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Physics Principles With Applications 6th Edition Giancoli Test Bank
Physics Principles With Applications 6th Edition Giancoli Test Bank Test Bank for Physics: Principles with Applications Sixth Edition Douglas Giancoli ... [Show More] Chapter 1 Introduction, Measurement, Estimating Conceptual Questions 1) Four students measure the mass of an object, each using a different scale. They record their results as follows: Student A B C D Mass (g ) 49.06 49 50 49.2 Which student used the least precise scale? A) A B)B C) C D) D Answer: C Diff: 1 Page Ref: Sec. 1.4 2) Four students measure the mass of an object, each using a different scale. They record their results as follows: Student A B C D Mass (g ) 49.06 49 50 49.2 Which student used the most precise scale? A) A B)B C) C D) D Answer: A Diff: 1 Page Ref: Sec. 1.4 3) A useful method of expressing very small or very large numbers is A) scientific notation. B) arabic numerals. C) the metric system. D) roman numerals. Answer: A Diff: 1 Page Ref: Sec. 1.4 2 Chapter 1: Introduction, Measurement, Estimating 4) All of the following are base units of the SI system except: A) kilogram. B) kelvin. C) meter. D) volt. Answer: D Diff: 1 Page Ref: Sec. 15-1.6 5) Select the list which contains only SI basic units. A) liter, meter, second, watt B)joule, kelvin, kilogram, watt C) candela, kelvin, meter, second D) joule, newton, second, watt Answer: C Diff: 1 Page Ref: Sec. 1.5-1.6 6) How many basic units does the SI system have? A) four B)five C) seven D) ten Answer: C Diff: 1 Page Ref: Sec. 1.5-1.6 7) The base SI unit of time is A) hour. B) minute. C) second. D) millisecond. Answer: C Diff: 1 Page Ref: Sec. 1.5-1.6 8) In the CGS system, what are the fundamental units? A) Newton, centimeter, second B)kilogram, meter, second C) gram, centimeter, minute D) gram, centimeter, second Answer: D Diff: 2 Page Ref: Sec. 1.5-1.6 3 Physics: Principles with Applications, Sixth Edition 9) The metric prefix for one one-thousandth is A) milli. B) centi. C) kilo. D) mega. Answer: A Diff: 1 Page Ref: Sec. 1.5-1.6 10) The metric prefix for one one-hundredth is A) milli. B) centi. C) kilo. D) mega. Answer: B Diff: 1 Page Ref: Sec. 1.5-1.6 11) The metric prefix for one thousand is A) milli. B) centi. C) kilo. D) mega. Answer: C Diff: 1 Page Ref: Sec. 1.5-1.6 12) Express the number 0.02 days using a prefix of Table 1-4. A) 2 decidays B) 2 centidays C) 2 millidays D) 2 microdays Answer: B Diff: 1 Page Ref: Sec. 1.5-1.6 13) A) 0.0278 m/s B) 0.278 m/s C) 3.60 m/s D) 16.7 m/s Answer: B Diff: 1 Page Ref: Sec. 1.5-1.6 What is the conversion factor between km/h and m/s? 4 Chapter 1: Introduction, Measurement, Estimating 14) What is the conversion factor between km/h2 and m/s2? A) 7.72 ˛ 10-6 m/s2 B) 2.78 ˛ 10-1 m/s2 C) 1.30 ˛ 104 m/s2 D) 3.60 m/s2 Answer: A Diff: 1 Page Ref: Sec. 1.5-1.6 15) What is the conversion factor between cm2 and m2? A) 0.01 m2/cm2 B) 0.0001 m2/cm2 C) 100 m2/cm2 D) 10000 m2/cm2 Answer: B Diff: 1 Page Ref: Sec. 1.5-1.6 16) The position, x, of an object is given by the equation x = A + Bt +Ct2, where t refers to time. What are the dimensions of A, B, and C? A) distance, distance, distance B)distance, time, time2 C) distance, distance/time, distance/time2 D) distance/time, distance/time2, distance/time3 Answer: C Diff: 2 Page Ref: Sec. 