Giambattista - College Physics 4th c2013 solutions ISM (1) - INSTRUCTOR SOLUTIONS MANUAL(parts 1-5 ch 1-30 Instructors Solutions Manual to accompany

Giambattista - College Physics 4th c2013 solutions ISM (1)

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Unformatted text preview: INSTRUCTOR SOLUTIONS MANUAL (parts 1-5, ch 1-30) Instructor’s Solutions Manual to accompany COLLEGE PHYSICS Table of Contents Chapter 1 Introduction Chapter 2 Force Chapter 3 Acceleration and Newton’s Second Law of Motion Chapter 4 Motion with Constant Acceleration Chapter 5 Circular Motion Review and Synthesis: Chapters 1–5 Chapter 6 Conservation of Energy Chapter 7 Linear Momentum Chapter 8 Torque and Angular Momentum Review and Synthesis: Chapters 6–8 Chapter 9 Fluids Chapter 10 Elasticity and Oscillations Chapter 11 Waves Chapter 12 Sound Review and Synthesis: Chapters 9–12 Chapter 13 Temperature and the Ideal Gas Chapter 14 Heat Chapter 15 Thermodynamics Review and Synthesis: Chapters 13–15 Chapter 16 Electric Forces and Fields Chapter 17 Electric Potential Chapter 18 Electric Current and Circuits Review and Synthesis: Chapters 16–18 Chapter 19 Magnetic Forces and Fields Chapter 20 Electromagnetic Induction Chapter 21 Alternating Current Review and Synthesis: Chapters 19–21 Chapter 22 Electromagnetic Waves Chapter 23 Reflection and Refraction of Light Chapter 24 Optical Instruments Chapter 25 Interference and Diffraction Review and Synthesis: Chapters 22–25 Chapter 26 Relativity Chapter 27 Early Quantum Physics and the Photon Chapter 28 Quantum Physics Chapter 29 Nuclear Physics Chapter 30 Particle Physics Review and Synthesis: Chapters 26–30 1 24 62 105 144 171 186 218 247 281 302 330 361 387 408 420 447 471 496 507 537 566 615 627 663 692 726 737 757 792 824 858 867 893 921 948 974 981 Chapter 1 INTRODUCTION Conceptual Questions 1. Knowledge of physics is important for a full understanding of many scientific disciplines, such as chemistry, biology, and geology. Furthermore, much of our current technology can only be understood with knowledge of the underlying laws of physics. In the search for more efficient and environmentally safe sources of energy, for example, physics is essential. Also, many study physics for the sense of fulfillment that comes with learning about the world we inhabit. 2. Without precise definitions of words for scientific use, unambiguous communication of findings and ideas would be impossible. 3. Even when simplified models do not exactly match real conditions, they can still provide insight into the features of a physical system. Often a problem would become too complicated if one attempted to match the real conditions exactly, and an approximation can yield a result that is close enough to the exact one to still be useful. 4. (a) 3 (b) 9 5. Scientific notation eliminates the need to write many zeros in very large or small numbers. Also, the appropriate number of significant digits is unambiguous when written this way. 6. In scientific notation the decimal point is placed after the first (leftmost) numeral. The number of digits written equals the number of significant figures. 7. Not all of the significant digits are precisely known. The least significant digit (rightmost) is an estimate and is less precisely known than the others. 8. It is important to list the correct number of significant figures so that we can indicate how precisely a quantity is known and not mislead the reader by writing digits that are not at all known to be correct. 9. The kilogram, meter, and second are three of the base units used in the SI system. 10. The SI system uses a well-defined set of internationally agreed upon standard units and makes measurements in terms of these units and their powers of ten. The U.S. Customary system contains units that are primarily of historical origin and are not based upon powers of ten. As a result of this international acceptance and the ease of manipulation that comes from dealing with powers of ten, scientists around the world prefer to use the SI system. 