Exam (elaborations) TEST BANK FOR University Phyiscs 12th Edition By Hugh D. Young, Roger A. Freedman, and Lewis Ford (Solution manual)

1.1. IDENTIFY: 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) 2 3 1.00 mi (1.00 mi) 5280 ft 12 in. 2.54 cm 1 m 1 km 1.61 km 1 mi 1 ft 1 in. 10 cm 10 m = ⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞ = ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠ (b) 3 2 1.00 km (1.00 km) 10 m 10 cm 1 in. 1 ft 3.28 103 ft 1 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 in.3 . SET UP: 1 L =1000 cm3 . 1 in. = 2.54 cm EXECUTE: 3 3 0.473 L 1000 cm 1 in. 28.9 in.3. 1 L 2.54 cm ⎛ ⎞ ⎛ ⎞ ×⎜ ⎟×⎜ ⎟ = ⎝ ⎠ ⎝ ⎠ EVALUATE: 1 in.3 is greater than 1 cm3 , so the volume in in.3 is a smaller number than the volume in cm3 , which is 473 cm3 . 1.3. IDENTIFY: We know the speed of light in m/s. t = d /v . Convert 1.00 ft to m and t from s to ns. SET UP: The speed of light is v = 3.00×108 m/s . 1 ft = 0.3048 m . 1 s =109 ns . EXECUTE: 9 8 0.3048 m 1.02 10 s 1.02 ns 3.00 10 m/s t = = × − = × EVALUATE: In 1.00 s light travels 3.00×108 m = 3.00×105 km =1.86×105 mi . 1.4. IDENTIFY: Convert the units from g to kg and from cm3 to m3 . SET UP: 1 kg =1000 g . 1 m =1000 cm . EXECUTE: 3 4 3 3 11.3 g 1 kg 100 cm 1.13 10 kg cm 1000 g 1m m ⎛ ⎞⎛ ⎞ ×⎜ ⎟×⎜ ⎟ = × ⎝ ⎠⎝ ⎠ EVALUATE: The ratio that converts cm to m is cubed, because we need to convert cm3 to m3 . 1.5. IDENTIFY: Convert volume units from in.3 to L. SET UP: 1 L =1000 cm3 . 1 in. = 2.54 cm . EXECUTE: (327 in.3 )×(2.54 cm in.) 3 ×(1 L 1000 cm3 ) = 5.36 L EVALUATE: The volume is 5360 cm3 . 1 cm3 is less than 1 in.3 , so the volume in cm3 is a larger number than the volume in in.3 . 1.6. IDENTIFY: Convert ft2 to m2 and then to hectares. SET UP: 1.00 hectare =1.00×104 m2 . 1 ft = 0.3048 m . EXECUTE: The area is 2 2 4 2 (12.0 acres) 43,600 ft 0.3048 m 1.00 hectare 4.86 hectares 1 acre 1.00 ft 1.00 10 m ⎛ ⎞⎛ ⎞ ⎛ ⎞ = ⎜ ⎟⎜ ⎟ ⎜ × ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠ . EVALUATE: Since 1 ft = 0.3048 m , 1 ft2 = (0.3048)2 m2 . 1.7. IDENTIFY: Convert seconds to years. SET UP: 1 billion seconds =1×109 s . 1 day = 24 h . 1 h = 3600 s . EXECUTE: 1.00 billion seconds (1.00 109 s) 1 h 1 day 1 y 31.7 y 3600 s 24 h 365 days ⎛ ⎞⎛ ⎞⎛ ⎞ = × ⎜ ⎟⎜ ⎟⎜ ⎟ = ⎝ ⎠⎝ ⎠⎝ ⎠ . 1 1-2 Chapter 1 EVALUATE: The conversion 1 y = 3.156×107 s assumes 1 y = 365.24 d , which is the average for one extra day every four years, in leap years. The problem 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: (180,000 furlongs fortnight) 0.125 mi 1 fortnight 1 day 67 mi/h 1 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) 55.0 miles/gallon (55.0 miles/gallon) 1.609 km 1 gallon 23.4 km/L 1 mi 3.788 L = ⎛ ⎞⎛ ⎞ = ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ . (b) The volume of gas required is 1500 km 64.1 L 23.4 km/L = . 