Instructor's Solution Manuals to College Physics PDF
Preview Instructor's Solution Manuals to College Physics
Chapter 1 Introduction to Physics Conceptual Questions 1.1 The meter (SI unit for length) is defined by the distance light travels in a vacuum in a tiny fraction (1/299792458) of a second. The second (SI unit for time) is defined as the time it takes for 9,192,631,770 periods of the transition between the two split levels of the ground state of the cesium-133 atom. The kelvin (SI unit for temperature) is defined in terms of the conditions under which water can exist as ice, liquid, and gas simultaneously. The kilogram (SI unit for mass) is defined by the mass of a carefully protected prototype block made of platinum and iridium that was manufactured in 1889. 1.2 Yes, it is possible to define a system of units where length is not one of the fundamental properties. For example, a system that has fundamental properties of speed and time can derive the quantity of length. 1.3 Use of the metric prefixes makes any numerical calculation much easier to follow. Instead of an obscure conversion (12 in/ft, 1760 yards/mi, 5280 ft/mi), simple powers of 10 make the transformations (10 mm/cm, 1000 m/km, 1026 m/m). 1.4 The answer should be written as 55.0. When dividing quantities, the number with the fewest significant figures dictates the number of significant figures in the answer. In this case 3411 has four significant figures and 62.0 has three significant figures, which means we are allowed three significant figures in our answer. 1.5 No. The equation “3 meters 5 70 meters” has consistent units but it is false. The same goes for “1 5 2,” which consistently has no units. 1.6 To be a useful standard of measurement, an object, system, or process should be unchanging, replicable, and possible to measure precisely so that errors in its measurement do not carry over into calibration errors in every other measurement. 1.7 The SI unit for length is the meter; the SI unit for time is the second. Therefore, the SI unit for acceleration is the meter/(second)2, or m/s2. 1.8 The fewest number of significant figures in 61,000 is two—the “6” and the “1.” If the period is acting as a decimal point, then the trailing zeros are significant and the quantity 61,000. would have five significant figures. When numbers are written in scientific notation, all of the digits before the power of 10 are significant. Therefore, 6.10 3 104 has three significant figures. Multiple-Choice Questions 1.9 B (length). Mass density, area, and resistance are all derived quantities. 2 Chapter 1 Introduction to Physics 1.10 E (1 m). It is easiest to answer this question by first converting all of the choices into meters: 1 m A) 10 nm 3 51028 m 109 nm 1 m B) 10 cm 3 51021 m 102 cm 1 m C) 102 mm 3 51021 m 103 mm D) 1022 m E) 1 m 1.11 C (1029s). The prefix nano- means 1029. 1.12 E (1026). 1 m 3 1 cm3 3 5(cid:31) 1026 m3 (cid:31) 100 cm a b 1.13 C (104). 