1-1 UNITS, PHYSICAL QUANTITIES AND VECTORS 1.1. I DENTIFY : Convert units from mi to km and from km to ft. SET UP: 1 in. 2.54 cm = , 1 km = 1000 m , 12 in. 1 ft = , 1 mi = 5280 ft . EXECUTE : (a) 235280 ft 12 in. 2.54 cm 1 m 1 km1.00 mi (1.00 mi) 1.61 km1 mi 1 ft 1 in. 10 cm 10 m⎛⎞ ⎛ ⎞ ⎛ ⎞ ⎛⎞ ⎛ ⎞== ⎜⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎟ ⎜ ⎟⎝⎠ ⎝ ⎠ ⎝ ⎠ ⎝⎠ ⎝ ⎠ (b) 32
3 10 m 10 cm 1 in. 1 ft1.00 km (1.00 km) 3.28 10 ft1 km 1 m 2.54 cm 12 in.⎛⎞ ⎛ ⎞ ⎛⎞ ⎛ ⎞== × ⎜⎟ ⎜ ⎟ ⎜⎟ ⎜ ⎟⎝⎠ ⎝ ⎠ ⎝⎠ ⎝ ⎠ EVALUATE : A mile is a greater distance than a kilometer. There are 5280 ft in a mile but only 3280 ft in a km. 1.2. IDENTIFY : Convert volume units from L to 3in.. SET UP: 31 L 1000 cm= . 1 in. 2.54 cm = EXECUTE : 3 3
3 1000 cm 1 in.0.473 L 28.9 in. .1 L 2.54 cm⎛⎞ ⎛⎞××=⎜⎟ ⎜⎟⎝⎠ ⎝⎠ EVALUATE : 31 in. is greater than 31 cm , so the volume in 3in. is a smaller number than the volume in 3cm , which is 3473 cm . 1.3. IDENTIFY : We know the speed of light in m/s. / td v= . Convert 1.00 ft to m and t from s to ns. SET UP: The speed of light is 83.00 10 m/sv=× . 1 ft 0.3048 m = . 91 s 10 ns= . EXECUTE : 9
80.3048 m1.02 10 s 1.02 ns3.00 10 m/st−== × =× EVALUATE : In 1.00 s light travels 85 53.00 10 m 3.00 10 km 1.86 10 mi ×= × = × . 1.4. IDENTIFY : Convert the units from g to kg and from 3cm to 3m. SET UP: 1 kg 1000 g = . 1 m 1000 cm = . EXECUTE : 3
4
33g 1 kg 100 cm kg11.3 1.13 10cm 1000 g 1 m m⎛⎞ ⎛ ⎞×× = ×⎜⎟ ⎜ ⎟
⎝⎠ ⎝ ⎠ EVALUATE : The ratio that converts cm to m is cubed, because we need to convert 3cm to 3m. 1.5. IDENTIFY : Convert volume units from 3in. to L. SET UP: 31 L 1000 cm= . 1 in. 2.54 cm = . EXECUTE : ( )() ( ) 3 33327 in. 2.54 cm in. 1 L 1000 cm 5.36 L ×× = EVALUATE : The volume is 35360 cm . 31 cm is less than 31 in. , so the volume in 3cm is a larger number than the volume in 3in. . 1.6. IDENTIFY : Convert 2ft to 2m and then to hectares. SET UP: 421.00 hectare 1.00 10 m =× . 1 ft 0.3048 m = . EXECUTE : The area is 2 2
4243,600 ft 0.3048 m 1.00 hectare(12.0 acres) 4.86 hectares1 acre 1.00 ft 1.00 10 m⎛⎞ ⎛⎞ ⎛ ⎞= ⎜⎟ ⎜⎟ ⎜ ⎟× ⎝⎠ ⎝ ⎠ ⎝⎠. EVALUATE : Since 1 ft 0.3048 m = , 22 21 ft (0.3048) m = . 1.7. IDENTIFY : Convert seconds to years. SET UP: 91 billion seconds 1 10 s =× . 1 day 24 h = . 1 h 3600 s = . EXECUTE : ()9 1 h 1 day 1 y1.00 billion seconds 1.00 10 s 31.7 y3600 s 24 h 365 days⎛⎞ ⎛⎞ ⎛ ⎞=× = ⎜⎟ ⎜⎟ ⎜ ⎟⎝⎠ ⎝ ⎠ ⎝⎠. 1 1-2 Chapter 1 EVALUATE : The conversion 71 y 3.156 10 s=× assumes 1 y 365.24 d = , which is the average for one extra day every four years, in leap years. The probl em says instead to assume a 365-day year. 1.8. IDENTIFY : Apply the given conversion factors. SET UP: 1 furlong 0.1250 mi and 1 fortnight 14 days. == 1 day 24 h. = EXECUTE : ()0.125 mi 1 fortnight 1 day180,000 furlongs fortnight 67 mi/h1 furlong 14 days 24 h⎛⎞ ⎛ ⎞ ⎛ ⎞= ⎜⎟ ⎜⎟ ⎜ ⎟⎝⎠ ⎝⎠ ⎝ ⎠ EVALUATE : A furlong is less than a mile and a fortnight is many hours, so the speed limit in mph is a much smaller number. 1.9. IDENTIFY : Convert miles/gallon to km/L. SET UP: 1 mi 1.609 km = . 1 gallon 3.788 L. = EXECUTE : (a) 1.609 km 1 gallon55.0 miles/gallon (55.0 miles/gallon) 23.4 km/L1 mi 3.788 L⎛⎞ ⎛ ⎞== ⎜⎟ ⎜ ⎟⎝⎠ ⎝ ⎠. (b) The volume of gas required is 1500 km64.1 L23.4 km/L= . 64.1 L1.4 tanks45 L/tank= . EVALUATE : 1 mi/gal 0.425 km/L = . A km is very roughly half a mile and there are roughly 4 liters in a gallon, so 2
