1 Chapter CHAPTER 1 - Introduction, Measurement, Estimating 1. (a) Assuming one significant figure, we have 10 billion yr = 10 109 yr = 1 1010 yr. 10 yr)(3 107 s/yr) = (b) (1 10 3 1017 s. (a) (b) (c) (d) (e) (f) (g) (a) (b) (c) (d) (e) (f) (a) (b) (c) (d) (e) 4 significant figures. Because the zero is not needed for placement, we have 4 significant figures. 3 significant figures. Because the zeros are for placement only, we have 1 significant figure. Because the zeros are for placement only, we have 2 significant figures. 4 significant figures. 2, 3, or 4 significant figures, depending on the significance of the zeros. 1,156 = 1.156 103. 21.8 = 2.18 101. 0.0068 = 6.8 103. 27.635 = 2.7635 101. 0.219 = 2.19 101. 22 = 2.2 101. 8.69 7.1 6.6 8.76 8.62 104 = 103 = 101 = 102 = 105 = 86,900. 7,100. 0.66. 876. 0.000 086 2.
2.
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5. 6.
% uncertainty = [(0.25 m)/(3.26 m)] 100 = 7.7%. Because the uncertainty has 2 significant figures, the % uncertainty has 2 significant figures. We assume an uncertainty of 1 in the last place, i. e., 0.01, so we have % uncertainty = [(0.01 m)/(1.28 m)] 100 = 0.8%. Because the uncertainty has 1 significant figure, the % uncertainty has 1 significant figure. We assume an uncertainty of 0.2 s. (a) % uncertainty = [(0.2 s)/(5 s)] 100 = 4%. Because the uncertainty has 1 significant figure, the % uncertainty has 1 significant figure. (b) % uncertainty = [(0.2 s)/(50 s)] 100 = 0.4%. Because the uncertainty has 1 significant figure, the % uncertainty has 1 significant figure. (c) % uncertainty = [(0.2 s)/(5 min)(60 s/min)] 100 = 0.07%. Because the uncertainty has 1 significant figure, the % uncertainty has 1 significant figure. For multiplication, the number of significant figures in the result is the least number from the multipliers; in this case 2 from the second value. (2.079 102 m)(0.072 101) = 0.15 101 m = 1.5 m. To add, we make all of the exponents the same: 9.2 103 s + 8.3 104 s + 0.008 106 s = 0.92 104 s + 8.3 104 s + 0.8 104 s = 10.02 104 s = 1.0 105 s. Because we are adding, the location of the uncertain figure for the result is the one furthest to the left. In this case, it is fixed by the third value.
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10. We assume an uncertainty of 0.1 104 cm. We compare the area for the specified radius to the area for
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Chapter 1 the extreme radius. A1 = pR12 = p(3.8 104 cm)2 = 4.54 109 cm2 ; A2 = pR22 = p[(3.8 + 0.1) 104 cm]2 = 4.78 109 cm2 , so the uncertainty in the area is ?A =A2 A1 = 0.24 109 cm2 = 0.2 109 cm2. We write the area as A = (4.5 0.2) 109 cm2. We could also treat the change as a differential: dA =2pR dR = 2p(3.8 104 cm)(0.1 104 cm) = 2 108 cm2. 11. We compare the volume with the specified radius to the volume for the extreme radius. V1 = )pR13 = )p(2.86 m)3 = 98.0 m3; V2 = )pR23 = )p(2.86 m + 0.08 m)3 = 106.45 m3, so the uncertainty in the volume is ?V =V2 V1 = 8.5 m3; and the % uncertainty is % uncertainty = [(8.5 m3)/(98.0 m3)] 100 = 9%. 12. (a) (b) (c) (d) (e) 106 volts = 1 megavolt = 1 Mvolt. 106 meters = 1 micrometer = 1 m. 3 days = 6 kilodays = 6 10 6 kdays. 18 102 bucks = 18 hectobucks = 1.8 kbucks. 8 109 pieces = 8 nanopieces = 8 npieces.
13. (a) 286.6 mm = 286.6 103 m = 0.286 6 m. (b) 85 V = 85 106 V = 0.000 085 V. (c) 760 mg = 760 103 g = 760 106 g = 0.000 760 kg. This assumes that the last zero is significant. (d) 60.0 picoseconds = 60.0 1012 s = 0.000 000 000 060 0 s. (e) 22.5 femtometers = 22.5 1015 m = 0.000 000 000 000 022 5 m. (f) 2.50 gigavolts = 2.50 109 volts = 2,500,000,000 volts. 14. 50 hectocars = 50 102 cars = 5,000 cars. 1 megabuck/yr = 1 106 bucks/yr = 1,000,000 bucks/yr 15. If we assume a height of 5 ft 10 in, we have height = 5 ft 10 in = 70 in = (70 in)[(1 m)/(39.37 in)] = (millionaire). 1.8 m. 1.5 1011 m.