11. Fathoms, kilometers, miles, and inches are units with dimensions of length. Grams and kilograms are units with dimensions of mass. Years, months, and seconds are units with dimensions of time. 12. The first step toward successfully solving almost any physics problem is to thoroughly read the question and obtain a precise understanding of the scenario. The second step is to visualize the problem, often making a quick sketch to outline the details of the situation and the known parameters. 13. Trends in a set of data are often the most interesting aspect of the outcome of an experiment. Such trends are more apparent when data is plotted graphically rather than listed in numerical tables. 14. The statement gives a numerical value for the speed of sound in air, but fails to indicate the units used for the measurement. Without units, the reader cannot relate the speed to one given in familiar units such as km/s. 15. After solving a problem, it is a good idea to check that the solution is reasonable and makes intuitive sense. It may also be useful to explore other possible methods of solution as a check on the validity of the first. 1 Chapter 1: Introduction College Physics Multiple-Choice Questions 1. (b) 2. (b) 3. (a) 4. (c) 5. (d) 6. (d) 7. (b) 8. (d) 9. (b) 10. (c) Problems 1. Strategy The new fence will be 100% + 37% = 137% of the height of the old fence. Solution Find the height of the new fence. 1.37 × 1.8 m = 2.5 m 2. Strategy There are 60 s 60 min 24 h × × = 86, 400 seconds in one day and 24 hours in one day. 1 min 1h 1d Solution Find the ratio of the number of seconds in a day to the number of hours in a day. 86, 400 24 × 3600 = = 3600 1 24 24 3. Strategy Relate the surface area S to the radius r using S = 4π r 2 . Solution Find the ratio of the new radius to the old. S1 = 4π r12 and S2 = 4π r22 = 1.160 S1 = 1.160(4π r12 ). 4π r22 = 1.160(4π r12 ) r22 = 1.160r12 ⎛ r2 ⎜⎜ ⎝ r1 2 ⎞ ⎟⎟ = 1.160 ⎠ r2 = 1.160 = 1.077 r1 The radius of the balloon increases by 7.7%. 4. Strategy Relate the surface area S to the radius r using S = 4π r 2 . Solution Find the ratio of the new radius to the old. S1 = 4π r12 and S2 = 4π r22 = 2.0S1 = 2.0(4π r12 ). 4π r22 = 2.0(4π r12 ) r22 = 2.0r12 ⎛ r2 ⎜⎜ ⎝ r1 2 ⎞ ⎟⎟ = 2.0 ⎠ r2 = 2.0 = 1.4 r1 The radius of the balloon increases by a factor of 1.4. 5. Strategy To find the factor by which the metabolic rate of a 70 kg human exceeds that of a 5.0 kg cat use a ratio. Solution Find the factor. ⎛ mh ⎜⎜ ⎝ mc ⎞ ⎟⎟ ⎠ 3/4 ⎛ 70 ⎞ =⎜ ⎟ ⎝ 5.0 ⎠ 3/4 = 7.2 2 College Physics Chapter 1: Introduction 6. Strategy To find the factor Samantha’s height increased, divide her new height by her old height. Subtract 1 from this value and multiply by 100 to find the percent increase. Solution Find the factor. 1.65 m = 1.10 1.50 m Find the percentage. 1.10 − 1 = 0.10, so the percent increase is 10 % . 7. Strategy Recall that area has dimensions of length squared. Solution Find the ratio of the area of the park as represented on the map to the area of the actual park. map length 1 map area = = 10−4 , so = (10−4 )2 = 10−8 . actual length 10, 000 actual area 8. Strategy Let X be the original value of the index. Solution Find the net percentage change in the index for the two days. (first day change) × (second day change) = [ X × (1 + 0.0500)] × (1 − 0.0500) = 0.9975 X The net percentage change is (0.9975 − 1) × 100% = −0.25%, or down 0.25% . 9. Strategy Use a proportion. Solution Find Jupiter’s orbital period. T 2 R3 T 2 ∝ R3 , so J = J = 5.193. Thus, TJ = 5.193/2 TE = 11.8 yr . TE2 RE3 10. Strategy The area of the circular garden is given by A = π r 2 . Let the original and final areas be A1 = π r12 and A2 = π r22 , respectively. Solution Calculate the percentage increase of the area of the garden plot. π r 2 − π r12 r2 − r2 1.252 r12 − r12 ∆A 1.252 − 1 × 100% = 2 × 100% = 2 1 × 100% = × 100% = × 100% = 56% A 1 r12 r12 π r12 11. Strategy The area of the poster is given by A = w. Let the original and final areas be A1 = A2 = 2 w2 , 1w1 and respectively. Solution Calculate the percentage reduction of the