41 mi/gal km/L ∼ , which is roughly our result. 1.10. IDENTIFY : Convert units. SET UP: Use the unit conversions given in the problem. Also, 100 cm 1 m = and 1000 g 1 kg = . EXECUTE : (a) mi 1h 5280 ft ft60 88h 3600s 1mi s⎛⎞ ⎛ ⎞ ⎛⎞= ⎜⎟ ⎜ ⎟ ⎜⎟⎝⎠ ⎝⎠ ⎝ ⎠ (b) 22ft 30.48cm 1 m m32 9.8s 1ft 100 cm s⎛⎞ ⎛⎞ ⎛ ⎞= ⎜⎟ ⎜⎟ ⎜ ⎟⎝⎠ ⎝ ⎠ ⎝⎠ (c) 3
3
33g 100 cm 1 kg kg1.0 10cm 1 m 1000 g m⎛⎞ ⎛⎞ ⎛⎞= ⎜⎟ ⎜⎟ ⎜⎟⎝⎠ ⎝⎠ ⎝⎠ EVALUATE : The relations 60 mi/h 88 ft/s = and 33 31 g/cm 10 kg/m = are exact. The relation 2232 ft/s 9.8 m/s = is accurate to only two significant figures. 1.11. IDENTIFY : We know the density and mass; thus we can find the volume using the relation density mass/volume / mV == . The radius is then found from the volume equation for a sphere and the result for the volume. SET UP: 3Density 19.5 g/cm = and critical 60.0 kg. m = For a sphere 3 4
3Vr π= . EXECUTE : 3
critical 360.0 kg 1000 g/ density 3080 cm19.5 g/cm 1.0 kgVm⎛⎞ ⎛ ⎞== =⎜⎟ ⎜ ⎟⎝⎠ ⎝ ⎠. ()333333080 cm 9.0 cm44Vrππ== = . EVALUATE : The density is very large, so th e 130 pound sphere is small in size. 1.12. IDENTIFY : Use your calculator to display 710π× . Compare that number to the number of seconds in a year. SET UP: 1 yr 365.24 days, = 1 day 24 h, = and 1 h 3600 s. = EXECUTE : 7 24 h 3600 s(365.24 days/1 yr) 3.15567... 10 s1 day 1 h⎛⎞ ⎛⎞=× ⎜⎟ ⎜⎟⎝⎠ ⎝⎠; 7710 s 3.14159... 10 sπ×= × The approximate expression is accurate to two significant figures. EVALUATE : The close agreement is a numerical accident. 1.13. IDENTIFY : The percent error is the error divided by the quantity. SET UP: The distance from Berlin to Paris is given to the nearest 10 km. EXECUTE : (a) 3
310 m1 . 11 0% .890 10 m−=×× (b) Since the distance was given as 890 km, the total distance should be 890,000 meters. We know the total distance to only three significant figures. EVALUATE : In this case a very small percentage error has disastrous consequences. 1.14. IDENTIFY : When numbers are multiplied or divided, the numbe r of significant figures in the result can be no greater than in the factor with the fewest significant figur es. When we add or subtract numbers it is the location of the decimal that matters. Units, Physical Quantities and Vectors 1-3 SET UP: 12 mm has two significant figures and 5.98 mm has three significant figures. EXECUTE : (a) () ( )212 mm 5.98 mm 72 mm ×= (two significant figures) (b) 5.98 mm0.5012 mm= (also two significant figures) (c) 36 mm (to the nearest millimeter) (d) 6 mm (e) 2.0 (two significant figures) EVALUATE : The length of the rectangle is known only to the nearest mm, so the answers in parts (c) and (d) are known only to the nearest mm. 1.15. IDENTIFY and SET UP: In each case, estimate the precision of the measurement. EXECUTE : (a) If a meter stick can measure to the neares t millimeter, the error will be about 0.13%. (b) If the chemical balance can measure to the nearest milligram, the error will be about 38.3 10 %.−× (c) If a handheld stopwatch (as opposed to electric timing devices ) can measure to the nearest tenth of a second, the error will be about 22.8 10 %.−× EVALUATE : The percent errors are those due only to the limit of precision of the measurement. 1.16. IDENTIFY : Use the extreme values in the pieces length and width to find the uncertainty in the area. SET UP: The length could be as large as 5.11 cm a nd the width could be as large as 1.91 cm. EXECUTE : The area is 9.69 ± 0.07 cm2. The fractional uncertainty in the area is 2