16. (a) 93 million mi = 93 106 mi = (93 106 mi)[(1610 m)/(1 mi)] = (b) 1.5 1011 m = 150 109 m = 150 Gm. 17. (a) 1 ft2 = (1 ft2)[(1 yd)/(3 ft)]2 = 0.111 yd2. 2 = (1 m2)[(3.28 ft)/(1 m)]2 = (b) 1 m 10.76 ft2.
18. We find the time from time = distance/speed = [(1.0 mi)(1.61 km/mi)]/[2300 km/h)/(3600 s/h)] =
2.5 s.
19. (a) 1.0 1010 m = (1.0 1010 m)[(1 in)/(2.54 cm)][(100 cm)/(1 m)] = 3.9 109 in. (b) We let the units lead us to the answer: (1.0 cm)[(1 m)/(100 cm)][(1 atom)/(1.0 1010 m)] = 1.0 108 atoms. 20. To add, we make all of the units the same: 2.00 m + 142.5 cm + 7.24 105 m = 2.00 m + 1.425 m + 0.724 m = 4.149 m = 4.15 m. Because we are adding, the location of the uncertain figure for the result is the one furthest to the left. In this case, it is fixed by the first value.
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Chapter 1 21. (a) 1 km/h = (1 km/h)[(0.621 mi)/(1 km)] = 0.621 mi/h. (b) 1 m/s = (1 m/s)[(1 ft)/(0.305 m)] = 3.28 ft/s. (c) 1 km/h = (1 km/h)[(1000 m)/(1 km)][(1 h)/(3600 s)] = 0.278 m/s. A useful alternative is 1 km/h = (1 km/h)[(1 h)/(3.600 ks)] = 0.278 m/s. 22. For the length of a one-mile race in m, we have 1 mi = (1 mi)[(1610 m)/(1 mi)] = 1610 m; so the difference is 110 m. The % difference is % difference = [(110 m)/(1500 m)] 100 = 7.3%. 23. (a) 1.00 ly = (2.998 108 m/s)(1.00 yr)[(365.25 days)/(1 yr)][(24 h)/(1 day)][(3600 s)/(1 h)] = 9.46 1015 m. (b) 1.00 ly = (9.46 1015 m)[(1 AU)/(1.50 108 km)][(1 km)/(1000 m)] = 6.31 104 AU. 8 m/s)[(1 AU)/(1.50 108 km)][(1 km)/(1000 m)][(3600 s)/(1 h)] (c) speed of light = (2.998 10 = 7.20 AU/h.
24. For the surface area of a sphere, we have AMoon = 4prMoon2 = 4p[! (3.48 106 m)]2 = 3.80 1013 m2. We compare this to the area of the earth by finding the ratio: AMoon /AEarth = 4prMoon2 /4prEarth2 = (rMoon/rEarth)2 = [(1.74 103 km)/(6.38 103 km)]2 = 7.44 x 102. Thus we have AMoon = 7.44 x 102 AEarth. 25. (a) (b) (c) (d) 2800 = 2.8 103 ~ 1 103 = 103. 2 ~ 100 102 = 4. 86.30 10 10 0.0076 = 0.76 102 ~ 102. 15.0 108 = 1.50 109 ~ 109.
26. We assume that a good runner can run 6 mi/h (equivalent to a 10-min mile) for 5 h/day. Using 3000 mi for the distance across the United States, we have time = (3000 mi)/(6 mi/h)(5 h/day) ~ 100 days. 27. We assume a rectangular house 40 ft 30 ft, 8 ft high; so the total wall area is Atotal = [2(40 ft) + 2(30 ft)](8 ft) ~ 1000 ft2. If we assume there are 12 windows with dimensions 3 ft 5 ft, the window area is Awindow = 12(3 ft)(5 ft) ~ 200 ft2. Thus we have % window area = [Awindow/Atotal ](100) = [(200 ft2)/(1000 ft2)](100) ~ 20%. 28. If we take an average lifetime to be 70 years and the average pulse to be 60 beats/min, we have N = (60 beats/min)(70 yr)(365 day/yr)(24 h/day)(60 min/h) ~ 2 109 beats. 29. If we approximate the body as a box with dimensions 6 ft, 1 ft, and 8 in, we have volume = (72 in)(12 in)(8 in)(2.54 cm/in)3 ~ 1 105 cm3.