100 cm 100 cm 1 m2 3 3 5(cid:31) 104 cm2 (cid:31) 1 m 1 m 1.14 E (32). When adding or subtracting quantities, the quantity with the fewest decimal places (not significant figures) dictates the number of decimal places in the final answer. That quantity is 15 in this case. 1.15 D (2.5). When dividing quantities, the quantity with the fewest significant figures, which is 0.28 in this case, dictates the number of significant figures in the final answer. L 1.16 B (v/t). Acceleration has dimensions of . To answer this question, we should first T 2 3 4 determine the dimensions of each of the choices: 3 4 L 2 1 L 2 A) # 5 T 2 T T 3 3 4 3 4 L 1 L B) 3 4 # 3 45 3 4 T T T 2 3 4 3 4 L 1 L C) 3 4 # 3 4 53 4 T T 2 T 3 3 4 3 4 L 1 1 D) 3 4 # 3 4 5 3 4 T L 2 L T 3 4 L 2 1 1 E) 3 4 #3 4 53 43 4 T 2 L 2 T 2 3 4 3 4 3 4 3 4 Chapter 1 Introduction to Physics 3 1.17 C (1810). Both 25.8 and 70.0 have three significant figures. When multiplying quantities, the quantity with the fewest significant figures dictates the number of significant figures in the final answer. Multiplying 25.8 by 70.0 gives 1806, which has four significant figures. Our final answer must have three significant figures, so we round 1806 to 1810. Estimation/Numerical A nalysis 1.18 There is no one answer to this question. When estimating, keep in mind that 1 meter is a little more than 3 feet. 1.19 The distance from home plate to the center field fence is about 100 m. A well-hit ball leaves the bat at around 100 mph, or 45 m/s. Assuming the ball comes off the bat 1 s horizontal to the ground, this gives an estimate of 100 m 3 5(cid:31) 2 s (cid:31). This is 45 m probably a little low but is still reasonable. 1.20 Because laptop computers are very common nowadays, most people (students, faculty, etc.) have at least one laptop. Various departments (for example, academic departments, campus IT) also have laptops available, so there is probably about one laptop per person on campus. 1.21 We can split an average student’s daily water use into four categories: showering, cooking/drinking/hand-washing, flushing the toilet, and doing laundry. A person uses about 100 L of water when showering, about 10 L for cooking/drinking/hand-washing, about 24 L when flushing the toilet, and about 40 L when doing two loads of laundry. This works out to about 150–200 L of water per day. 1.22 There is no one answer to this question. The Environmental Protection Agency estimated that an average American produced about 2 kg (4.6 lb) of garbage a day in 2006 (http://www.epa.gov/wastes/nonhaz/). 1.23 There is no one answer to this question. When estimating, keep in mind that one storey is about 10 feet. 1.24 The footprint of Chicago is around 600 km2. A city block of intermediate size is around 100 m by 100 m. Sidewalk all around the perimeter would translate to 400 m of sidewalk per city block. In each square kilometer, there would be about 100 city blocks. The length of the sidewalks is 100 blocks 400 m sidewalk 1 km 600 km2 3 3 3 5(cid:31) 20,000 km (cid:31). 1 km2 1 block 1000 m 1.25 We can estimate the number of cells in the human body by determining the mass of a cell and comparing it to the mass of a human. An average human male has a mass of 80 kg. A person is mainly water, so we can approximate the density of a human body (and its cells) as 1000 kg/m3. We are told that the volume of a cell is the same as a sphere with a radius of 1025 m, or approximately 4 3 10215 m3; the mass of a single 1000 kg cell is 4 3 10215 m3 3 54 3 10212 kg. The number of cells in