Exam (elaborations) INSTRUCTOR’S SOLUTIONS MANUAL FOR_SERWAY AND VUILLE’S_COLLEGE PHYSICS NINTH EDITION, VOLUME 1

Using a calculator to multiply the length by the width gives a raw answer of 6783 m2, but this answer must be rounded to contain the same number of signifi cant fi gures as the least accurate factor in the product. The least accurate factor is the length, which contains either 2 or 3 signifi cant fi gures, depending on whether the trailing zero is signifi cant or is being used only to locate the decimal point. Assuming the length contains 3 signifi cant fi gures, answer (c) correctly expresses the area as 6.78 ×103 m2 . However, if the length contains only 2 signifi cant fi gures, answer (d) gives the correct result as 6.8 ×103 m2 . 2. Both answers (d) and (e) could be physically meaningful. Answers (a), (b), and (c) must be meaningless since quantities can be added or subtracted only if they have the same dimensions. 3. According to Newton’s second law, Force = mass×acceleration. Thus, the units of Force must be the product of the units of mass (kg) and the units of acceleration (m s2 ). This yields kg⋅m s2, which is answer (a). 4. The calculator gives an answer of 57.573 for the sum of the 4 given numbers. However, this sum must be rounded to 58 as given in answer (d) so the number of decimal places in the result is the same (zero) as the number of decimal places in the integer 15 (the term in the sum containing the smallest number of decimal places). 5. The required conversion is given by: h =( )⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 00 1 000 1 00 . . m mm 1.00 m cubitus 445 mm ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 4.49 cubiti This result corresponds to answer (c). 6. The given area (1 420 ft2 ) contains 3 signifi cant fi gures, assuming that the trailing zero is used only to locate the decimal point. The conversion of this value to square meters is given by: A = ( × )⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 42 103 = 1 32 ×10 2 . ft . 2 1.00 m 3.281 ft 2 m2 = 132 m2 Note that the result contains 3 signifi cant fi gures, the same as the number of signifi cant fi gures in the least accurate factor used in the calculation. This result matches answer (b). 7. You cannot add, subtract, or equate a number apples and a number of days. Thus, the answer is yes for (a), (c), and (e). However, you can multiply or divide a number of apples and a number of days. For example, you might divide the number of apples by a number of days to fi nd the number of apples you could eat per day. In summary, the answers are (a) yes, (b) no, (c) yes, (d) no, and (e) yes. 1 2 Chapter 1 8. The given Cartesian coordinates are x = −5.00, and y = 12.00, with the least accurate containing 3 signifi cant fi gures. Note that the specifi ed point (with x 0 and y 0) is in the second quadrant. The conversion to polar coordinates is then given by: r = x2 + y2 = (− )2 + ( )2 = 5.00 12.00 13.0 tan . . θ = = . θ . − y = − = − (− x 12 00 5 00 2 40 and tan 1 2 40) = −67.3°+180° = 113° Note that 180° was added in the last step to yield a second quadrant angle. The correct answer is therefore (b) (13.0, 113°). 9. Doing dimensional analysis on the fi rst 4 given choices yields: (a) [v] ⎡⎣ ⎤⎦ = = t 2 2 3 L T T L T (b) [v] ⎡⎣ ⎤⎦ = = − − x2 2 L T 1 1 L L T (c) v2 2 2 2 3 L T T L T T L T ⎡⎣ ⎤⎦ [ ] =( )= = t 2 (d) v2 2 2 2 L T L L T L L T ⎡⎣ ⎤⎦ [ ] =( )= = x 2 Since acceleration has units of length divided by time squared, it is seen that the relation given in answer (d) is consistent with an expression yielding a value for acceleration. 