area. A2 = 2 w2 = (0.800 1 )(0.800 w1 ) = 0.640 1w1 = 0.640 A1 A1 − A2 A − 0.640 A1 × 100% = 1 × 100% = 36.0% A1 A1 12. Strategy The volume of the rectangular room is given by V = wh. Let the original and final volumes be V1 = 1w1h1 and V2 = 2 w2 h2 , respectively. Solution Find the factor by which the volume of the room increased. V2 w h (1.50 1 )(2.00w1 )(1.20h1 ) = 2 2 2 = = 3.60 V1 1w1h1 1w1h1 3 Chapter 1: Introduction College Physics 13. Strategy Assuming that the cross section of the artery is a circle, we use the area of a circle, A = π r 2. Solution A1 = π r12 and A2 = π r22 = π (2.0r1 ) 2 = 4.0π r12 . Form a proportion. A2 4.0π r12 = = 4.0 A1 π r12 The cross-sectional area of the artery increases by a factor of 4.0. 14. (a) Strategy The diameter of the xylem vessel is one six-hundredth of the magnified image. Solution Find the diameter of the vessel. d magnified 3.0 cm dactual = = = 5.0 × 10−3 cm 600 600 (b) Strategy The area of the cross section is given by A = π r 2 = π (d 2)2 = (1 4)π d 2 . Solution Find by what factor the cross-sectional area has been increased in the micrograph. Amagnified 14 π d magnified 2 ⎛ 3.0 cm ⎞2 = =⎜ ⎟ = 360,000 . 2 −3 1πd Aactual × 5.0 10 cm ⎝ ⎠ actual 4 15. Strategy Use the fact that RB = 1.42 RA . Solution Calculate the ratio of PB to PA . PB PA = V2 RB V2 RA = RA RB = RA 1.42 RA = 1 = 0.704 1.42 16. Strategy Recall that each digit to the right of the decimal point is significant. Solution Comparing the significant figures of each value, we have (a) 5, (b) 4, (c) 2, (d) 2, and (e) 3. From fewest to greatest we have c = d, e, b, a. 17. (a) Strategy Rewrite the numbers so that the power of 10 is the same for each. Then add and give the answer with the number of significant figures determined by the less precise of the two numbers. Solution Perform the operation with the appropriate number of significant figures. 3.783 × 106 kg + 1.25 × 108 kg = 0.03783 × 108 kg + 1.25 × 108 kg = 1.29 × 108 kg (b) Strategy Find the quotient and give the answer with the number of significant figures determined by the number with the fewest significant figures. Solution Perform the operation with the appropriate number of significant figures. (3.783 × 106 m) ÷ (3.0 × 10−2 s) = 1.3 × 108 m s 4 College Physics Chapter 1: Introduction 18. (a) Strategy Move the decimal point eight places to the left and multiply by 108. Solution Write the number in scientific notation. 310,000,000 people = 3.1× 108 people (b) Strategy Move the decimal point 15 places to the right and multiply by 10−15. Solution Write the number in scientific notation. 0.000 000 000 000 003 8 m = 3.8 × 10−15 m 19. (a) Strategy Rewrite the numbers so that the power of 10 is the same for each. Then subtract and give the answer with the number of significant figures determined by the less precise of the two numbers. Solution Perform the calculation using an appropriate number of significant figures. 3.68 × 107 g − 4.759 × 105 g = 3.68 × 107 g − 0.04759 × 107 g = 3.63 × 107 g (b) Strategy Find the quotient and give the answer with the number of significant figures determined by the number with the fewest significant figures. Solution Perform the calculation using an appropriate number of significant figures. 6.497 × 104 m 2 = 1.273 × 102 m 5.1037 × 102 m 20. (a) Strategy Rewrite the numbers so that the power of 10 is the same for each. Then add and give the answer with the number of significant figures determined by the less precise of the two numbers. Solution Write your answer using the appropriate number of significant figures. 6.85 × 10−5 m + 2.7 × 10−7 m = 6.85 × 10−5 m + 0.027 × 10−5 m = 6.88 × 10−5 m (b) Strategy Add and give the answer with the number of significant figures determined by the less precise of the two numbers. Solution Write your answer using the appropriate number of significant figures. 702.35 km + 1897.648 km = 2600.00 km (c) Strategy Multiply and give the answer with the number of significant figures determined by the number with the fewest significant figures. Solution Write your answer using the appropriate number of significant figures. 