net force is zero. 16. 9.8 N. Holding one end of the string before the pulley, the scale would measure only one mass.

Fr 2 N m2 Gm1m2 G = = , so, 17. F = m1m2 kg 2 r2 18. Because F  1 / r 2 , then F approaches zero only as r approaches infinity. 19. Smaller mass and size (radius). 20. Weight zero (gravity essentially zero). Same mass, 70 kg. 21. Only if the resultant of two or more gravitational forces of an object is zero. 22. Denser liquid, greater buoyant force 23. Enough volume for displacing sufficient water for floating. 24. Zero. Just another liter of water. 25. Same displacement, same force. 26. No. Large person displaces more water, but more buoyant force needed because of greater weight. 27. Balloon buoyant in air and would float away. 28. kg · m/s 29. In the absence of an unbalanced external force, an object or system has a constant velocity and hence a constant momentum. 30. (a) Gravity (the weight force) and the normal upward force of the surface on the blocks. (b) These forces cancel each other, and there is no net external force on the block.

Introduction to Physical Sciences 14e

31. Through the conservation of angular momentum. Tucking reduces the r of the mass distribution and the rotational speed increases.

ANSWERS TO VISUAL CONNECTION a. inertia, b. mass, c. constant velocity, d. net force, e. acceleration, f. m/s2, g. action, h. equal and opposite reaction, i. different objects

ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. Gravity is less (g/6), and in walking, the leg muscles are accustomed to applying a greater force for the Earth’s gravity. 2. Opposite reactions in both cases affects weight. (Downward force of arms, upward force on scale and more weight. 3. Set mg equal to Eq. 3.4. Moon has smaller M and R. 4. (a) To increase the amount of water displaced and increase the buoyancy. (b) Salt water is denser than regular water, and the average density of a person is less than that of salt water. 5. The adhesion between the water and the clothes is not sufficient to provide the necessary centripetal force for the water to rotate with the clothes, and hence the water becomes separated. 6. Shorter lever arm, less torque. 7. Shorter lever arm, more torque.

ANSWERS TO EXERCISES 1. (a) 3.0 N in direction of 8.0 N force. (b) 13.0 N in direction of forces. 2. 350 N to equal fmax. 3. F = m a = (3.0 kg)(5.0 m/s2) = 15 N 4. a = F/m = 2.1 N/(7.0  10–3 kg) = 300 m/s2 5. a = F/m = 950 N/1000 kg = 0.95 m/s2 6. a = F/m = 6.0 N/15 kg = 4.0 m/s2 7. w = mg = (6.0 kg)(9.8 m/s2) = 59 N 8. w = mg = (4.0 kg) (9.8 m/s2) = 39 N 9. (a) 120 1b (4.45 N/lb) = 534 N (b) Personal 10. (a) w = mg = (75 kg)(9.8 m/s2) = 74 N. (b) Same, zero acceleration. 11. F = Gm1m2/r2= (6.67  10–11 N-m2 /kg2)(3.0 kg)(3.0 kg)/(0.15 m)2 = 2.7  10–8 N

Introduction to Physical Sciences 14e

12. (a) F = Gm1m2/r 2= (6.67  10–11 N-m2/kg2)(103 kg)(103 kg)/(25 m)2 = 1.1  10–7 N (b) Much, much smaller; w = mg = (103 kg)(9.8 m/s2) = 9.8  103 N 13. (a) r2 = 2r1, and F 2 / F 1 = (r1/2r1)2= (1/2)2= ¼ (b) r2 = r1/2, and F 2/ F 1 = (2)2 = 4 14. (a) r2 = 2r 1/3, and F 2/ F 1 r1/r2)2 = (3/2)2 = 9/4 = 2.25 (b) r2 = 3r1 and F 2 F 1 = (1/3)2 = 1/9 15. (a) wM = wE /6 = 180 lb/6 = 30 lb (b) Personal. 16. g = F/m = 49 N/125 kg = 0.39 m/s2 17. Float. Density = m/V = 120 g/125 cm3 = 0.96 g/cm3, less than water. 18. Float. Density = 0.28 g/cm3 19. p = mv = (103 kg)(20 m/s) = 2.6  104 kg-m/s, east. 20. p = mv = (9.0  10 2 kg)(30 m/s) = 2.7  104 kg-m/s, north. Less, about 3/4 times less. Direction not a factor. 21. pg = –pb , or mgvg = –mbvb, and with m = w/g, vg = (-mb/mg)vb = (–wb/wg)vg = (–735 N/490 N)(0.50 m/s) = –0.75 m/s 22. mg = (0.75) mb , and vg = (–mb/mg)vb = (–1/0.75)(0.50 m/s) = –0.67 m/s 23. v2 = rlvl/r2 = (600  106 mi)(15,000 mi/h)/(100  106 mi) = 90,000 mi/h 24. (a) 36 km/h = 10 m/s. p = mv = (1.2 kg)(10 m/s) = 12 kg-m/s. (b) 76 mi/h = 33 m/s. p = mv = (1.2 kg)(33 m/s) = 40 kg-m/s.