the body is then 1 m3 m 80 kg body n 5 5 5(cid:31) 2 3 1013 (cid:31). cell m 4 3 10212 kg cell 4 Chapter 1 Introduction to Physics 1.26 In order to estimate the volume flow rate of air that fills your lungs as you take a deep breath, we need to estimate the volume of your lungs and the time it takes to take a deep breath. Let’s assume that your lungs fill your rib cage, which has dimensions of 10 in 3 6 in 3 4 in. This is a volume of 240 in3, or 3.9 3 1023 m3. We can time how long it takes to take a deep breath; it is about 4 s. Putting these estimates together, we find that the volume flow rate when taking a deep breath is about (cid:31) 1023 m3/s (cid:31). Problems 1.27 Set Up Scientific notation is a simple, compact way of expressing large and small numbers. The numbers are written as a coefficient multiplied by a power of 10. The coefficient should contain all of the significant figures in the quantity and be written as a nonzero digit in the ones place, a decimal place, and then the remaining significant digits. Solve A) 2.37 3 102 E) 1.487 3 104 B) 2.23 3 1023 F) 2.1478 3 102 C) 4.51 3 101 G) 4.42 3 1026 D) 1.115 3 103 H) 1.2345678 3 107 RefleCt Scientific notation easily shows the significant figures in a quantity. 1.28 Set Up We are given eight numbers written using a power of 10 and asked to write them as decimals. For numbers smaller than one, it is customary to include the zero before the decimal place. Solve A) 0.00442 E) 456,000 B) 0.00000709 F) 0.0224 C) 828 G) 0.0000375 D) 6,020,000 H) 0.000138 RefleCt The answers to this problem show the advantage scientific notation offers in easily reading numbers. Chapter 1 Introduction to Physics 5 1.29 Set Up A list of powers of 10 is given. We can use Table 1-3 in the text to determine the correct metric prefix associated with each factor. Eventually, knowing some of the more common prefixes will become second nature. Solve A) kilo (k) E) milli (m) B) giga (G) F) pico (p) C) mega (M) G) micro () D) tera (T) H) nano (n) RefleCt Whether or not the prefix is capitalized is important. Mega- and milli- both use the letter “m,” but mega- is “M” and milli- is “m.” Confusing these two will introduce an error of 109! 1.30 Set Up A list of metric prefix symbols is given. We are asked to write the power of 10 associated with each prefix. We can use Table 1-3 in the text to determine the correct factor associated with each metric prefix. Solve A) pico 5 10212 E) femto 5 10215 B) milli 5 1023 F) giga 5 109 C) mega 5 106 G) tera 5 1012 D) micro 5 1026 H) centi 5 1022 RefleCt It will be useful to memorize some of the more common prefixes, such as milli-, mega-, micro-, and centi-. 1.31 Set Up This problem provides practice converting between some of the most common metric units. The prefixes we will need are centi- (1022), kilo- (103), and milli- (1023). Don’t forget to convert each factor when dealing with areas and volumes. Solve 1 m A) 125 cm 3 5(cid:31) 1.25 m (cid:31) 100 cm 1 kg B) 233 g 3 5(cid:31) 0.233 kg (cid:31) 1000 g 6 Chapter 1 Introduction to Physics 1 s C) 786 ms 3 5(cid:31) 0.786 s (cid:31) 1000 ms 106 mg D) 454 kg 3 5(cid:31) 4.54 3 108 mg (cid:31) 1 kg 1 m 2 E) 208 cm2 3 5(cid:31) 0.0208 m2 (cid:31) 100 cm a b 100 cm 2 F) 444 m2 3 5(cid:31) 4.44 3 106 cm2 (cid:31) 1 m a b 1 m 3 G) 12.5 cm3 3 5(cid:31) 1.25 3 1025 m3 (cid:31) 100 cm a b 100 cm 3 H) 144 m3 3 5(cid:31) 1.44 3 108 cm3 (cid:31) 1 m a b RefleCt Although it is an extra step to convert from, say, kilograms to grams to milligrams, it is easier (and more useful) to memorize how many grams are in a kilogram and milligrams in a gram than to memorize how many milligrams are in a kilogram. 