30. We assume the distance from Beijing to Paris is 10,000 mi.
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Chapter 1 (a) If we assume that today's race car can travel for an extended period at an average speed of 40 mi/h, we have time = [(10,000 mi)/(40 mi/h)](1 day/24 h) ~ 10 days. (b) If we assume that in 1906 a race car could travel for an extended period at an average speed of 5 mi/h, we have time = [(10,000 mi)/(5 day/24 mi/h)](1 h) ~ 80 days. 31. We assume that each dentist sees 10 patients/day, 5 days/wk for 48 wk/yr, for a total number of visits of Nvisits = (10 visits/day)(5 days/wk)(48 wk/yr) ~ 2400 visits/yr. We assume that each person sees a dentist 2 times/yr. (a) We assume that the population of San Francisco is 700,000. We let the units lead us to the answer: N = (700,000)(2 visits/yr)/(2400 visits/yr) ~ 600 dentists. (b) Left to the reader to estimate the population. 32. If we assume that the person can mow at a speed of 1 m/s and the width of the mower cut is 0.5 m, the rate at which the field is mown is (1 m/s)(0.5 m) = 0.5 m 2/s. If we take the dimensions of the field to be 110 m by 50 m, we have time = [(110 m)(50 m)/(0.5 m2/s)]/(3600 s/h) ~ 3 h. 33. We assume an average time of 3 yr for the tire to wear d = 1 cm and the tire has a radius r = 30 cm and a width w = 10 cm. Thus the volume of rubber lost by a tire in 3 yr is V = wd2pr. If we assume there are 100 million vehicles, each with 4 tires, we have m = (100 106 vehicles)(4 tires/vehicle)(0.1 m)(0.01 m)2p(0.3 m/tire)(1200 kg/m3)/(3 yr) ~ 3 108 kg/yr. 34. We assume that one-third of the floor space can be used for shelves and there are five shelves, each with a depth of 25 cm, in a stack. If the average book has a width of 4.0 cm, we have N = [@(1500 m2)(5 shelves)/(4.0 cm/book)(25 cm/shelf)](100 cm/m)2 = 2.5 105 books. 35. For the right triangle shown on the diagram we have r2 + d2 = (r + h)2; d2 = 2rh + h2 = 2(6.4 106 m)(200 m) + (200 m)2, which gives d = 5.1 104 m = 51 km.
36. Because every term must have the same dimensions, we have v = At3 Bt; [L/T] = [A][T]3 [B][T]. Thus we get [A] = [L/T]/[T3] = [L/T4], and [B] = [L/T]/[T] = [L/T2]. 37. For the units we have A = [L/T4] = m/s4, and B = [L/T2] = m/s2.
38. We test to see if each term has the same dimensions.
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Chapter 1 (a) x = vt2 + 2at; [L] =? [L/T][T]2 + [L/T2][T]; [L] ? [LT] + [L/T]; therefore, this is not correct. 2; (b) x = v0t + ! at [L] =? [L/T][T] + [L/T2][T]2; [L] = [L] + [L]; therefore, this may be correct. (c) x = v0t + 2at2; [L] =? [L/T][T] + [L/T2][T]2; [L] = [L] + [L]; therefore, this may be correct. 39. (a) (b) (c) (d) 1.0 = (1.0 1010 m)/(109 m/nm) = 0.10 nm. 1.0 = (1.0 1010 m)/(1015 m/fm) = 1.0 105 fm. 1.0 m = (1.0 m)/(1010 m/) = 1.0 1010 . From the result for Problem 23, we have 1.0 ly = (9.5 1015 m)/(1010 m/) = 9.5 1025 .
40. We use the values for the masses from Table 13. (a) Nbacterium = (1015 kg)/(1027 kg/proton) = 1012 protons (or neutrons). 17 kg)/(1027 kg/proton) = 10 protons (or neutrons). (b) NDNA = (10 10 2 kg)/(1027 kg/proton) = (c) Nhuman = (10 1029 protons (or neutrons). 41 kg)/(1027 kg/proton) = (d) Ngalaxy = (10 1068 protons (or neutrons). 41. (a) 1.00 yr = (365.25 days)(24 h/day)(3600 s/h) = 3.16 107 s (b) 1.00 yr = (3.16 107 s)/(109 s/ns) = 3.16 1016 ns. (c) 1.00 s = (1.00 s)/(3.16 107 s/yr) = 3.17 108 yr. 42. 1 hectare = (104 m2)[(3.28 ft)/(1 m)]2[(1 acre)/(4 104 ft2)] = ~ p 107 s.
2.69 acres.