Introduction to Physical Sciences 14e

Chapter 4

WORK AND ENERGY This chapter should be covered thoroughly in lecture and assignment. The relationship of work and energy is of the greatest importance in understanding many daily activities. Also, the development of the physical environment is closely associated with the control of energy. The law of conservation of energy is one of the most important general laws and has been a key to many of nature's secrets. Thus it is important for the student to know the meanings of work and energy and to be familiar with various forms of energy. Although this chapter deals primarily with general concepts and mechanical energy, it should be pointed out how easy it is to change other types of energy, such as chemical and electrical energy, to other forms and use them to do work. The textbook tries to get the student to think in terms of symbols, and this is a good chapter to stress this kind of thinking. For example, when referring to kinetic energy, think ½ mv2, and when thinking of gravitational potential energy, think mgh. In addition, the status of energy consumption and resources is covered, including alternative and renewable sources.

DEMONSTRATIONS A simple pendulum can be used to display the transformation of potential energy to kinetic energy, and vice versa. As the pendulum swings back and forth, ask the students at what stages the velocity, acceleration, potential energy, and kinetic energy have their minimum and maximum values. Demonstrate the examples of work as shown in the illustrations in the textbook. A radiometer can be used to show that light can do work. (General references to teaching aids are given in the Teaching Aids section.)

ANSWERS TO MATCHING QUESTIONS a. 14 b. 8, c. 13 d. 5

e. 9 f. 11 g. 1

j. 6

n. 10 o. 4

k. 7

l. 12

m. 2

h. 3 i. 15

ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1.a

2. d

3. d

4. c

5. b

6. d 7. c 8. a 9. b

10. a 11. d

12. c

13. b 14. c

Introduction to Physical Sciences 14e

ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. parallel

2. scalar

7. transferring

3. joule

8. isolated

4. work

5. motion, position

9. power, watt

10. 0.75

6. square

11. energy

12. coal 13. exhausted 14.

ethanol (alcohol)

ANSWERS TO SHORT-ANSWER QUESTIONS 1. A force moving an object a distance. 2. No, there must be motion. No work is done in holding an object stationary, but work is done in lifting. 3. Friction. Dust reduces friction. 4. No work while stationary. Work was done on the weights in lifting. 5 Kinetic energy transferred to frictional heat. 6. B. vB = 2vA, mb = ma/3. KA = ½ mavA2

KB = ½ (ma/3)(2vA)2 = ½(4/3)(mavA2)

7. To take advantage of potential energy. 8. W2/W1 = (v2/v1)2 (2)2 = 4 9. Both may be correct, depending on the zero reference point chosen. 10. (a) The height depends on the initial kinetic energy, mgh = ½ mv2. (b) By the conservation of energy, it would have the same speed as it had initially. 11. Yes, relocate the arbitrary zero reference position. 12. Total energy includes all forms of energy. Mechanical energy is the sum of the kinetic and potential energies. 13. 50 J. Potential energy converted to kinetic energy. 14. Total energy: when energy neither enters or leaves a system and thus has a constant value. Mechanical energy: no energy loss. 15. (a) a and e. (b) c. (c) c. (d) a and e. (e) a and e. (f) c. (g) a and e. (h) c. (i) c. (j) a and e. 16. Same initial speed from same height. Both will have same speed on striking the ground. Conservation of energy. 17. (a) b and c. (b) a. (c) a. (d) b and c. (e) b and c. (f) a. (g) b, c; and e, d when going toward h = 0. (h) a; and e, d, when spring is compressing. (i) a; and e, d, when spring is compressing. (j) b, c; and e, d when going toward h = 0. 18. Yes, kinetic energy into potential energy into kinetic energy.

Introduction to Physical Sciences 14e

19. Yes, run the ¼-hp three times as long, or run larger motor one-third the time. 20. P = W/t, so person A, with the shorter time, expends more power. 21. (a) More work done in a given time. (b) Doing a given amount of work faster. 22. Energy, kilowatt-hour (kWh). 23. 100 W/60 W = 1.67 J (same time) 24. Oil. 25. Coal. 26. 100 J/s 27. Radiant, chemical, nuclear, sound, and heat. 28. Examples of alternative energy sources: electricity, natural gas, biodiesel, methane, ethanol, hydrogen, propane 29. Examples of renewable energy sources: hydropower, wind power, solar power, geothermal, tides 30. Solar and wind.

ANSWERS TO VISUAL CONNECTION a. work, b. energy, c. kinetic energy, d. power

ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. Yes, this is numerically possible. If there is no motion, the kinetic energy is zero. If the student selects his or her position as the zero reference point, then the potential energy is zero. 2. Same speed. Upward ball has same speed as other ball on return to initial height. 3. Body energy is used to increase height, but there is a limit. 4. Piecework involves power because the more energy expended per unit time, the greater the output and the more pay. An hourly rate implies a more constant power output, or at least compensation for same. 5. We can feel heat and vibrations, see energetic phenomena and light, and hear sound. However, it is doubtful that you could smell or taste energy. 6. Turn off lights when not needed, turn off appliances when not being used, keep thermostats properly set, and have good home insulation. (Limit children’s TV watching?)