20.07 cm0.72%,9.69 cm= and the fractional uncertainties in the length and width are 0.01 cm0.20%5.10 cm= and 0.01 cm0.53%.1.9 cm= The sum of these fractional uncertainties is 0.20% 0.53% 0.73% += , in agreement with the fractional uncertainty in the area. EVALUATE : The fractional uncertainty in a product of numbers is greater than the fractional uncertainty in any of the individual numbers. 1.17. IDENTIFY : Calculate the average volume and diameter and the uncertainty in these quantities. SET UP: Using the extreme values of the input data gives us th e largest and smallest values of the target variables and from these we get the uncertainty. EXECUTE : (a) The volume of a disk of diameter d and thickness t is 2(/ 2 ). Vd tπ= The average volume is 23(8.50 cm/2) (0.50 cm) 2.837 cm . Vπ== But t is given to only two significant figures so the answer should be expressed to two significant figures: 32.8 cm .V= We can find the uncertainty in the volume as follows. The volume could be as large as 23(8.52 cm/2) (0.055 cm) 3.1 cm , Vπ== which is 30.3 cm larger than the average value. The volume could be as small as 23(8.52 cm/2) (0.045 cm) 2.5 cm , Vπ== which is 30.3 cm smaller than the average value. The uncertainty is 30.3 cm ,± and we express the volume as 32.8 0.3 cm .V=± (b) The ratio of the average diameter to the average thickness is 8.50 cm/0.050 cm 170. = By taking the largest possible value of the diameter and the smallest possible thickness we get the largest possible value for this ratio: 8.52 cm/0.045 cm 190. = The smallest possible value of the ratio is 8.48/ 0.055 150. = Thus the uncertainty is 20± and we write the ratio as 170 20. ± EVALUATE : The thickness is uncertain by 10% and the percentage uncertainty in the diameter is much less, so the percentage uncertainty in the volume and in the ratio should be about 10%. 1.18. IDENTIFY : Estimate the number of people and then use the estimates given in the problem to calculate the number of gallons. SET UP: Estimate 831 0× people, so 821 0× cars. EXECUTE : () ( ) Number of cars miles/car day / mi/gal gallons/day ×= ( )()8 82 10 cars 10000 mi/yr/car 1 yr/365 days / 20 mi/gal 3 10 gal/day×× × = × EVALUATE : The number of gallons of gas used each day a pproximately equals the population of the U.S. 1.19. IDENTIFY : Express 200 kg in pounds. Express each of 200 m, 200 cm and 200 mm in inches. Express 200 months in years. SET UP: A mass of 1 kg is equivalent to a weight of about 2.2 lbs. 1 in. 2.54 cm = . 1 y 12 months = . EXECUTE : (a) 200 kg is a weight of 440 lb. This is much larger than the typical weight of a man. (b) 43 1 in.200 m (2.00 10 cm) 7.9 10 inches2.54 cm⎛⎞=× = × ⎜⎟⎝⎠. This is much greater than the height of a person. (c) 200 cm 2.00 m 79 inches 6.6 ft == = . Some people are this tall, but not an ordinary man. 1-4 Chapter 1 (d) 200 mm 0.200 m 7.9 inches == . This is much too short. (e) 1 y200 months (200 mon) 17 y12 mon⎛⎞== ⎜⎟⎝⎠. This is the age of a teenager; a middle-aged man is much older than this. EVALUATE : None are plausible. When specifying the value of a measured quantity it is essential to give the units in which it is being expressed. 1.20. IDENTIFY : The number of kernels can be calculated as bottle kernel/. NV V= SET UP: Based on an Internet search, Iowan corn farmers use a sieve having a hole size of 0.3125 in. ≅ 8 mm to remove kernel fragments. Therefore estimate the average kernel length as 10 mm, the width as 6 mm and the depth as 3 mm. We must also apply the conversion factors 31 L 1000 cm and 1 cm 10 mm.== EXECUTE : The volume of the kernel is: ( )( )( )3