1.8 Quantitative Problems 1) What is the percent uncertainty in the measurement 2.58 ± 0.15 cm? A) 2.9% B) 5.8% C) 8.7% D) 12% Answer: B Diff: 2 Page Ref: Sec. 1.4 2) What, approximately, is the percent uncertainty for the measurement 5.2? A) 1% B) 2% C) 3% D) 4% Answer: B Diff: 2 Page Ref: Sec. 1.4 5 Physics: Principles with Applications, Sixth Edition 3) What is the percent uncertainty in the area of a circle whose radius is 1.8 ˛ 104 cm? A) 1.1% B) 5.6% C) 11% D) 56% Answer: C Diff: 3 Page Ref: Sec. 1.4 4) What is the volume, and its approximate uncertainty, of a sphere of radius 1.96 ± 0.01 m? A) 31.5 ± 0.2 m2 B) 31.5 ± 0.3 m2 C) 31.5 ± 0.4 m2 D) 31.5 ± 0.5 m2 Answer: D Diff: 3 Page Ref: Sec. 1.4 5) The number of significant figures in 10001 is A) two. B)three. C) five. D) six. Answer: C Diff: 1 Page Ref: Sec. 1.4 6) The number of significant figures in 0.01500 is A) two. B)three. C) four. D) five. Answer: C Diff: 1 Page Ref: Sec. 1.4 7) The number of significant figures in 0.040 is A) one. B)two. C) three. D) four. Answer: B Diff: 1 Page Ref: Sec. 1.4 6 Chapter 1: Introduction, Measurement, Estimating 8) Which of the following has three significant figures? A) 305.0 cm B) 0.0500 mm C) 1.00081 kg D) 8.060 ˛ 1011 m2 Answer: B Diff: 1 Page Ref: Sec. 1.4 9) What is the sum of 2.67 + 1.976 + 2.1? A) 6.7 B) 6.75 C) 6.746 D) 6.7460 Answer: A Diff: 1 Page Ref: Sec. 1.4 10) What is the difference between 103.5 and 102.24? A) 1.3 B) 1.26 C) 1.260 D) 1.2600 Answer: A Diff: 1 Page Ref: Sec. 1.4 11) What is the product of 12.56 and 2.12? A) 27 B) 26.6 C) 26.23 D) 26.627 Answer: B Diff: 1 Page Ref: Sec. 1.4 12) What is the result of 2.43 d 4.561? A) 5.3278 ˛ 10-1 B) 5.328 ˛ 10-1 C) 5.33 ˛ 10-1 D) 5.3 ˛ 10-1 Answer: C Diff: 1 Page Ref: Sec. 1.4 7 Physics: Principles with Applications, Sixth Edition 13) What is the cosine of 55e? A) 0.6 B) 0.57 C) 0.574 D) 0.5736 Answer: B Diff: 1 Page Ref: Sec. 1.4 14) The length and width of a rectangle are 1.125 m and 0.606 m, respectively. Multiplying, your calculator gives the product as 0.68175. Rounding properly to the correct number of significant figures, the area should be written as A) 0.68 m2. B) 0.682 m2. C) 0.6818 m2. D) 0.68175 m2. Answer: B Diff: 1 Page Ref: Sec. 1.4 15) The length and width of a rectangle are 1.125 m and 0.606 m, respectively. You calculate the rectangle's perimeter by adding these and multiplying by two. Your calculator's display reads 3.462. To the correct number of significant figures, this should be written as A) 3.5 m. B) 3.46 m. C) 3.462 m. D) 3.4620 m. Answer: C Diff: 1 Page Ref: Sec. 1.4 16) A rectangle is 3.25 m long and 1.5 m wide. What is its area? A) 4.875 m2 B) 4.87 m2 C) 4.80 m2 D) 4.9 m2 Answer: D Diff: 2 Page Ref: Sec. 1.4 8 Chapter 1: Introduction, Measurement, Estimating 17) A rectangular garden measures 15 m long and 13.7 m wide. What is the length of a diagonal from one corner of the garden to the other? A) 18 m B) 19 m C) 20 m D) 4.1 ˛ 102 m Answer: C Diff: 2 Page Ref: Sec. 1.4 18) Select the smallest value. A) 15 ˛ 10-3 B) 0.15 ˛ 100 C) 0.00015 ˛ 103 D) 0.00000015 ˛ 106 Answer: A Diff: 1 Page Ref: Sec. 1.4 19) Write the number 0.00045 in power of ten notation. A) 4.5 ˛ 10-4 B) 4.5 ˛ 10-3 C) 4.5 ˛ 10-2 D) 4.5 ˛ 10-1 Answer: A Diff: 1 Page Ref: Sec. 1.4 20) 0.0001776 can also be expressed as A) 1.776 ˛ 10-4. B) 17.72 ˛ 104. C) 1772 ˛ 105. D) 177.2 ˛ 107. Answer: A Diff: 1 Page Ref: Sec. 1.4 21) 4567.89 is properly expressed in scientific notation as A) 4.56789 ˛ 103. B) 45.6789 ˛ 102. C) 456.789 ˛ 101. D) 4567.89 ˛ 100. Answer: A Diff: 1 Page Ref: Sec. 1.4 9 Physics: Principles with Applications, Sixth Edition 22) Convert 1.2 ˛ 10-3 to decimal notation. A) 1.200 B) 0.1200 C) 0.0120 D) 0.0012 Answer: D Diff: 1 Page Ref: Sec. 1.4 23) Write out the number 8.42 ˛ 10-5 in full with a decimal point and correct number of zeros. A) 0.00000842 B) 0.0000842 C) 0.000842 D) 0.00842 Answer: B Diff: 1 Page Ref: Sec. 1.4 24) What is the result of (0.410 + 0.021) ˛ (2.20 ˛ 103)? A) 880 B) 946 C) 948 D) 950 Answer: C Diff: 2 Page Ref: Sec. 1.4 25) Write the number 13.5 gigameters as full (decimal) numbers with standard units. A) 135,000 m B) 135,000,000 m C) 135,000,000,000 m D) 13,500,000,000 m Answer: D Diff: 1 Page Ref: Sec. 1.5-1.6 26) 100 mL is equivalent to which of the following? A) 1 kL B) 10-6 mL C) 0.1 L D) 0.01 ML Answer: C Diff: 1 Page Ref: Sec. 1.5-1.6 10 Chapter 1: Introduction, Measurement, Estimating 27) How many grams is forty milligrams? A) 0.000040 g B) 0.00040 g C) 0.040 g D) 40000 g Answer: C Diff: 1 Page Ref: Sec. 1.5-1.6 28) How many meters is sixty kilometers? A) 600,000 m B) 60,000 m C) 60 m D) 0.06 m Answer: B Diff: 1 Page Ref: Sec. 1.5-1.6 29) 1 angstrom = 10-10 m and 1 fermi = 10-15 m, what is the relationship between these units? A) 1 angstrom = 105 fermi B) 1 angstrom = 10-5 fermi C) 1 angstrom = 10-25 fermi D) 1 angstrom = 10+25 fermi Answer: A Diff: 1 Page Ref: Sec. 1.5-1.6 30) 0.00325 ˛ 10-8 cm can also be expressed in mm as A) 3.25 ˛ 10-12 mm. B) 3.25 ˛ 10-11 mm. C) 3.25 ˛ 10-10 mm. D) 3.25 ˛ 10-9 mm. Answer: C Diff: 2 Page Ref: Sec. 1.5-1.6 31) Which one of the following is not equivalent to 2.50 miles? (1 mi = 1.609 km = 5280 ft, 1 ft = 12 in.) A) 1.32 ˛ 104 ft B) 1.58 ˛ 105 in. C) 4.02 ˛ 103 km D) 4.40 ˛ 103 yd Answer: C Diff: 1 Page Ref: Sec. 1.5-1.6 [Show Less]
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Physics Principles With Applications 7th Edition Giancoli Test Bank (With Instructor's Solution Man
Physics Principles With Applications 7th Edition Giancoli Test Bank (With Instructor's Solution Manual) INTRODUCTION, MEASUREMENT, ESTIMATING 1 1-2 Chap... [Show More] ter 1 No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. (a) There are about 800,000 people in San Francisco, as estimated in 2009 by the U.S. Census Bureau. Assume that half of them have cars. If each of these 400,000 cars needs servicing twice a year, then there are 800,000 visits to mechanics in a year. If mechanics typically work 50 weeks a year, then about 16,000 cars would need to be seen each week. Assume that on average, a mechanic can work on 4 cars per day, or 20 cars a week. The final estimate, then, is 800 car mechanics in San Francisco. (b) Answers will vary. Responses to MisConceptual Questions 1. (d) One common misconception, as indicated by answers (b) and (c), is that digital measurements are inherently very accurate. A digital scale is only as accurate as the last digit that it displays. 2. (a) The total number of digits present does not determine the accuracy, as the leading zeros in (c) and (d) are only placeholders. Rewriting the measurements in scientific notation shows that (d) has two-digit accuracy, (b) and (c) have three-digit accuracy, and (a) has four-digit accuracy. Note that since the period is shown, the zeros to the right of the numbers are significant. 