Introduction to Physical Sciences 14e

ANSWERS TO EXERCISES 1. W = Fd = (250 N)(3.0 m) = 750 J 2. F = W/d = 400 J/2.0 m = 200 N 3. W = mgh = (5.0 kg)(9.8 m/s2)(0.45 m) = 22 J 4. W = mgh = (6.0 kg)(9.8 m/s2)(1.5 m) = 88 J 5. W = (60%) Fd = (0.60)(200 N)(6.0 m) = 7.2  102 J 6. W = (40%) Fd = (0.40)(200N)(6.0 m) = 4.8  102J 7. 14.7 J = W = mgh = (0.150 kg)(9.80 m/s2)(10.0 m) 8. 11.0 J = W = mgh = (0.150 kg)(9.80 m/s2)(2.50 m) 9. (a) Ek = ½ mv 2 = ½ (1000 kg)(25 m/s)2 =3.1  105 J (b) W = Ek = 3.1  105 J 10. Ek= ½ mv2, or v =

2E / m =

2(1.2x104J / 60kg = 20 m/s [36 km/h/(m/s)] = 72 km/h

[[ED. Space on each side of x and after 60.]] 11. Ek = ½ mv2 =½ (20 kg)(9.0 m/s)2 = 8.1  102 J 12. Ekb = ½mv2 = ½ (2.0  10-3 kg)(4.0  10 2 m/s)2 = 1.6  102 J. Eko = ½ mv 2= ½ (6.4  107 kg)(10 m/s)2= 3.2  109 J. Ocean liner has greater kinetic energy. 13. 32 J = K = W = Fd = (8.0 N)(4.0 m) 14. d = W/F = 42 J/8.0 N = 5.3 m, so 1.3 m farther. 15. Ep = mgh = (3.00 kg)(9.80 m/s2)(-10.0 m) = –294 J. Below ground zero point. 16. 294 J Ep = +294 J (same magnitude as in Exercise 15.) 17. ½ E p̂ (lost) = ½ mgh = ½ mg (12 m), and h = 6.0 m 18. (0.33E p̂ ) = (0.33) mgh = (0.33) mg (6.0 m), and h = 2.0 m 19. (a) E = mgh = (60 kg)(9.8 m/s2)(12 m) = 7.1  103 J. (b) Same. 20. mgh = ½ mv2, and v =

2gh (h from top) [[Ed. Square root sign over all]]

(a) v =

2 g (5.00 m) = 9.90 m/s

(b) v =

2 g (7.50 m) = 12.1 m/s

21. P = W/t = 7.2  10 2 J/10 s = 72 W 22. t = W/P = 7.2  102 J)/18W = 40 s 23. (a) W = Fh = (556 N)(4.0 m) = 2.2  103 J (b) P = W/t = 2.2  103 J/25 s = 89 W

Introduction to Physical Sciences 14e

24. P = W/t = mgh/t = 125 lb(1 kg/2.2 b)(9.8 m/s2)(4.0 m)/5.0 s = 4.5  102 W 25. E = Pt = (1.60 kW)(1/6 h) + (1.10 kW)(4/6 h) = 0.34 kWh (8¢/kWh) = 2.7¢ 26. (a) E = Pt = (1.25 kW)(4.0/60 h) = 0.083 kWh (b) 0.083 kWh (12¢/kWh) = 1¢

Introduction to Physical Sciences 14e

Chapter 5

TEMPERATURE AND HEAT Chapter 5 is an important chapter because temperature and heat are two of the most common physical concepts that students experience. In general, temperature measurements are given, and we say that heat is a form of energy. Hence it is important that a basic understanding and distinction of temperature and heat is obtained. In large part, the chapter is concerned with the measurement of macroscopic quantities of heat, such as specific heat and latent heat. The general trend is to express these heats in joules (J). However, calculations will be done primarily in kilocalories (kcal) because of the difficulty of adding numbers expressed in powers of 10, which is necessary when using joules. Calculations are much easier when done in kilocalories, and the results can be converted to joules if so desired. Because heat transfer has many applications in daily life, this is an important and interesting topic that should be covered in some detail. The chapter contains interesting Highlights: Freezing from the Top Down, and Hot Gases: Aerosol Cans and Popcorn. Finally, the basics of thermodynamics are discussed in Section 5.7.

DEMONSTRATIONS A thermometer may be calibrated in class by using boiling water for the steam point and ice water for the ice point. Uncalibrated thermometers are available, and students find it interesting and obtain a grasp for temperature scales when the interval between the ice and steam points is divided into degrees. (How should it be done?) Also, have a calibrated thermometer on hand so that you can check and see how accurate your calibration is. The bulb of one of two thermometers may be painted black and exposed to sunlight or a heat lamp to show the difference in radiation absorption.