kernel 10 mm 6 mm 3 mm 180 mm V == . The bottles volume is: () () () () ()33 36 3
bottle 2.0 L 1000 cm 1.0 L 10 mm 1.0 cm 2.0 10 mm V ⎡⎤ ⎡⎤ == × ⎣⎦ ⎣ ⎦ . The number of kernels is then () ( )63 3
kernels bottle kernels / 2.0 10 mm 180 mm 11,000 kernels NV V =≈ × = . EVALUATE : This estimate is highly dependent upon your estimat e of the kernel dimensions. And since these dimensions vary amongst the different available type s of corn, acceptable answers could range from 6,500 to 20,000. 1.21. IDENTIFY : Estimate the number of pages and the number of words per page. SET UP: Assuming the two-volume edition, there are approximately a thousand pages, and each page has between 500 and a thousand words (counting captions and the smaller print, such as the end-of-chapter exercises and problems). EXECUTE : An estimate for the number of words is about 610 . EVALUATE : We can expect that this estimate is accurate to within a factor of 10. 1.22. IDENTIFY : Approximate the number of breaths per minute. Convert minutes to years and 3cm to 3m to find the volume in 3m breathed in a year. SET UP: Assume 10 breaths/min . 5 24 h 60 min1 y (365 d) 5.3 10 min1 d 1 h⎛⎞ ⎛ ⎞== × ⎜⎟ ⎜ ⎟⎝⎠ ⎝ ⎠. 210 cm 1 m = so 63 310 cm 1 m = . The volume of a sphere is 33 41
36Vr d π π == , where r is the radius and d is the diameter. Dont forget to account for four astronauts. EXECUTE : (a) The volume is 5
63 43 5.3 10 min(4)(10 breaths/min)(500 10 m ) 1 10 m / yr1 y− ⎛⎞ ××= × ⎜⎟
⎝⎠. (b) 1/3 1/3 4366 [ 1 1 0 m ]27 mVdππ⎛⎞ × ⎛⎞== = ⎜⎟ ⎜⎟⎝⎠ ⎝⎠ EVALUATE : Our estimate assumes that each 3cm of air is breathed in only once, where in reality not all the oxygen is absorbed from the air in each breath. Therefore, a somewhat smaller volume would actually be required. 1.23. IDENTIFY : Estimate the number of blinks per minute. Convert mi nutes to years. Estimate the typical lifetime in years. SET UP: Estimate that we blink 10 times per minute. 1 y 365 days = . 1 day 24 h = , 1 h 60 min = . Use 80 years for the lifetime. EXECUTE : The number of blinks is 8 60 min 24 h 365 days(10 per min) (80 y/lifetime) 4 101 h 1 day 1 y⎛⎞ ⎛ ⎞ ⎛⎞=× ⎜⎟ ⎜ ⎟ ⎜⎟⎝⎠ ⎝⎠ ⎝ ⎠ EVALUATE : Our estimate of the number of blinks per minute can be off by a factor of two but our calculation is surely accurate to a power of 10. 1.24. IDENTIFY : Estimate the number of beats per minute and the duration of a lifetime. The volume of blood pumped during this interval is then the volume per beat multiplied by the total beats. SET UP: An average middle-aged (40 year-old) adult at rest has a heart rate of roughly 75 beats per minute. To calculate the number of beats in a lifetime, use the current average lifespan of 80 years. EXECUTE : ()9
beats60 min 24 h 365 days 80 yr75 beats/min 3 10 beats/lifespan1 h 1 day yr lifespanN⎛⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞== × ⎜⎟ ⎜⎟ ⎜ ⎟ ⎜ ⎟⎝⎠ ⎝⎠ ⎝ ⎠ ⎝ ⎠ ()9
3 7
blood 31 L 1 gal 3 10 beats50 cm /beat 4 10 gal/lifespan1000 cm 3.788 L lifespanV⎛⎞× ⎛⎞ ⎛ ⎞== × ⎜⎟ ⎜ ⎟ ⎜⎟⎝⎠ ⎝ ⎠ ⎝⎠ EVALUATE : This is a very large volume.