3. (b) The leading zeros are not significant. Rewriting this number in scientific notation shows that it only has two significant digits. 4. (b) When you add or subtract numbers, the final answer should contain no more decimal places than the number with the fewest decimal places. Since 25.2 has one decimal place, the answer must be rounded to one decimal place, or to 26.6. 5. (b) The word “accuracy” is commonly misused by beginning students. If a student repeats a measurement multiple times and obtains the same answer each time, it is often assumed to be accurate. In fact, students are frequently given an “ideal” number of times to repeat the experiment for “accuracy.” However, systematic errors may cause each measurement to be inaccurate. A poorly working instrument may also limit the accuracy of your measurement. 6. (d) This addresses misconceptions about squared units and about which factor should be in the numerator of the conversion. This error can be avoided when students treat the units as algebraic symbols that must be cancelled out. 7. (e) When making estimates, students frequently believe that their answers are more significant than they actually are. This question helps the student realize what an order-of-magnitude estimation is NOT supposed to accomplish. 8. (d) This addresses the fact that the generic unit symbol, like [L], does not indicate a specific system of units. Solutions to Problems 1. (a) 214 3 significant figures (b) 81 60 4 significant figures . (c) 7 03 3 significant figures . Introduction, Measurement, Estimating 1-3 © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. (d) 0 03 1 significant figure . (e) 0 0086 2 significant figures . (f) 3236 4 significant figures (g) 8700 2 significant figures 2. (a) 0 1 156 1.156 10 .= × (b) 1 21.8 2.18 10 = × (c) 3 0.0068 6.8 10− = × (d) 2 328.65 3.2865 10 = × (e) 1 0.219 2.19 10− = × (f) 2 444 4.44 10 = × 3. (a) 4 8.69 10 86,900 × = (b) 3 9.1 10 9100 × = (c) 1 8.8 10 0.88 − × = (d) 2 4.76 10 476 × = (e) 5 3.62 10 0.0000362 − × = 4. (a) 10 14 billion years 1.4 10 years = × (b) 7 10 17 3.156 10 s (1.4 10 yr) 4.4 10 s 1 yr ⎛ ⎞ × × = ⎜ ⎟ × ⎝ ⎠ 5. 0.25 m % uncertainty 100% 4.6% 5.48 m = ×= 6. (a) 0.2 s % uncertainty 100% 3.636% 4% 5.5 s =× = ≈ (b) 0.2 s % uncertainty 100% 3636% 0.4% 55 s = × = 0. ≈ (c) The time of 5.5 minutes is 330 seconds. 0.2 s % uncertainty 100% 0.0606% 0.06% 330 s =×= ≈ 1-4 Chapter 1 © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 7. To add values with significant figures, adjust all values to be added so that their exponents are all the same. 3 4 6 3 33 3 35 (9.2 10 s) (8.3 10 s) (0.008 10 s) (9.2 10 s) (83 10 s) (8 10 s) (9.2 83 8) 10 s 100.2 10 s 1.00 10 s × + × + × = × + × +× = + +× = × = × When you add, keep the least accurate value, so keep to the “ones” place in the last set of parentheses. 8. When you multiply, the result should have as many digits as the number with the least number of significant digits used in the calculation. 2 1 22 (3.079 10 m)(0.068 10 m) 2.094 m 2.1 m − × ×= ≈ 9. The uncertainty is taken to be 0.01 m. 2 2 0.01 m % uncertainty 100% 0.637% 1% 1.57 m = ×= ≈ 10. To find the approximate uncertainty in the volume, calculate the volume for the minimum radius and the volume for the maximum radius. Subtract the extreme volumes. The uncertainty in the volume is then half of this variation in volume. 