ANSWERS TO MATCHING QUESTIONS a. 21 b. 19 n. 5

c.6

d. 22 e. 17

f. 9

g. 15

h. 25 i. 4

j. 11 k. 2

o. 13 p. 20

q. 18 r. 3

s. 24

t. 1

u. 12 v. 23

w. 14

l. 16 m. 7

x. 8 y. 10

ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. c

2. a

3. a

4. c 5. b

6. c

7. a

8. b

9. c

10. c 11. c

12. b

Introduction to Physical Sciences 14e

ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. greater 2. temperature 7. conduction

8. gas

3. 1000 4. J/ kg  C

9. Kelvin (absolute)

5. seven

10. inversely

6. pressure

11. direction

12. pump

ANSWERS TO SHORT ANSWER QUESTIONS 1. Fahrenheit 2. Celsius or Kelvin 3. Alcohol, low; mercury, high 4. Because of the thermal expansion of the bimetallic coil on which it sits. 5. Heat is energy in transit 6. Cold in, hot out 7. No 8. It is a measure of J/ kg C  for a particular substance and is characteristic of or specific for that substance. 9. Water in the filling has high specific heat 10. Half the water for the double temperature change 11. Condensation of moisture 12. Specific heat, c = J/ kg C  ; latent heat, J/kg. The latent heat process occurs at a particular temperature, hence there is no temperature change. 13. Thermal conductors: silver, copper, aluminum (metals). Thermal insulators: cloth, Styrofoam, wood. A difference in electron mobility and air space. 14. The tile floor which has a greater thermal conductivity. 15. Cold air underneath causes freezing 16. Loose fitting gives air insulation. Some have loose knitting for air spaces. 17. Temperature and pressure. 18. Solid: definite shape and volume. Liquid: definite volume, assumes shape of container. Gas: no definite shape or volume. (Volume may be restricted to a rigid container.) 19. (a) Sublimation. (b) Deposition. 20. Consists of molecules moving independently at high speeds in all directions. 21. A gas made up of point particles that interact only by collisions.

Introduction to Physical Sciences 14e

22. When the pressure is such that the space between molecules is small relative to the size of the molecules or the temperature drops to where attractions among the molecules are significant. 23. Molecular collisions with the walls of the gas container. 24. Frequent collisions exert a steady average force per unit area on the inside of the ball. 25. Heat is removed from the system (balloon), and negative work is done as the balloon collapses. 26. First law: Energy is conserved in thermodynamic processes. Second law: The direction of a process and whether or not a process will take place spontaneously. 27. First law (conservation of energy) and second law (entropy increases in every natural process). 28. It may be reduced by energy input, which requires a similar increase of entropy, so never destroyed. 29. It never decreases—the universe is the largest closed system of which we can think. 30. No, according to the third law of thermodynamics. 31. Zero.

ANSWERS TO VISUAL CONNECTION (a) deposition, (b) sublimation, (c) melting, (d) freezing, (e) condensation, (f) vaporization ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. Hot air flows out by convection. 2. When steam condenses, latent heat is given up. 3. No, there is conduction (by coolant and through metal) and convection (by fan). 4. Initially the glass expands making the bore slightly larger. 5. Hole becomes larger. Would behave as circular cut out piece. 6. In a sense, the Earth absorbs energy (sunlight) and which does work (processes) in the environment. 7. No. Although there is more order and less entropy in the ice tray system, there is more disorder and a bigger increase in entropy somewhere else in the universe. 8. Answers may vary depending on approximations. Check and see how close.

Introduction to Physical Sciences 14e

ANSWERS TO EXERCISES 1. TF = (9/5)Tc + 32 = (9/5)17° + 32 = 63°F 2. (a) 245oF because of smaller degrees. (b) 375oF (200oC = 392oF) 3. TC = 5/9 (TF – 32) = 5/9(68° – 32) = 20°C 4. TC = 5/9 (TF – 32) = 5/9(103° – 32) = 39oC 5. (a) TC = 5/9(– 40° – 32) = – 40°C (b) TK = TC + 273 = 233 K 6. (a) TC = TK – 273 = 3 – 273 = –270 K (b) TF = (9/5)(–270) + 32 = –454°F 7. 100 kcal/h (4186 J/kcal) = 4.2 x 105 J/h = 420 kJ/h 8. 250 kJ (1 kcal/4.2 x 103 kJ) = 6.0 x 10-2 kcal 9. 0.45 Cal/ h·lb (8 h)(150 lb) = 540 Cal 10. 3500 Cal / (0.45 Cal/h) = 7.8  103 h (approx. 324 days, almost a year). 11. 3500 Cal / (325 Cal/h) = 11 h 12. 4.0 mi/h  11 h = 44 mi 13. H = mc  T = (0.50 kg)(1.0 kcal/ kg∙oC )(10OC) = 5.0 kcal 14. H = mc  T = (1.0 kg)(4186 kg∙oC )(100C°) = 4.2  105 J 15. (a) H = mc  T = (1.0 kg)(1.0 kcal/ kg∙oC ) (80 C°) = 80 kcal (b) (80 kcal) (0.00116 kWh/kcal)(12¢/kWh) = 1.1¢ 16. . H = mc  T. With H and m equal,  Ti/  Tal = cal/ci = (0.22 kcal/kg-oC)/(0.105 kcal/kg-oC) = 2.l.