3 10 m 10 cm 1 in. 1 ft1.00 km (1.00 km) 3.28 10 ft1 km 1 m 2.54 cm 12 in.⎛⎞ ⎛ ⎞ ⎛⎞ ⎛ ⎞== × ⎜⎟ ⎜ ⎟ ⎜⎟ ⎜ ⎟⎝⎠ ⎝ ⎠ ⎝⎠ ⎝ ⎠ EVALUATE : A mile is a greater distance than a kilometer. There are 5280 ft in a mile but only 3280 ft in a km. 1.2. IDENTIFY : Convert volume units from L to 3in.. SET UP: 31 L 1000 cm= . 1 in. 2.54 cm = EXECUTE : 3 3
3 1000 cm 1 in.0.473 L 28.9 in. .1 L 2.54 cm⎛⎞ ⎛⎞××=⎜⎟ ⎜⎟⎝⎠ ⎝⎠ EVALUATE : 31 in. is greater than 31 cm , so the volume in 3in. is a smaller number than the volume in 3cm , which is 3473 cm . 1.3. IDENTIFY : We know the speed of light in m/s. / td v= . Convert 1.00 ft to m and t from s to ns. SET UP: The speed of light is 83.00 10 m/sv=× . 1 ft 0.3048 m = . 91 s 10 ns= . EXECUTE : 9
80.3048 m1.02 10 s 1.02 ns3.00 10 m/st−== × =× EVALUATE : In 1.00 s light travels 85 53.00 10 m 3.00 10 km 1.86 10 mi ×= × = × . 1.4. IDENTIFY : Convert the units from g to kg and from 3cm to 3m. SET UP: 1 kg 1000 g = . 1 m 1000 cm = . EXECUTE : 3
4
33g 1 kg 100 cm kg11.3 1.13 10cm 1000 g 1 m m⎛⎞ ⎛ ⎞×× = ×⎜⎟ ⎜ ⎟
⎝⎠ ⎝ ⎠ EVALUATE : The ratio that converts cm to m is cubed, because we need to convert 3cm to 3m. 1.5. IDENTIFY : Convert volume units from 3in. to L. SET UP: 31 L 1000 cm= . 1 in. 2.54 cm = . EXECUTE : ( )() ( ) 3 33327 in. 2.54 cm in. 1 L 1000 cm 5.36 L ×× = EVALUATE : The volume is 35360 cm . 31 cm is less than 31 in. , so the volume in 3cm is a larger number than the volume in 3in. . 1.6. IDENTIFY : Convert 2ft to 2m and then to hectares. SET UP: 421.00 hectare 1.00 10 m =× . 1 ft 0.3048 m = . EXECUTE : The area is 2 2
4243,600 ft 0.3048 m 1.00 hectare(12.0 acres) 4.86 hectares1 acre 1.00 ft 1.00 10 m⎛⎞ ⎛⎞ ⎛ ⎞= ⎜⎟ ⎜⎟ ⎜ ⎟× ⎝⎠ ⎝ ⎠ ⎝⎠. EVALUATE : Since 1 ft 0.3048 m = , 22 21 ft (0.3048) m = . 1.7. IDENTIFY : Convert seconds to years. SET UP: 91 billion seconds 1 10 s =× . 1 day 24 h = . 1 h 3600 s = . EXECUTE : ()9 1 h 1 day 1 y1.00 billion seconds 1.00 10 s 31.7 y3600 s 24 h 365 days⎛⎞ ⎛⎞ ⎛ ⎞=× = ⎜⎟ ⎜⎟ ⎜ ⎟⎝⎠ ⎝ ⎠ ⎝⎠. 1 1-2 Chapter 1 EVALUATE : The conversion 71 y 3.156 10 s=× assumes 1 y 365.24 d = , which is the average for one extra day every four years, in leap years. The probl em says instead to assume a 365-day year. 1.8. IDENTIFY : Apply the given conversion factors. SET UP: 1 furlong 0.1250 mi and 1 fortnight 14 days. == 1 day 24 h. = EXECUTE : ()0.125 mi 1 fortnight 1 day180,000 furlongs fortnight 67 mi/h1 furlong 14 days 24 h⎛⎞ ⎛ ⎞ ⎛ ⎞= ⎜⎟ ⎜⎟ ⎜ ⎟⎝⎠ ⎝⎠ ⎝ ⎠ EVALUATE : A furlong is less than a mile and a fortnight is many hours, so the speed limit in mph is a much smaller number. 1.9. IDENTIFY : Convert miles/gallon to km/L. SET UP: 1 mi 1.609 km = . 1 gallon 3.788 L. = EXECUTE : (a) 1.609 km 1 gallon55.0 miles/gallon (55.0 miles/gallon) 23.4 km/L1 mi 3.788 L⎛⎞ ⎛ ⎞== ⎜⎟ ⎜ ⎟⎝⎠ ⎝ ⎠. (b) The volume of gas required is 1500 km64.1 L23.4 km/L= . 64.1 L1.4 tanks45 L/tank= . EVALUATE : 1 mi/gal 0.425 km/L = . A km is very roughly half a mile and there are roughly 4 liters in a gallon, so 2
41 mi/gal km/L ∼ , which is roughly our result. 1.10. IDENTIFY : Convert units. SET UP: Use the unit conversions given in the problem. Also, 100 cm 1 m = and 1000 g 1 kg = . EXECUTE : (a) mi 1h 5280 ft ft60 88h 3600s 1mi s⎛⎞ ⎛ ⎞ ⎛⎞= ⎜⎟ ⎜ ⎟ ⎜⎟⎝⎠ ⎝⎠ ⎝ ⎠ (b) 22ft 30.48cm 1 m m32 9.8s 1ft 100 cm s⎛⎞ ⎛⎞ ⎛ ⎞= ⎜⎟ ⎜⎟ ⎜ ⎟⎝⎠ ⎝ ⎠ ⎝⎠ (c) 3