4 4 3 33 specified specified 3 3 V r == = π π (0.84 m) 2.483 m 4 4 3 33 min min 3 3 4 4 3 33 max max 3 3 (0.80 m) 2.145 m (0.88 m) 2.855 m V r V r π π π π == = == = 1 1 33 3 2 2 max min Δ= − = − = VVV ( ) (2.855 m 2.145 m ) 0.355 m The percent uncertainty is 3 3 specified 0.355 m 100 14.3 14% . 2.483 m V V Δ = ×= ≈ 11. To find the approximate uncertainty in the area, calculate the area for the specified radius, the minimum radius, and the maximum radius. Subtract the extreme areas. The uncertainty in the area is then half this variation in area. The uncertainty in the radius is assumed to be 4 0.1 10 cm. × 2 4 2 9 2 specified specified A r = =× =× π π (3.1 10 cm) 3.019 10 cm 2 4 2 9 2 min min A r ==× =× π π (3.0 10 cm) 2.827 10 cm 2 4 2 9 2 max max A r = =× =× π π (3.2 10 cm) 3.217 10 cm 1 1 92 92 9 2 2 2 max min Δ= − = × − × = × AA A ( ) (3.217 10 cm 2.827 10 cm ) 0.195 10 cm Thus the area should be quoted as 9 2 A =±× (3.0 0.2) 10 cm . 12. (a) 286.6 mm 3 286.6 10 m − × 0.2866 m (b) 85 Vμ 6 85 10 V − × 0.000085 V (c) 760 mg 6 760 10 kg − × 0.00076 kg (if last zero is not significant) Introduction, Measurement, Estimating 1-5 © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. (d) 62.1 ps 12 62.1 10 s − × 0.0000000000621 s (e) 22.5 nm 9 22.5 10 m − × 0.0000000225 m (f) 2.50 gigavolts 9 2.50 10 volts × 2,500,000,000 volts Note that in part (f ) in particular, the correct number of significant digits cannot be determined when you write the number in this format. 13. (a) 6 1 10 volts × 1 megavolt 1 MV = (b) 6 2 10 meters − × 2 micrometers 2 m = μ (c) 3 6 10 days × 6 kilodays 6 kdays = (d) 2 18 10 bucks × 18 hectobucks 18 hbucks = or 1.8 kilobucks (e) 7 7 10 seconds − × 700 nanoseconds 700 ns or 0.7 s = μ 14. 2 4 2 4 2 1.000 10 m 3.281 ft 1 acre 1 hectare (1 hectare) 2.471 acres 1 hectare 1 m 4.356 10 ft ⎛ ⎞ × ⎛ ⎞ ⎛ ⎞ = = ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠ × 15. (a) 6 11 93 million miles (93 10 miles)(1610 m/1 mile) 1.5 10 m =× =× (b) 11 11 3 8 1.5 10 m (1.5 10 m)(1 km/10 m) 1.5 10 km × = × =× 16. To add values with significant figures, adjust all values to be added so that their units are all the same. 5 1.80 m 142.5 cm 5.34 10 m 1.80 m 1.425 m 0.534 m 3.759 m 3.76 m + +× = + + = = μ When you add, the final result is to be no more accurate than the least accurate number used. In this case, that is the first measurement, which is accurate to the hundredths place when expressed in meters. 17. (a) 10 10 9 1.0 10 m 1.0 10 m 39.37 in/1 m 3.9 10 in − − − × = ( × )( ) = × (b) 8 10 1 m 1 atom (1.0 cm) 1.0 10 atoms 100 cm 1.0 10 m − ⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟ = × ⎝ ⎠⎝ ⎠ × 18. (a) 0.621 mi (1 km/h) 0.621 mi/h, 1 km ⎛ ⎞ ⎜ ⎟ = ⎝ ⎠ so the conversion factor is 0.621 mi/h . 1 km/h (b) 3.28 ft (1 m/s) 3.28 ft/s, 1 m ⎛ ⎞ ⎜ ⎟ = ⎝ ⎠ so the conversion factor is 3.28 ft/s . 1 m/s (c) 1000 m 1 h (1 km/h) 0.278 m/s, 1 km 3600 s ⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟ = ⎝ ⎠⎝ ⎠ so the conversion factor is 0.278 m/s . 1 km/h 1-6 Chapter 1 No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Note that if more significant figures were used in the original factors, such as 0.6214 miles per kilometer, more significant figures could have been included in the answers. 