So iron will have the higher temperature, 2.1 times higher. 17. HL = mc  T = (0.500 kg)(0.50 kcal/ kg∙oC )(10 C°) = 2.5 kcal kg)(80 kcal/kg) = 40.0 kcal,

H2 = mLf = (0.500

H3 = (0.500)(1.00 kcal/ kg∙oC )(20 C°) = 10.0 kcal

Total=

52.5 kcal 18. H1 = (0.200 kg)(0.50 kcal/ kg∙oC )(10 C°) = 1.0 kcal H2= (0.200 kg)(540 kcal/kg) = 108 kcal H3 = (0.200 kg)(1.0 kcal/ kg∙oC ) (100 C°) = 20 kcal

H4 = (0.200 kg)(80 kcal/kg) =

16 kcal Total = 145 kcal 19. With V2 = 4 Vij p2 = (V1/V2) p1 = (1/4) p1 20. p2 = (T2/T1) p2 = (1.2) p1 21. T1 = 20° + 273 = 293 K, T2 = 40° + 273 = 313 K

Introduction to Physical Sciences 14e

p2 = (T2/T1) p1 = (313 K/293 K)

p1 = (1.07) p1

22. T1 = 20° + 273 = 293 K T2 = (p2/p1) T1 = (1.5)293 K = 440 K; 440 K – 273 = 167°C 23. p2 = (V1/V2) p1 = (0.500 m3/0.150 m3)(200 Pa) = 667 Pa 24. T1 = 20° + 273 = 293 K T2 = (V2/V1) T1 = (0.600 m3)/0.500 m3)(293 K) = 352 K; 352 K – 273 = 79°C

Introduction to Physical Sciences 14e

Chapter 6

WAVES AND SOUND The concepts of waves, sound, and light are discussed in this chapter. Because most information about our environment comes to us by means of waves (see Section 1.3, “The Senses”), the general properties of waves are studied to prepare the student for the many physical concepts that involve waves. The general properties of waves are considered. Light is treated generally as waves, not electromagnetic waves since electric and magnetic fields have not been studied (Chapter 8). Major emphasis is given to sound waves because of their relationship to the environment. This includes the Doppler effect with its many practical applications. A qualitative discussion of resonance and standing waves without mathematics is presented.

DEMONSTRATIONS There are many demonstrations that illustrate waves, sound, and the Doppler effect. For the concept of transverse waves, a long length of rubber hose is very useful. The demonstration of longitudinal waves can be made with a toy Slinky. A number of demonstrations are available for sound and electromagnetic waves. Consult the Teaching Aids section (Appendix) for sources of these. Sound demonstrations are particularly easy and interesting. Illustrate the production of sound by several methods. Perform the ringingbell demonstration in a bell jar from which the air can be evacuated. The Doppler effect can be demonstrated using an ordinary hair dryer. Use an oscilloscope for graphic representation of different sounds, including speech. Standing waves can be demonstrated using a mechanical vibrator and a piece of string with a suspended weight.

ANSWERS TO MATCHING QUESTIONS a. 4

b. 7 c. 18

d. 15 e. 13 f. 9

g. 20

h. 12

i. 1

k. 14

l. 6 m. 16

n. 11 o. 3

j. 21

p. 8 q. 19

r. 5

s. 10 t. 17

u. 2

ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. b

2. a

3. a

4. a

5. d

6. b

7. c 8. b

9. d 10. a

11. b

12. a

13. d

Introduction to Physical Sciences 14e

ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. energy

2. perpendicular

electromagnetic 9. intensity

3. wavelength

4. wavelength 5. light or 3.00  108 m/s

7. longitudinal 8. 20 kHz

10. 3 11. higher 12. approaching 13. natural or characteristic

ANSWERS TO SHORT-ANSWER QUESTIONS 1. A propagation of energy. 2. No, electromagnetic waves need no medium. Travels in vacuum. 3.(a) parallel to wave velocity direction. (b) perpendicular to wave velocity direction 4. (a) m (b) Hz (or 1/s)

(d) m

5. Two, the maximum wave displacement from the equilibrium position, and E  A2 6. (a) 2 cm (b) f = 1/0.6 s = 1.7 Hz 7. No, light travels at c in a vacuum. 8. Longer wavelength: red end. Higher frequency: blue end. 9. No, they are electromagnetic waves. 10. Light:  = 400 nm to 700 nm; sound:  = 10–1 m to 1 m. (Calculate, or from table.) 11. The lowest density part of a sound wave. 12. Loses energy, smaller amplitude. 13. 25,