3
33g 100 cm 1 kg kg1.0 10cm 1 m 1000 g m⎛⎞ ⎛⎞ ⎛⎞= ⎜⎟ ⎜⎟ ⎜⎟⎝⎠ ⎝⎠ ⎝⎠ EVALUATE : The relations 60 mi/h 88 ft/s = and 33 31 g/cm 10 kg/m = are exact. The relation 2232 ft/s 9.8 m/s = is accurate to only two significant figures. 1.11. IDENTIFY : We know the density and mass; thus we can find the volume using the relation density mass/volume / mV == . The radius is then found from the volume equation for a sphere and the result for the volume. SET UP: 3Density 19.5 g/cm = and critical 60.0 kg. m = For a sphere 3 4
3Vr π= . EXECUTE : 3
critical 360.0 kg 1000 g/ density 3080 cm19.5 g/cm 1.0 kgVm⎛⎞ ⎛ ⎞== =⎜⎟ ⎜ ⎟⎝⎠ ⎝ ⎠. ()333333080 cm 9.0 cm44Vrππ== = . EVALUATE : The density is very large, so th e 130 pound sphere is small in size. 1.12. IDENTIFY : Use your calculator to display 710π× . Compare that number to the number of seconds in a year. SET UP: 1 yr 365.24 days, = 1 day 24 h, = and 1 h 3600 s. = EXECUTE : 7 24 h 3600 s(365.24 days/1 yr) 3.15567... 10 s1 day 1 h⎛⎞ ⎛⎞=× ⎜⎟ ⎜⎟⎝⎠ ⎝⎠; 7710 s 3.14159... 10 sπ×= × The approximate expression is accurate to two significant figures. EVALUATE : The close agreement is a numerical accident. 1.13. IDENTIFY : The percent error is the error divided by the quantity. SET UP: The distance from Berlin to Paris is given to the nearest 10 km. EXECUTE : (a) 3
310 m1 . 11 0% .890 10 m−=×× (b) Since the distance was given as 890 km, the total distance should be 890,000 meters. We know the total distance to only three significant figures. EVALUATE : In this case a very small percentage error has disastrous consequences. 1.14. IDENTIFY : When numbers are multiplied or divided, the numbe r of significant figures in the result can be no greater than in the factor with the fewest significant figur es. When we add or subtract numbers it is the location of the decimal that matters. Units, Physical Quantities and Vectors 1-3 SET UP: 12 mm has two significant figures and 5.98 mm has three significant figures. EXECUTE : (a) () ( )212 mm 5.98 mm 72 mm ×= (two significant figures) (b) 5.98 mm0.5012 mm= (also two significant figures) (c) 36 mm (to the nearest millimeter) (d) 6 mm (e) 2.0 (two significant figures) EVALUATE : The length of the rectangle is known only to the nearest mm, so the answers in parts (c) and (d) are known only to the nearest mm. 1.15. IDENTIFY and SET UP: In each case, estimate the precision of the measurement. EXECUTE : (a) If a meter stick can measure to the neares t millimeter, the error will be about 0.13%. (b) If the chemical balance can measure to the nearest milligram, the error will be about 38.3 10 %.−× (c) If a handheld stopwatch (as opposed to electric timing devices ) can measure to the nearest tenth of a second, the error will be about 22.8 10 %.−× EVALUATE : The percent errors are those due only to the limit of precision of the measurement. 1.16. IDENTIFY : Use the extreme values in the pieces length and width to find the uncertainty in the area. SET UP: The length could be as large as 5.11 cm a nd the width could be as large as 1.91 cm. EXECUTE : The area is 9.69 ± 0.07 cm2. The fractional uncertainty in the area is 2