19. (a) Find the distance by multiplying the speed by the time. 8 7 15 15 1.00 ly (2.998 10 m/s)(3.156 10 s) 9.462 10 m 9.46 10 m = × × = × ≈× (b) Do a unit conversion from ly to AU. 15 4 11 9.462 10 m 1 AU (1.00 ly) 6.31 10 AU 1.00 ly 1.50 10 m ⎛ ⎞ × ⎛ ⎞ ⎜ ⎟⎜ ⎟ = × ⎝ ⎠⎝ ⎠ × 20. One mile is 1609 m, according to the unit conversions in the front of the textbook. Thus it is 109 m longer than a 1500-m race. The percentage difference is calculated here. 109 m 100% 7.3% 1500 m × = 21. Since the meter is longer than the yard, the soccer field is longer than the football field. soccer football 1.094 yd 100.0 m 100.0 yd 9.4 yd 1 m A A −= × − = soccer football 1 m 100.0 m 100.0 yd 8.6 m 1.094 yd A A −= − × = Since the soccer field is 109.4 yd compared with the 100.0-yd football field, the soccer field is 9.4% longer than the football field. 22. (a) # of seconds in 1.00 yr: 7 3.156 10 s 7 1.00 yr (1.00 yr) 3.16 10 s 1 yr ⎛ ⎞ × = =× ⎜ ⎟ ⎝ ⎠ (b) # of nanoseconds in 1.00 yr: 7 9 3.156 10 s 1 10 ns 16 1.00 yr (1.00 yr) 3.16 10 ns 1 yr 1 s ⎛ ⎞ × × ⎛ ⎞ = = ⎜ ⎟⎜ ⎟ × ⎝ ⎠⎝ ⎠ (c) # of years in 1.00 s: 8 7 1 yr 1.00 s 1.00 s 3.17 10 yr 3.156 10 s − ⎛ ⎞ =( ) = × ⎜ ⎟ ⎝ ⎠ × 23. (a) 15 12 27 10 kg 1 proton or neutron 10 protons or neutrons 1 bacterium 10 kg − − ⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟ = ⎝ ⎠⎝ ⎠ (b) 17 10 27 10 kg 1 proton or neutron 10 protons or neutrons 1 DNA molecule 10 kg − − ⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟ = ⎜ ⎟ ⎝ ⎠⎝ ⎠ (c) 2 29 27 10 kg 1 proton or neutron 10 protons or neutrons 1 human 10 kg − ⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟ = ⎝ ⎠⎝ ⎠ (d) 41 68 27 10 kg 1 proton or neutron 10 protons or neutrons 1 Galaxy 10 kg − ⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟ = ⎝ ⎠⎝ ⎠ Introduction, Measurement, Estimating 1-7 No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 24. The radius of the ball can be found from the circumference (represented by “c” in the equations below), and then the volume can be found from the radius. Finally, the mass is found from the volume of the baseball multiplied by the density ( mass/volume) ρ = of a nucleon. ( ) 3 ball ball 4 4 3 ball ball ball ball ball 3 3 nucleon nucleon nucleon ball ball nucleon ball ball ball 4 3 3 4 1 nucleon 3 nucleon 3 2 nucleon 2 4 ball n 3 2 ; 2 2 2 c c c rr V r mm m mV V V V V r d c m π π π π π ρ π π π π ⎛ ⎞ = →= = = ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ == = = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ ( ) 3 3 ucleon ball 27 nucleon 3 15 4 1 nucleon 3 2 nucleon 14 14 0.23 m (10 kg) (10 m) 3.9 10 kg 4 10 kg c m d d π π π − − ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ = = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ = × ≈× 25. (a) 3 33 2800 2.8 10 1 10 10 = × ≈× = (b) 3 4 45 86.30 10 8.630 10 10 10 10 × = × ≈× = (c) 3 32 0.0076 7.6 10 10 10 10 − −− = × ≈× = (d) 8 9 99 15 0 10 1 5 10 1 10 10 .× =.× ≈× = 26. The textbook is approximately 25 cm deep and 5 cm wide. With books on both sides of a shelf, the shelf would need to be about 50 cm deep. If the aisle is 1.5 m wide, then about 1/4 of the floor space is covered by shelving. The number of books on a single shelf level is then 1 2 4 4 1 book (3500 m ) 7.0 10 books. (0.25 m)(0.05 m) ⎛ ⎞ ⎜ ⎟ = × ⎝ ⎠ With 8 shelves of books, the total number of books stored is as follows: ( ) 4 5 books 7 0 10 8 shelves 6 10 books shelf level ⎛ ⎞ .