5.0 m × 4.3 m = 22 m 2 (d) Strategy Find the quotient and give the answer with the number of significant figures determined by the number with the fewest significant figures. Solution Write your answer using the appropriate number of significant figures. ( 0.04 π ) cm = 0.01 cm 5 Chapter 1: Introduction College Physics (e) Strategy Find the quotient and give the answer with the number of significant figures determined by the number with the fewest significant figures. Solution Write your answer using the appropriate number of significant figures. ( 0.040 π ) m = 0.013 m 21. Strategy Multiply and give the answer in scientific notation with the number of significant figures determined by the number with the fewest significant figures. Solution Solve the problem. (3.2 m) × (4.0 × 10−3 m) × (1.3 × 10−8 m) = 1.7 × 10−10 m3 22. Strategy Follow the rules for identifying significant figures. Solution (a) All three digits are significant, so 7.68 g has 3 significant figures. (b) The first zero is not significant, since it is used only to place the decimal point. The digits 4 and 2 are significant, as is the final zero, so 0.420 kg has 3 significant figures. (c) The first two zeros are not significant, since they are used only to place the decimal point. The digits 7 and 3 are significant, so 0.073 m has 2 significant figures. (d) All three digits are significant, so 7.68 × 105 g has 3 significant figures. (e) The zero is significant, since it comes after the decimal point. The digits 4 and 2 are significant as well, so 4.20 × 103 kg has 3 significant figures. (f) Both 7 and 3 are significant, so 7.3 × 10−2 m has 2 significant figures. (g) Both 2 and 3 are significant. The two zeros are significant as well, since they come after the decimal point, so 2.300 × 104 s has 4 significant figures. 23. Strategy Divide and give the answer with the number of significant figures determined by the number with the fewest significant figures. Solution Solve the problem. 3.21 m 3.21 m = = 459 m s 7.00 ms 7.00 × 10−3 s 24. Strategy Convert each length to meters. Then, rewrite the numbers so that the power of 10 is the same for each. Finally, add and give the answer with the number of significant figures determined by the less precise of the two numbers. Solution Solve the problem. 3.08 × 10−1 km + 2.00 × 103 cm = 3.08 × 102 m + 2.00 × 101 m = 3.08 × 102 m + 0.200 × 102 m = 3.28 × 102 m 6 College Physics Chapter 1: Introduction 25. Strategy Use the rules for determining significant figures and for writing numbers in scientific notation. Solution (a) 0.00574 kg has three significant figures, 5, 7, and 4. The zeros are not significant, since they are used only to place the decimal point. To write this measurement in scientific notation, we move the decimal point three places to the right and multiply by 10−3. (b) 2 m has one significant figure, 2. This measurement is already written in scientific notation (c) 0.450 × 10−2 m has three significant figures, 4, 5, and the 0 to the right of 5. The zero is significant, since it comes after the decimal point and is not used to place the decimal point. To write this measurement in scientific notation, we move the decimal point one place to the right and multiply by 10−1. (d) 45.0 kg has three significant figures, 4, 5, and 0. The zero is significant, since it comes after the decimal point and is not used to place the decimal point. To write this measurement in scientific notation, we move the decimal point one place to the left and multiply by 101. (e) 10.09 × 104 s has four significant figures, 1, 9, and the two zeros. The zeros are significant, since they are between two significant figures. To write this measurement in scientific notation, we move the decimal point one place to the left and multiply by 101. (f) 0.09500 × 105 mL has four significant figures, 9, 5, and the two zeros to the right of 5. The zeros are significant, since they come after the decimal point and are not used to place the decimal point. To write this measurement