20.07 cm0.72%,9.69 cm= and the fractional uncertainties in the length and width are 0.01 cm0.20%5.10 cm= and 0.01 cm0.53%.1.9 cm= The sum of these fractional uncertainties is 0.20% 0.53% 0.73% += , in agreement with the fractional uncertainty in the area. EVALUATE : The fractional uncertainty in a product of numbers is greater than the fractional uncertainty in any of the individual numbers. 1.17. IDENTIFY : Calculate the average volume and diameter and the uncertainty in these quantities. SET UP: Using the extreme values of the input data gives us th e largest and smallest values of the target variables and from these we get the uncertainty. EXECUTE : (a) The volume of a disk of diameter d and thickness t is 2(/ 2 ). Vd tπ= The average volume is 23(8.50 cm/2) (0.50 cm) 2.837 cm . Vπ== But t is given to only two significant figures so the answer should be expressed to two significant figures: 32.8 cm .V= We can find the uncertainty in the volume as follows. The volume could be as large as 23(8.52 cm/2) (0.055 cm) 3.1 cm , Vπ== which is 30.3 cm larger than the average value. The volume could be as small as 23(8.52 cm/2) (0.045 cm) 2.5 cm , Vπ== which is 30.3 cm smaller than the average value. The uncertainty is 30.3 cm ,± and we express the volume as 32.8 0.3 cm .V=± (b) The ratio of the average diameter to the average thickness is 8.50 cm/0.050 cm 170. = By taking the largest possible value of the diameter and the smallest possible thickness we get the largest possible value for this ratio: 8.52 cm/0.045 cm 190. = The smallest possible value of the ratio is 8.48/ 0.055 150. = Thus the uncertainty is 20± and we write the ratio as 170 20. ± EVALUATE : The thickness is uncertain by 10% and the percentage uncertainty in the diameter is much less, so the percentage uncertainty in the volume and in the ratio should be about 10%. 1.18. IDENTIFY : Estimate the number of people and then use the estimates given in the problem to calculate the number of gallons. SET UP: Estimate 831 0× people, so 821 0× cars. EXECUTE : () ( ) Number of cars miles/car day / mi/gal gallons/day ×= ( )()8 82 10 cars 10000 mi/yr/car 1 yr/365 days / 20 mi/gal 3 10 gal/day×× × = × EVALUATE : The number of gallons of gas used each day a pproximately equals the population of the U.S. 1.19. IDENTIFY : Express 200 kg in pounds. Express each of 200 m, 200 cm and 200 mm in inches. Express 200 months in years. SET UP: A mass of 1 kg is equivalent to a weight of about 2.2 lbs. 1 in. 2.54 cm = . 1 y 12 months = . EXECUTE : (a) 200 kg is a weight of 440 lb. This is much larger than the typical weight of a man. (b) 43 1 in.200 m (2.00 10 cm) 7.9 10 inches2.54 cm⎛⎞=× = × ⎜⎟⎝⎠. This is much greater than the height of a person. (c) 200 cm 2.00 m 79 inches 6.6 ft == = . Some people are this tall, but not an ordinary man. 1-4 Chapter 1 (d) 200 mm 0.200 m 7.9 inches == . This is much too short. (e) 1 y200 months (200 mon) 17 y12 mon⎛⎞== ⎜⎟⎝⎠. This is the age of a teenager; a middle-aged man is much older than this. EVALUATE : None are plausible. When specifying the value of a measured quantity it is essential to give the units in which it is being expressed. 1.20. IDENTIFY : The number of kernels can be calculated as bottle kernel/. NV V= SET UP: Based on an Internet search, Iowan corn farmers use a sieve having a hole size of 0.3125 in. ≅ 8 mm to remove kernel fragments. Therefore estimate the average kernel length as 10 mm, the width as 6 mm and the depth as 3 mm. We must also apply the conversion factors 31 L 1000 cm and 1 cm 10 mm.== EXECUTE : The volume of the kernel is: ( )( )( )3