43. (a) The maximum number of buses is needed during rush hour. If we assume that at any time there are 40,000 persons commuting by bus and each bus has 30 passengers, we have N = (40,000 commuters)/(30 passengers/bus) ~ 1000 buses ~ 1,000 drivers. (b) Left to the reader. 44. If we ignore any loss of material from the slicing, we find the number of wafers from (30 cm)(10 mm/cm)/(0.60 mm/wafer) = 500 wafers. For the maximum number of chips, we have (500 wafers)(100 chips/wafer) = 50,000 chips. 45. If we assume there is 1 automobile for 2 persons and a U. S. population of 250 million, we have 125 million automobiles. We estimate that each automobile travels 15,000 miles in a year and averages 20 mi/gal. Thus we have N = (125 106 automobiles)(15,000 mi/yr/automobile)/(20 mi/gal) ~ 1 1011 gal/yr. 46. We let D represent the diameter of a gumball. Because there are air gaps around the gumballs, we estimate the volume occupied by a gumball as a cube with volume D3 . The machine has a square crosssection with sides equivalent to 10 gumballs and is about 14 gumballs high, so we have N = volume of machine/volume of gumball = (14D)(10D)2/D3 ~ 1.4 103 gumballs.
47. We will convert all units to meters. The volume used in one year is
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Chapter 1
V = [(40,000 persons)/(4 persons/family)](1200 L/family day) (365 days/yr)(103 m3/L) = 4.4 106 m3/yr. If we let d represent the loss in depth, we have d = V/area = (4.4 106 m3/yr)/(50 km2)(103 m/km)2 ~ 0.09 m ~ 9 cm/yr. 48. For the volume of a 1-ton rock, we have V = (2000 lb)/(3)(62 lb/ft3) ~ 11 ft3. If we assume the rock is a sphere, we find the radius from 11 ft3 = )pr3, which gives r ~ 1.4 ft, so the diameter would be
~ 3 ft ~ 1 m. 4 105 t.
49. We find the amount of water from its volume: m = (5 km)(8 km)(1.0 cm)(105 cm/km)2(103 kg/cm3)/(103 kg/t) = 40 104 t ~ 50. We will use a pencil with a diameter of 5 mm and assume that it is held 0.5 m from the eye. Because the triangles AOD and BOC are similar, we can equate the ratio of distances: BC/AD = OQ/OP; BC/(0.005 m) = (3.8 105 km)/(0.5 m), which gives BC ~ 4 103 km.
51. We assume that we can walk an average of 15 miles a day. If we ignore the impossibility of walking on water and travel around the equator, the time required is time = 2prEarth/speed = 2p(6 103 km)(0.621 mi/km)/(15 mi/day)(365 days/yr) ~ 4 yr. 52. If we use 0.5 m for the cubit, for the dimensions we have 150 m long, 25 m wide, and 15 m high. 53. From the diagram we see that d/L = tan ; (120 yd)/L = tan 30, which gives L = 2.1 102 yd = (2.1 102 yd)(3 ft/yd)/(3.28 ft/m) =
1.9 102 m.
54. We assume the oil slick is circular with diameter D and thickness d. For the volume we have V = d(pr2) = #dpD2; (1 L)(1000 cm3/L)/(1 102 cm/m)3 = #(2 1010 m)pD2, which gives D ~ 3 103 m.
55. The trigonometric function is not a linear function of the angle, so we can find the uncertainty in the sine by calculating two values.
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Chapter 1 (a) Percent uncertainty in = (0.5/15.0) 100 = 3%. We find the percent uncertainty in the sine from sin 15.0 = 0.2588; sin 15.5 = 0.2672; Percent uncertainty in sin = [(0.2672 0.2588)/0.2588] 100 = 3%. (b) Percent uncertainty in = (0.5/75.0) 100 = 0.7%. We find the percent uncertainty in the sine from sin 75.0 = 0.9659; sin 75.5 = 0.9681; Percent uncertainty in sin = [(0.9681 0.9659)/0.9659] 100 = 0.2%. Note that it is possible to approximate uncertainties with differential quantities, if the angle is not very small (i. e., sin is not small). We have d(sin ) = cos d ; d(sin )/(sin ) = d /(tan ). The angle must be in radians, so we get d(sin )/(sin ) = [(0.5)(p/180)]/tan 15.0 = 0.03 = 3%; d(sin )/(sin ) = [(0.5)(p/180)]/tan 75.0 = 0.002 = 0.2%. 56. Because you are lying on the sand, your line of sight is tangent to the surface of the Earth. If the height of the hull is h, from the triangle on the diagram we have r2 + d2 = (r + h)2; d2 = 2rh + h2 = 2(6.38 106 m)(2.5 m) + (2.5 m)2, which gives d = 5.6 103 m = 5.6 km.
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