1 25

14. (a) Frequency. (b) Intensity, or amplitude. (c) Harmonics, or overtones. 15. Sound from different parts of the band interfere. Interference distorts sound. 16. Energy is in joules (J). Intensity is J/s per m2 or W/m2. 17. No. The dB scale is not linear. (An increase of 3 dB doubles intensity.) 18. Light travels faster than sound. 19. (a) Shortened. (b) Lengthened. 20. (a) Shift to a higher frequency. (b) Shift to a lower frequency. 21. (680 m/s)/340 m/s = (Mach) 2 22. Detection involves the reflection of waves from an object. Speed and ranging involve the Doppler shift of the reflected waves. 23. Maximum energy transfer to the system when driven at a resonance frequency.

Introduction to Physical Sciences 14e

24. Node 25. The tension applied to a string. By varying the length (with a finger on the string) and/or tension (tuning).

ANSWERS TO VISUAL CONNECTION (a) transverse, (b) electromagnetic, (c) longitudinal, (d) sound, (e) inverse, f = 1/T, (f) wave speed ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. Greater speed downwind. Slower speed against the wind. Strong wind only you may hear yourself. 2. No. There is no atmosphere on the Moon. Communication is by radio waves, which are electromagnetic waves that require no medium for propagation. 3. Through sunburns and suntans. UV also can cause cataracts and blindness (not a good detection method). 4. Faster than the speed of sound in water. 5. Because of standing waves (overtones) set up between the shower walls. 6. Open: L = 1/2, 3/2, and 5/2 wavelengths Closed: L = 1/4, 3/4, and 5/4 wavelengths

ANSWERS TO EXERCISES 1. T = 1/f = 1/5,0 Hz =0.20 s 2. T = 1/f = 1/(0,25  103 Hz) = 4.0 x10–3 s 3. (a) f = v /  = (2.0 m/s)/(1.5 m) = 1.3 Hz

(b) T = 1/f = 1/(1.3 Hz) = 0.77 s

4.  = c/f = (344 m/s)/(3000 Hz) = 0.115 m 5. (a)  = c/f = (3.00  108 m/s)/(6.50  10 5 Hz) = 4.62  102 m (b)  = (3.00  108 m/s)/(9.51  106 Hz) = 3.15 m 6.  = (3.00  l08 m/s)/(1018 Hz) = 3.00  10–10 m, or 0.300 nm 7. f = c/  = (3.00  108 m/s)/(420  10–9 m) = 7.14  1014 Hz 8. f = c/  = (3.00  108 m/s)/6.00  10–6 m = 0.500  1014 Hz From Figure 6.7 (in text), infrared region. 9. 9.48 x 1012 km . d = ct = (3.00 x 105 km/s)(3.16 x 107 s)

Introduction to Physical Sciences 14e

10. (a) d = 3000 mi = 4.8 x 106 m, t = d/c = (4.8 x 106 m)/3.0 x 108 m/s =1.6 x 10-2 s. (b) d = 3.50 x 105 m, t = d/c = (3.50 x 105 m)/3.0 x 108 m/s = 1.2 x 10-3 s 11.  = v/f = (344 m/s)/(50  103 Hz) = 6.9  10–3 m 12.  = v/f = (344 m/s)/(20 Hz) = 17 m  2 = 344 m/s/(2.0  104 Hz) = 0.017 m

13. vm =15(3344 m/s) = 5.2 x 103 m/s  = v m /f = (5.2 x 10 3 m/s)/20 x l0 3 Hz = 0.26 m 14. vm =  f =(0.333 m)(15.0 x 10 3 ) = 5.00 x 10 3 m/s. v m /v s = (5.00 x 10 3 m/s)/340 m/s = 150 times. 15. d = vt = (1/3 km/s)(4.5 s) = 1.5 km = (1/5 mi/s)(4.5 s) =0.90 mi 16. d = vt = (1/3 km/s)(9.0 s) = 3.0 km t = d/v s = (3.0 km)/(15 km/h) = 1/4 h = 12 min 17. 100 dB – 90 dB = 20 dB or 2 B, then10 2 or 100 times greater. 18. 10,000 = 104, and 4 B or 40 dB increase to 120 dB.