kernel 10 mm 6 mm 3 mm 180 mm V == . The bottles volume is: () () () () ()33 36 3
bottle 2.0 L 1000 cm 1.0 L 10 mm 1.0 cm 2.0 10 mm V ⎡⎤ ⎡⎤ == × ⎣⎦ ⎣ ⎦ . The number of kernels is then () ( )63 3
kernels bottle kernels / 2.0 10 mm 180 mm 11,000 kernels NV V =≈ × = . EVALUATE : This estimate is highly dependent upon your estimat e of the kernel dimensions. And since these dimensions vary amongst the different available type s of corn, acceptable answers could range from 6,500 to 20,000. 1.21. IDENTIFY : Estimate the number of pages and the number of words per page. SET UP: Assuming the two-volume edition, there are approximately a thousand pages, and each page has between 500 and a thousand words (counting captions and the smaller print, such as the end-of-chapter exercises and problems). EXECUTE : An estimate for the number of words is about 610 . EVALUATE : We can expect that this estimate is accurate to within a factor of 10. 1.22. IDENTIFY : Approximate the number of breaths per minute. Convert minutes to years and 3cm to 3m to find the volume in 3m breathed in a year. SET UP: Assume 10 breaths/min . 5 24 h 60 min1 y (365 d) 5.3 10 min1 d 1 h⎛⎞ ⎛ ⎞== × ⎜⎟ ⎜ ⎟⎝⎠ ⎝ ⎠. 210 cm 1 m = so 63 310 cm 1 m = . The volume of a sphere is 33 41
36Vr d π π == , where r is the radius and d is the diameter. Dont forget to account for four astronauts. EXECUTE : (a) The volume is 5
63 43 5.3 10 min(4)(10 breaths/min)(500 10 m ) 1 10 m / yr1 y− ⎛⎞ ××= × ⎜⎟
⎝⎠. (b) 1/3 1/3 4366 [ 1 1 0 m ]27 mVdππ⎛⎞ × ⎛⎞== = ⎜⎟ ⎜⎟⎝⎠ ⎝⎠ EVALUATE : Our estimate assumes that each 3cm of air is breathed in only once, where in reality not all the oxygen is absorbed from the air in each breath. Therefore, a somewhat smaller volume would actually be required. 1.23. IDENTIFY : Estimate the number of blinks per minute. Convert mi nutes to years. Estimate the typical lifetime in years. SET UP: Estimate that we blink 10 times per minute. 1 y 365 days = . 1 day 24 h = , 1 h 60 min = . Use 80 years for the lifetime. EXECUTE : The number of blinks is 8 60 min 24 h 365 days(10 per min) (80 y/lifetime) 4 101 h 1 day 1 y⎛⎞ ⎛ ⎞ ⎛⎞=× ⎜⎟ ⎜ ⎟ ⎜⎟⎝⎠ ⎝⎠ ⎝ ⎠ EVALUATE : Our estimate of the number of blinks per minute can be off by a factor of two but our calculation is surely accurate to a power of 10. 1.24. IDENTIFY : Estimate the number of beats per minute and the duration of a lifetime. The volume of blood pumped during this interval is then the volume per beat multiplied by the total beats. SET UP: An average middle-aged (40 year-old) adult at rest has a heart rate of roughly 75 beats per minute. To calculate the number of beats in a lifetime, use the current average lifespan of 80 years. EXECUTE : ()9
beats60 min 24 h 365 days 80 yr75 beats/min 3 10 beats/lifespan1 h 1 day yr lifespanN⎛⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞== × ⎜⎟ ⎜⎟ ⎜ ⎟ ⎜ ⎟⎝⎠ ⎝⎠ ⎝ ⎠ ⎝ ⎠ ()9
3 7
blood 31 L 1 gal 3 10 beats50 cm /beat 4 10 gal/lifespan1000 cm 3.788 L lifespanV⎛⎞× ⎛⎞ ⎛ ⎞== × ⎜⎟ ⎜ ⎟ ⎜⎟⎝⎠ ⎝ ⎠ ⎝⎠ EVALUATE : This is a very large volume.