× ≈ × ⎜ ⎟ ⎝ ⎠ 27. The distance across the U.S. is about 3000 miles. (3000 mi)(1 km/0.621 mi)(1 h/10 km) 500 h ≈ Of course, it would take more time on the clock for a runner to run across the U.S. The runner obviously could not run for 500 hours non-stop. If he or she could run for 5 hours a day, then it would take about 100 days to cross the country. 28. A commonly accepted measure is that a person should drink eight 8-oz. glasses of water each day. That is about 2 quarts, or 2 liters of water per day. Approximate the lifetime as 70 years. 4 (70 yr)(365 d/1 yr)(2 L/1 d) 5 10 L ≈ × 29. An NCAA-regulation football field is 360 feet long (including the end zones) and 160 feet wide, which is about 110 meters by 50 meters, or 2 5500 m . We assume the mower has a cutting width of 1-8 Chapter 1 No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 0.5 meters and that a person mowing can walk at about 4.5 km/h, which is about 3 mi/h. Thus the distance to be walked is as follows: 2 area 5500 m 11000 m 11 km width 0.5 m d == = = At a speed of 4.5 km/h, it will take about 1 h 11 km 2.5 h 4.5 km × ≈ to mow the field. 30. There are about 8 3 10 × people in the U.S. Assume that half of them have cars, that they drive an average of 12,000 miles per year, and that their cars get an average of 20 miles per gallon of gasoline. 8 1 automobile 12,000 mi/auto 1 gallon 11 (3 10 people) 1 10 gal/yr 2 people 1 yr 20 mi ⎛ ⎞⎛ ⎞⎛ ⎞ × ≈ ⎜ ⎟⎜ ⎟⎜ ⎟ × ⎝ ⎠⎝ ⎠⎝ ⎠ 31. In estimating the number of dentists, the assumptions and estimates needed are: • the population of the city • the number of patients that a dentist sees in a day • the number of days that a dentist works in a year • the number of times that each person visits the dentist each year We estimate that a dentist can see 10 patients a day, that a dentist works 225 days a year, and that each person visits the dentist twice per year. (a) For San Francisco, the population as of 2010 was about 800,000 (according to the U.S. Census Bureau). The number of dentists is found by the following calculation: 5 2 visits/yr 1 yr 1 dentist (8 10 people) 700 dentists 1 person 225 workdays 10 visits/workday ⎛ ⎞⎛ ⎞⎛ ⎞ × ≈ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠ (b) For Marion, Indiana, the population is about 30,000. The number of dentists is found by a calculation similar to that in part (a), and would be about 30 dentists . There are about 40 dentists (of all types, including oral surgeons and orthodontists) listed in the 2012 Yellow Pages. 32. Consider the diagram shown (not to scale). The balloon is a distance h = 200 m above the surface of the Earth, and the tangent line from the balloon height to the surface of the Earth indicates the location of the horizon, a distance d away from the balloon. Use the Pythagorean theorem. ( ) 222 2 222 2 2 2 6 4 2 4 () 2 2 2 2(6.4 10 m)(200 m) 200 m 5.1 10 m 5 10 m ( 80 mi) r h r d r rh h r d rh h d d rh h d + =+ → + +=+ += →= + = × + = × ≈× ≈ 33. At $1,000 per day, you would earn $30,000 in the 30 days. With the other pay method, you would get 1 $0.01(2 ) t− on the tth day. On the first day, you get 1 1 $0.01(2 ) $0.01. − = On the second day, you get 2 1 $0.01(2 ) $0.02. − = On the third day, you get 3 1 $0.01(2 ) $0.04. − = On the 30th day, you get 30 1 6 $0.01(2 ) $5.4 10 , − = × which is over 5 million dollars. Get paid by the second method. h r r d Introduction, Measurement, Estimating 1-9 No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. To 1st sunset To 2nd sunset A B Earth center R R d h θ θ 34. In the figure in the textbook, the distance d is perpendicular to the radius that is drawn approximately vertically. Thus there is a right triangle, with legs of d and R, and a hypotenuse of R h + . Since 2 h R h Rh , 2. 