Introduction to Physical Sciences 14e

Chapter 7

OPTICS AND WAVE EFFECTS The six major properties of waves (reflection, refraction, dispersion, polarization, diffraction, and interference) are presented and discussed in this chapter. It is convenient to introduce the properties to the students by means of demonstrations. Reflection can be illustrated with a plane mirror. Also, note that everything in the classroom is seen by reflection except the light source. A pencil or a small ruler in a glass of water is a good demonstration of refraction, and a prism can be used to illustrate dispersion. Viewing a candle through the slit between two fingers is good for illustrating diffraction, as is viewing a candle flame through a feather. Interference of light waves is best illustrated by Young’s experiment. If an experiment with a spectroscope is not done in the laboratory, the instrument should be presented and demonstrated in class. Polarized light is easily demonstrated with crossed Polaroids. (Polarizing sunglasses may be used.) Considerable class time should be spent in constructing ray diagrams and locating real and virtual images for different types of spherical mirrors and lenses. To simplify matters, only raw diagrams are used to locate images. The mirror and lens equations have been omitted.

DEMONSTRATIONS There are many demonstrations for illustrating waves, sound, and the basic laws of optics. A few have been mentioned in the introduction above. The ripple tank is an excellent piece of apparatus for the production and projection of water waves. The apparatus can be used to study reflection, refraction, diffraction, and interference of waves. Distribute replica diffraction gratings to the students in class and let them see the spectrum from an incandescent light source. Demonstrate polarization with linear polarizers and double refraction with an Iceland spar crystal. If available, a laser is a spectacular demonstration in itself, and many optical properties can be demonstrated with commercial laser kits.

ANSWERS TO MATCHING QUESTIONS a. 15

b. 3 c. 18

d. 9 e. 7

f. 19

g. 20

h. 1 i. 11

j. 22 k. 5

l. 16

m. 10 n. 8

o. 17 p. 2

q. 6 r. 13

s. 21 t. 12

u. 14 v. 4

ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. d

2. d

3. b

4. a 5. b

6. c

7. b

8. c 9. c

10. a

11. a

12. a

Introduction to Physical Sciences 14e

ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. geometrical or ray 6. converging 11. greater

2. specular, diffuse

7. cannot

8. thicker

3. vacuum 4. toward

9. concave or diverging

5. total internal

10. transverse

12. principle of superposition

ANSWERS TO SHORT-ANSWER QUESTIONS 1. (a) perpendicular back reflection to surface. (b) parallel to surface, no reflection. 2. Left, seen as right in mirror. (self-portrait) 3. Walk toward you at same speed. In-step but opposite feet (right-left reversal). 4. 12 in, but becomes smaller in appearance the farther from the mirror. 5. In mirror with right-left reversal 6. (a) no, perpendicular to surface. (b) no, parallel to surface. 7. No, above atmosphere 8. Refraction at water-glass and glass-air surfaces. 9. Because n = c/v, and c is greater than v. 10. Internal reflection and dispersion 11. Yes, sunlight refracted over horizon by atmosphere. No atmosphere, no refraction – shorter. 12. The distance between the vertex and the center of curvature is the radius of curvature R, and half this distance is the focal length, which defines the focal point at R/2. 13. Real images are formed on the object side of the mirror and can be displayed or focused on a screen. Virtual images are formed “behind” or “inside” the mirror, and cannot be formed on a screen. 14. (a) Concave: real, Do > f. (b) Convex: always virtual. 15. Reflected parallel to axis. 16. Through the focal point 17. To get a reflected beam 18. Concave 19. At the edges. 20. (a) Convex: real, Do > f; virtual, Do < f. (b) Concave: always virtual.

Introduction to Physical Sciences 14e

21. Because the lens or lens system gives an inverted image. Focusing involves adjusting the object distance so that a sharp image is formed on the screen. 22. The image is that of the Sun. The sunlight is concentrated on the spot. 23. To correct for different distances 24. Sunglasses and LCDs (for example). 25. No, sound waves are longitudinal and cannot be polarized. 26. Longitudinal waves cannot be polarized; transverse waves can be. 27. Yes, the sheets would be “crossed” and darken. 28. Use long wavelength of light compared to the size of an opening or object. Also use a diffraction grating. 29. The wavelengths of sound satisfy the diffraction condition. Light waves have very short wavelengths. 30. AM, longer wavelengths. 31. Interference.

ANSWERS TO VISUAL CONNECTION (a) object; (b) convex (converging); (c) real, (d) inverted, (e) reduced ANSWERS TO APPLYING-YOUR-KNOWLEDGE QUESTIONS 1. A spherical convex mirror. 2. Reflections from different surfaces of glass. 3. The convex mirror produces a reduced image, and the smaller image may be interpreted as being more distant than it actually is. 4. (a) At infinity. (b) At the opposite focal point. 5. The fish would see the 360° above-water panorama in a circular cone defined by the critical angle. 6. Using polarizing sunglasses to see if you can darken the sunglasses with cross-polarization. 7. Darken (90°), (lighten 180°), darken (270°), and lighten (360°).