in scientific notation, we move the decimal point two places to the right and multiply by 10−2. The results of parts (a) through (f) are shown in the table below. Measurement Significant Figures Scientific Notation (a) 0.00574 kg 3 5.74 × 10−3 kg (b) 2m 1 2m (c) 0.450 × 10−2 m 3 4.50 × 10−3 m (d) 45.0 kg 3 4.50 × 101 kg (e) 10.09 × 104 s 4 1.009 × 105 s (f) 0.09500 × 105 mL 4 9.500 × 103 mL 26. Strategy Convert each length to scientific notation. Solution In scientific notation, the lengths are: (a) 1 µm = 1× 10−6 m, (b) 1000 nm = 1× 103 × 10−9 m = 1× 10−6 m, (c) 100 000 pm = 1× 105 × 10−12 m = 1× 10−7 m, (d) 0.01 cm = 1× 10−2 × 10−2 m = 1× 10−4 m, and (e) 0.000 000 0001 km = 1× 10−10 × 103 m = 1× 10−7 m. From smallest to greatest, we have c= e, a = b, d . 7 Chapter 1: Introduction College Physics 27. Strategy Convert each length to meters and each time to seconds. Recall that 1.0 mi = 1600 m. Solution In scientific notation, we have: (a) 55 mi h × 1600 m mi × 1 h 3600 s = 24 m s , (b) 82 km h × 1 h 3600 s × 1000 m km = 23 m s , (c) 33 m s , (d) 3.0 cm ms × 1 m 100 cm × 1000 ms s = 30 m s , and (e) 1.0 mi min × 1 min 60 s × 1600 m mi = 27 m s. From smallest to greatest, we have b, a, e, d, c . 28. Strategy Recall that 1 kg = 1000 g and 100 cm = 1 m. Solution Convert the density of body fat from g cm3 to kg m3 . 3 0.9 g cm3 × 1 kg 1000 g × (100 cm m ) = 900 kg m3 29. Strategy There are approximately 39.37 inches per meter. Solution Find the thickness of the cell membrane in inches. 7.0 × 10−9 m × 39.37 inches m = 2.8 × 10−7 inches 30. (a) Strategy There are approximately 3.785 liters per gallon and 128 ounces per gallon. Solution Find the number of fluid ounces in the bottle. 128 fl oz 1 gal 1L × × 355 mL × = 12.0 fluid ounces 1 gal 3.785 L 103 mL (b) Strategy From part (a), we have 355 mL = 12.0 fluid ounces. Solution Find the number of milliliters in the drink. 355 mL 16.0 fl oz × = 473 mL 12.0 fl oz 31. Strategy There are approximately 3.281 feet per meter. Solution Convert to meters. (a) 1595.5 ft × (b) 6016 ft × 1m = 4.863 × 102 m 3.281 ft 1m = 1.834 × 103 m 3.281 ft 32. Strategy For (a), convert milliliters to liters; then convert liters to cubic centimeters using the conversion 1 L = 103 cm3. For (b), convert cubic centimeters to cubic meters using the fact that 100 cm = 1 m. Solution Convert each volume. (a) 255 mL × 10−3 L 103 cm × = 255 cm3 1 mL 1L 3 1 m3 ⎛ 1m ⎞ 3 = 2.55 × 10−4 m (b) 255 cm3 × ⎜ ⎟ = 255 cm × 6 3 100 cm ⎝ ⎠ 10 cm 8 College Physics Chapter 1: Introduction 33. Strategy For (a), convert meters per second to miles per hour using the conversion 1 mi/h = 0.4470 m/s. For (b), convert meters per second to centimeters per millisecond using the conversions 1 m = 100 cm and 1 s = 1000 ms. Solution Convert each speed. (a) 80 m s × 1 mi h = 180 mi h 0.4470 m s (b) 80 m s × 102 cm 1s × = 8.0 cm ms 1m 103 ms 34. Strategy There are 0.6214 miles in 1 kilometer. Solution Find the length of the marathon race in miles. 0.6214 mi 42.195 km × = 26.22 mi 1 km 35. Strategy Calculate the change in the exchange rate and divide it by the original price to find the drop. Solution Find the actual drop in the value of the dollar over the first year. 1.27 − 1.45 −0.18 = = −0.12 1.45 1.45 The actual drop is 0.12 or 12% . 36. Strategy There are 1000 watts in one kilowatt and 100 centimeters in one meter. Solution Convert 1.4 kW m 2 to W cm 2 . 2 1.4 kW 1000 W ⎛ 1 m ⎞ 2 × ×⎜ ⎟ = 0.14 W cm 1 kW ⎝ 100 cm ⎠ 1 m2 37. Strategy Convert the radius to centimeters; then use the conversions 1 L = 103 cm3 and 60 s = 1 min. Solution Find the volume rate of blood flow 2 volume rate of blood flow = π r 2 v = π (1.2 cm ) (18 cm s ) × 1L 103 cm3 × 60 s = 4.9 L min 1 min 38. Strategy The distance traveled d is equal to the rate of travel r times the time of travel t. There are 1000 milliseconds in one second. Solution Find the distance the molecule would move. 459 m 1s d = rt = × 7.00 ms × = 3.21 m 1s 1000 ms 39. Strategy There are 1000 meters in a kilometer a...
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