10. The number of gallons of gasoline she can purchase is # gallons total expenditure cost per gallon E = ≈ 33 uros Euros L L quart quarts 1 5 1 1 4 . ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ 1 5 gal gal ⎛ ⎝ ⎜⎜ ⎞ ⎠ ⎟⎟ ≈ so the correct answer is (b). 11. The situation described is shown in the drawing at the right. From this, observe that tan 26 45 ° = h m , or h = (45 m) tan 26° = 22 m Thus, the correct answer is (a). 12. Note that we may write 1.365 248 0 ×107 as 136.524 80 ×105. Thus, the raw answer, including the uncertainty, is x = (136.524 80 ± 2) ×105. Since the fi nal answer should contain all the digits we are sure of and one estimated digit, this result should be rounded and displayed as 137 ×105 = 1.37 ×107 (we are sure of the 1 and the 3, but have uncertainty about the 7). We see that this answer has three signifi cant fi gures and choice (d) is correct. ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 2. Atomic clocks are based on the electromagnetic waves that atoms emit. Also, pulsars are highly regular astronomical clocks. 45 m h 26 Introduction 3 4. (a) ~0.5 lb ≈ 0.25 kg or ~10 −1 kg (b) ~4 lb ≈ 2 kg or ~100 kg (c) ~4000 lb ≈ 2000 kg or ~10 3 kg 6. Let us assume the atoms are solid spheres of diameter 10−10 m. Then, the volume of each atom is of the order of 10−30 m3. (More precisely, volume = 4πr3 3 = πd3 6.) Therefore, since 1 cm3 = 10−6 m3, the number of atoms in the 1 cm3 solid is on the order of 10−6 10−30 = 1024 atoms. A more precise calculation would require knowledge of the density of the solid and the mass of each atom. However, our estimate agrees with the more precise calculation to within a factor of 10. 8. Realistically, the only lengths you might be able to verify are the length of a football fi eld and the length of a housefl y. The only time intervals subject to verifi cation would be the length of a day and the time between normal heartbeats. 10. In the metric system, units differ by powers of ten, so it’s very easy and accurate to convert from one unit to another. ANSWERS TO EVEN NUMBERED PROBLEMS 2. (a) L T2 (b) L 4. All three equations are dimensionally incorrect. 6. (a) kg⋅m s (b) Ft = p 8. (a) 22.6 (b) 22.7 (c) 22.6 is more reliable 10. (a) 3.00 ×108 m s (b) 2.997 9 ×108 m s (c) 2.997 925×108 m s 12. (a) 346 m2 ±13 m2 (b) 66.0 m ±1.3 m 14. (a) 797 (b) 1.1 (c) 17.66 16. 3.09 cm s 18. (a) 5.60 ×102 km = 5.60 ×105 m = 5.60 ×107 cm (b) 0.491 2 km = 491.2m = 4.912 ×104 cm (c) 6.192 km = 6.192 ×103 m = 6.192 ×105 cm (d) 2.499 km = 2.499 ×103 m = 2.499 ×105 cm 20. 10 6. kmL 22. 9 2. nms 24. 2.9×102 m3 = 2.9×108 cm3 26. 2.57×106 m3 4 Chapter 1 28. ∼ 108 steps 30. ~108 people with colds on any given day 32. (a) 4.2 ×10−18 m3 (b) ~10−1 m3 (c) ~1016 cells 34. (a) ∼ 1029 prokaryotes (b) ~1014 kg (c) The very large mass of prokaryotes implies they are important to the biosphere. They are responsible for fi xing carbon, producing oxygen, and breaking up pollutants, among many other biological roles. Humans depend on them! 36. 2.2 m 38. 8.1 cm 40. Δs = r + r − r r ( − ) 1 2 2 2 1 2 1 2 2 cosθ θ 42. 2.33 m 44. (a) 1.50 m (b) 2.60 m 46. 8.60 m 48. (a) and (b) (c) y x = tan12.0°, y (x −1.00 km) = tan14.0° (d) 1.44×103 m 50. y d = ⋅ ⋅ − tan tan tan tan θ φ φ θ 52. (a) 1.609 km h (b) 88 km h (c) 16 km h 54. Assumes population of 300 million, average of 1 can week per person, and 0.5 oz per can. (a) ∼ 1010 cans yr (b) ∼ 105 tons yr 56. (a) 7.14×10−2 gal s (b) 2.70×10−4 m3 s (c) 1.03 h 58. (a) A A 2 1 = 4 (b) V V 2 1 = 8 60. (a) ∼ 102 yr (b) ∼ 104 times 62. ∼ 104 balls yr. Assumes 1 lost ball per hitter, 10 hitters per inning, 9 innings per game, and 81 games per year. Introduction 5 PROBLEM SOLUTIONS 1.1 Substituting dimensions into the given equation T = 2π