ANSWERS TO EXERCISES 1. θi = θr = 30° 2. θr = 90° – 30° = 60°

Introduction to Physical Sciences 14e

3. Bisecting triangles in the figure gives one-half height. Same for any distance. 4. 76 in./2 = 38 in., and 62 in./2 = 31 in. So,  = 7 in. 5. cm = c/ n = (3.00  108 m/s)/2.42 = 1.24  108 m/s 6. n = c/cm = (3.00  108 m/s)/(1.40  108 m/s) = 2.14 7. cm /c = 1/n = 1/1.52 = 0.658 (  100%) = 65.8% 8. cm /c = 0.413 = 1/n, and n = 1/0.413 = 2.42, diamond. (See Table 7.1.) 9. Sketch ray diagram. Real, inverted, same size. 10. Sketch ray diagrams. Reduced image becomes larger as objects move toward the mirror approaching the center of curvature, with M = 1 at that point. Continues to become larger as object approaches focal point, and image becomes virtual inside focal point. 11. Sketch ray diagram. 60 cm, virtual, upright, and smaller. 12. Sketch ray diagram. 5.5 cm, virtual, upright, and smaller. 13. Sketch ray diagram. Real, inverted, same size. 14. Sketch ray diagrams. Reduced image becomes larger as object moves toward lens approaching the 2f position, with M = 1 at that point. Then M > 1 as object approaches the focal point, and image becomes virtual inside the focal point. 15. Sketch a ray diagram. Di = 36 cm, real, inverted, and reduced. 16. Sketch ray diagrams. Di = 2f. 17. Parallel rays never meet or meet at infinity. No image. 18. Sketch a ray diagram. Di  60 cm, virtual, upright, and enlarged.

Introduction to Physical Sciences 14e

Chapter 8

ELECTRICITY AND MAGNETISM This is an important chapter because students should have a basic understanding of electric circuits because electricity plays such an important role in our lives. Also, knowledge of the fundamental concepts of electricity and magnetism is necessary in the study of atomic physics. Electric and magnetic laws and fields should be stressed so as to provide an understanding of some of the underlying concepts in the study of modern physics and chemistry. The discussion of electric fields has been expanded in the Thirteenth Edition, so as to make electromagnetism and waves better understood. This chapter is somewhat long, but the study of electric and magnetic phenomena is well worth the effort.

DEMONSTRATIONS Lecture demonstrations with an electroscope that can be projected on a large screen are very good and create interest. Demonstrate electromagnetic induction. (Commercial apparatuses are available.) Demonstrate the electric generator and electric motor. (May be constructed or commercially available.) Demonstrate the compass and dip needle. Pass around a cheap compass and bar magnets.

ANSWERS TO MATCHING QUESTIONS a. 6

b. 16 c. 8

m. 19

d. 24

n. 14 o. 7

e. 13 f. 20

p. 17

q. 9

g. 25

r. 15

h. 11 i. 23

s. 10

t. 1

u. 22

j. 2 k. 18

l. 5

v. 4 w. 12

x. 21

y. 3

ANSWERS TO MULTIPLE-CHOICE QUESTIONS 1. b

2. c

3. a

4. b

5. a

6. a 7. a 8. b

9. c

10. d 11. c

12. b 13. c 14. b

ANSWERS TO FILL-IN-THE-BLANK QUESTIONS 1. positively 7. I 2 R

2. amp-volt or (amp)2ohm

8. direct or dc 9. smallest

3. semiconductors

10. Curie

11. south

4. charge

5. open

6. ohm

12. secondary

ANSWERS TO SHORT-ANSWER QUESTIONS 1. Protons and neutrons. Protons and electrons.

Introduction to Physical Sciences 14e

2.Equal and opposite, Newton’s third law. 3. Because of attractive electrical forces arising from charging by induction and polarized molecules. 4. Static cling. 5. Electrical potential energy arises due to work done against an electric force. Voltage is electrical energy (or work) per unit charge, V = W/q. 6. A voltage source and a closed circuit. 7. P = V 2/R Small resistance, large joule heat, and vice versa. 8. Electric field. 9. Direct current has a voltage with constant polarity and flows in only one direction. Alternating current results from a changing polarity, and the current alternates in direction. 10. So that all will have the same voltage and independent paths so they can be operated independently. 11. (a) Opens a circuit by the melting of a metal strip when current is too large. (b) Opens a circuit by magnetic or thermal means when current is too large. (c) Dedicated grounding wire to prevent object being at high voltage. (d) Use of ground wire as a grounding wire. 12. Yes, less likely to strike. 13. Head-to-tail, series; all heads to tails, parallel. 14. Form small magnets that line up with the field. 15. Similar. Likes repel, unlikes attract. 16. (a) One that is easily magnetized. Iron, nickel, and cobalt. (b) The ferromagnetic material becomes an induced magnet. Above the Curie temperature, ferromagnetic materials loses their magnetism. 17. A current gives rise to a magnetic field in an iron core. 18. (a) The magnetic field of a bar magnet with the bar’s north magnetic pole near the Earth’s geographic south pole. (b) Declination is the horizontal angular distance between the Earth’s magnetic field lines or compass direction (magnetic north) and true north. Maps are based on true north, so it is necessary to know the declination to navigate properly. 19. Based on torque on a current-carrying wire. See text for description.