64.1 L 1.4 tanks 45 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 24 1 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) 60mi 1h 5280 ft 88 ft h 3600s 1mi s ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟= ⎝ ⎠⎝ ⎠⎝ ⎠ (b) 2 2 32 ft 30.48cm 1 m 9.8m s 1ft 100 cm s ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ (c) 3 3 3 3 1.0 g 100 cm 1 kg 10 kg cm 1 m 1000 g m ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = ⎝ ⎠⎝ ⎠⎝ ⎠ EVALUATE: The relations 60 mi/h = 88 ft/s and 1 g/cm3 =103 kg/m3 are exact. The relation 32 ft/s2 = 9.8 m/s2 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 = m/V . The radius is then found from the volume equation for a sphere and the result for the volume. SET UP: Density =19.5 g/cm3 and critical m = 60.0 kg. For a sphere 4 3 3 V = π r . EXECUTE: 3 critical 3 / density 60.0 kg 1000 g 3080 cm 19.5 g/cm 1.0 kg V = m = ⎛⎜ ⎞⎟⎛⎜ ⎞⎟ = ⎝ ⎠⎝ ⎠ . 3 3 ( 3 ) 3 3 3080 cm 9.0 cm 4 4 r V π π = = = . EVALUATE: The density is very large, so the 130 pound sphere is small in size. 1.12. IDENTIFY: Use your calculator to display π ×107 . 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: (365.24 days/1 yr) 24 h 3600 s 3.15567... 107 s 1 day 1 h ⎛ ⎞⎛ ⎞ = × ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ ; π ×107 s = 3.14159...×107 s 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 3 10 m 1.1 10 %. 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 number of significant figures in the result can be no greater than in the factor with the fewest significant figures. 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) (12 mm)×(5.98 mm) = 72 mm2 (two significant figures) (b) 5.98 mm 0.50 12 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 nearest millimeter, the error will be about 0.13%. (b) If the chemical balance can measure to the nearest milligram, the error will be about 8.3×10−3%. (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 2.8×10−2%. 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 piece’s length and width to find the uncertainty in the area. SET UP: The length could be as large as 5.11 cm and 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 2 0.07 cm 0.72%, 9.69 cm = and the fractional uncertainties in the length and width are 0.01 cm 0.20% 5.10 cm = and 0.01 cm 0.53%. 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 the 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 V =π (d / 2)2 t. The average volume is V =π (8.50 cm/2)2 (0.50 cm) = 2.837 cm3. But t is given to only two significant figures so the answer should be expressed to two significant figures: V = 2.8 cm3. We can find the uncertainty in the volume as follows. The volume could be as large as V =π (8.52 cm/2)2 (0.055 cm) = 3.1 cm3, which is 0.3 cm3 larger than the average value. The volume could be as small as V =π (8.52 cm/2)2 (0.045 cm) = 2.5 cm3, which is 0.3 cm3 smaller than the average value. The uncertainty is ±0.3 cm3 , and we express the volume as V = 2.8 ± 0.3 cm3. (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 3×108 people, so 2×108 cars. EXECUTE: (Number of cars×miles/car day)/ (mi/gal) = gallons/day (2×108 cars×10000 mi/yr/car ×1 yr/365 days)/ (20 mi/gal) = 3×108 gal/day EVALUATE: The number of gallons of gas used each day approximately 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) 200 m (2.00 104 cm) 1 in. 7.9 103 inches 2.54 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) 200 months (200 mon) 1 y 17 y 12 mon = ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ . This is the age of a teenager; a middle-aged man is much older than this. EVALUATE: None are plausible. When specifying