1.32 Set Up We are asked to convert a list of quantities from U.S. units to metric units. The conversions we will need are 1 in 5 2.54 cm, 1 L 5 33.8 oz, 1 kg 5 2.205 lb, and 1 mi 5 1.609344 km. We will write each answer with the correct number of significant figures. Solve 12 in 2.54 cm 1 m A) 238 ft 3 3 3 5(cid:31) 72.5 m (cid:31) 1 ft 1 in 100 cm 2.54 cm B) 772 in 3 5(cid:31) 1960 cm (cid:31) 1 in 2.54 cm 2 C) 1220 in2 3 5(cid:31) 7870 cm2 (cid:31) 1 in a b 1 L D) 559 oz 3 5(cid:31) 16.5 L (cid:31) 33.8 oz 1 kg 1000 g E) 973 lb 3 3 5(cid:31) 4.41 3 105 g (cid:31) 2.205 lb 1 kg 12 in 3 2.54 cm 3 1 m 3 F) 122 ft3 3 3 3 5(cid:31) 3.45 m3 (cid:31) 1 ft 1 in 100 cm a b a b a b Chapter 1 Introduction to Physics 7 1.609344 km 2 G) 1.28 mi2 3 5(cid:31) 3.32 km2 (cid:31) 1 mi a b 2.54 cm 3 H) 442 in3 3 5(cid:31) 7240 cm3 (cid:31) 1 in a b RefleCt Learning some common conversions between U.S. units and metric units will help you determine whether an answer you calculate is reasonable. For example, most Americans have a better handle on 3 feet versus 1 meter. Some simple conversions are 1 in 5 2.54 cm, 1 kg 5 2.2 lb, and 1 mi 5 1.6 km. 1.33 Set Up We are asked to perform some unit conversions. Some conversion factors we will need are 1 in 5 2.54 cm, 1 min 5 60 s, and 1 h 5 60 min. Solve 1 in A) 125 cm 3 5(cid:31) 49.2 in (cid:31) 2.54 cm 2.54 cm B) 233 in 3 5(cid:31) 592 cm (cid:31) 1 in 1 s 1 min 1 h C) 553 ms 3 3 3 5(cid:31) 1.54 3 1024 h (cid:31) 1000 ms 60 s 60 min 2.54 cm 1 m D) 454 in 3 3 5(cid:31) 11.5 m (cid:31) 1 in 100 cm 1 in 2 E) 355 cm2 3 5(cid:31) 55.0 in2 (cid:31) 2.54 cm a b 100 cm 2 1 in 2 1 ft 2 F) 333 m2 3 3 3 5(cid:31) 3580 ft2 (cid:31) 1 m 2.54 cm 12 in a b a b a b 2.54 cm 3 G) 424 in3 3 5(cid:31) 6950 cm3 (cid:31) 1 in a b 100 cm 3 1 in 3 H) 172 m3 3 3 5(cid:31) 1.05 3 107 in3 (cid:31) 1 m 2.54 cm a b a b RefleCt Don’t forget to convert each factor when dealing with areas and volumes. 8 Chapter 1 Introduction to Physics 1.34 Set Up We are asked to perform some volume unit conversions. Some useful conversion factors are 1 gal 5 231 in3 5 3.785 L, 1 L 5 33.8 fl oz, and 1 pint 5 0.4732 L. Solve 231 in3 1 ft 3 A) 118 gal 3 3 5(cid:31) 15.8 ft3 (cid:31) 1 gal 12 in a b 12 in 3 1 gal B) 1.3 ft3 3 3 5(cid:31) 9.7 gal (cid:31) 1 ft 231 in3 a b 1 L C) 14,400 fl oz 3 5(cid:31) 426 L (cid:31) 33.8 fl oz 1 L 1 m3 D) 128 fl oz 3 3 5(cid:31) 3.79 3 1023 m3 (cid:31) 33.8 fl oz 1000 L 2.54 cm 3 E) 487 in3 3 5(cid:31) 7980 cm3 (cid:31) 1 in a b 3.785 L F) 0.0032 gal 3 5(cid:31) 0.012 L (cid:31) 1 gal 2.54 cm 3 1 m 3 G) 129 in3 3 3 5(cid:31) 2.11 3 1023 m3 (cid:31) 1 in 100 cm a b a b 0.4732 L H) 324 pint 3 5(cid:31) 153 L (cid:31) 1 pint RefleCt Don’t forget to convert each factor when dealing with volume units. 1.35 Set Up This problem provides practice with metric unit conversions. We need to look up (and/or memorize) various conversions between SI units and hectares and liters. One hectare is equal to 104 m2, and 1000 L is equal to 1 m3. Another useful conversion is that 1 mL equals 1 cm3. Solve 1 mL 1 L A) 328 cm3 3 3 5(cid:31) 0.328 L (cid:31) 1 cm3 1000 mL 1 m3 B) 112 L 3 5(cid:31) 0.112 m3 (cid:31) 1000 L Chapter 1 Introduction to Physics 9 104 m2 C) 220 hectares 3 5(cid:31) 2.2 3 106 m2 (cid:31) 1 hectare 1 hectare D) 44300 m2 3 5(cid:31) 4.43 hectares (cid:31) 104 m2 1 m3 E) 225 L 3 5(cid:31) 0.225 m3 (cid:31) 1000 L 104 m2 103 L F) 17.2 hectare # m 3 3 5(cid:31) 1.72 3 108 L (cid:31) 1 hectare 1 m3 1 m3 1 hectare G) 2.253 3 105 L 3 3 5(cid:31) 2.253 3 1022 hectare # m (cid:31) 103 L 104 m2 1000 L 1000 mL H) 2000 m3 3 3 5(cid:31) 2 3 109 mL (cid:31) 1 m3 1 L RefleCt Rewriting the measurements and conversions in scientific notation makes the calculations simpler and helps give some physical intuition. 1.36 Set Up We are asked to perform some unit conversions. Some useful conversion factors for this problem are 1 gal 5 3.785 L, 1 acre 5 43,560 ft2, 1 ft 5 0.3048 m, 1 L 5 33.814 fl oz, 1 cm3 5 1 mL, 1 cup 5 8 fl oz, 1 hectare 5 104 m2, 1 pint 5 473.2 mL, and 1 quart 5 2 pints. Solve 3.785 L A) 33.5 gal 3 5(cid:31) 127 L (cid:31) 1 gal 1 gal B) 62.8 L 3 5(cid:31) 16.6 gal (cid:31) 3.785 L 43,560 ft2 0.3048 m 3 1000 L C) 216 acre # ft 3 3 3 5(cid:31) 2.66 3 108 L (cid:31) 1 acre 1 ft 1 m3 a b 3.785 L 1 m3 D) 1770 gal 3 3 5(cid:31) 6.70 m3 (cid:31) 1 gal 1000 L 1 L 1000 mL 1 cm3 E) 22.8 fl oz 3 3 3 5(cid:31) 674 cm3 (cid:31) 33.814 fl oz 1 L 1 mL 10 Chapter 1 Introduction to Physics 1 mL 1 L 33.814 fl oz 1 cup F) 54.2 cm3 3 3 3 3 5(cid:31) 0.229 cup (cid:31) 1 cm3 1000 mL 1 L 8 fl oz 104 m2 1 ft 2 1 acre G) 1.25 hectares 3 3 3 5(cid:31) 3.09 acres (cid:31) 1 hectare 0.3048 m 43,560 ft2 a b 1 cm 3 1 mL 1 pint 1 quart H) 644 mm3 3 3 3 3 5(cid:31) 6.80 3 1024 qt (cid:31) 10 mm 1 cm3 473.2 mL 2 pints a b RefleCt The conversion 1 cm3 5 1 mL is very useful to know. 1.37 Set Up We are given three quantities and asked to rewrite them in scientific notation without prefixes. The prefixes correspond to kilo- (103), micro- (1026), and giga- (109), respectively. Be sure to include all of the significant figures in the quantity. Solve A) 300 km 5 300 3 103 m 5 (cid:31) 3 3 105 m (cid:31) B) 33.7 m 5 33.7 3 1026 m 5 (cid:31) 3.37 3 1025 m (cid:31) C) 77.5 GW 5 77.5 3 109 W 5 (cid:31) 7.75 3 1010 W (cid:31) RefleCt Writing numbers using powers of 10 allows you to perform calculations quickly and without a calculator. 1.38 Set Up We are given quantities in scientific notation and asked to rewrite these quantities using metric prefixes. Since the metric prefixes correspond to factors of 103, it is easiest to first change the scientific notation to a power of 103 and then replace it with a prefix. Solve A) 3.45 3 1024 s 5 345 3 1026 s 5 (cid:31) 345 s (cid:31) B) 2.00 3 10211 W 5 20.0 3 10212 W 5 (cid:31) 20.0 pW (cid:31) C) 2.337 3 108 m 5 233.7 3 106 m 5 (cid:31) 233.7 Mm (cid:31) D) 6.54 3 104 g 5 65.4 3 103 g 5 (cid:31) 65.4 kg (cid:31) RefleCt There are actually many different answers for each part. For example, we can rewrite 345 s as 0.345 ms.
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