2 22 2 2 2 2 2 2 2 6 () 2 2 2 2 (4400 m) 6.5 10 m 2(1.5 m) d d R R h R Rh h d Rh h d Rh R h +=+ =+ + = + ≈ = → →→ = =× A better measurement gives 6 R = × 6.38 10 m. 35. For you to see the Sun “disappear,” your line of sight to the top of the Sun must be tangent to the Earth’s surface. Initially, you are lying down at point A, and you see the first sunset. Then you stand up, elevating your eyes by the height h = 130 cm. While you stand, your line of sight is tangent to the Earth’s surface at point B, so that is the direction to the second sunset. The angle θ is the angle through which the Sun appears to move relative to the Earth during the time to be measured. The distance d is the distance from your eyes when standing to point B. Use the Pythagorean theorem for the following relationship: 22 2 2 2 2 2 d R R h R Rh h d Rh h + =+ = + + → = + () 2 2 The distance h is much smaller than the distance R, so 2 h Rh 2 which leads to 2 d Rh ≈ 2 . We also have from the same triangle that d R/ tan , = θ so d R = tan . θ Combining these two relationships gives 2 22 d Rh R ≈ = 2 tan , θ so 2 2 . tan h R θ = The angle θ can be found from the height change and the radius of the Earth. The elapsed time between the two sightings can then be found from the angle, because we know that a full revolution takes 24 hours. 1 1 2 o 2 6 o 2 o o o 2 2 2(1.3 m) tan tan (3 66 10 ) tan 6.38 10 m sec 3600 s 360 24 h 1 h 3600 s (3 66 10 ) 3600 s 24 h 24 h 8.8 s 360 1 h 360 1 h h h R R t t θ θ θ θ − − − − = → = = =.× × = → × ⎛ ⎞⎛⎞ ⎛⎞ ⎛ ⎞ . × = ×= ×= ⎜⎟ ⎜⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝⎠ ⎝⎠ ⎝ ⎠ 36. 3 mass units Density units volume units M L ⎡ ⎤ = = ⎢ ⎥ ⎣ ⎦ 1-10 Chapter 1 © Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 37. (a) For the equation 3 υ = − At Bt, the units of 3 At must be the same as the units of . υ So the units of A must be the same as the units of 3 υ/ , t which would be 4 L T/ . Also, the units of Bt must be the same as the units of . υ So the units of B must be the same as the units of υ/ ,t which would be 2 L T/ . (b) For A, the SI units would be 4 m/s , and for B, the SI units would be 2 m/s . 38. (a) The quantity 2 υt has units of 2 (m/s)(s ) m s, = i which do not match with the units of meters for x. The quantity 2at has units 2 (m/s )(s) m/s, = which also do not match with the units of meters for x. Thus this equation cannot be correct . (b) The quantity 0 υ t has units of (m/s)(s) m, = and 1 2 2 at has units of 2 2 (m/s )(s ) m. = Thus, since each term has units of meters, this equation can be correct . (c) The quantity 0 υ t has units of (m/s)(s) m, = and 2 2at has units of 2 2 (m/s )(s ) m. = Thus, since each term has units of meters, this equation can be correct . 39. Using the units on each of the fundamental constants (c, G, and h), we find the dimensions of the Planck length. We use the values given for the fundamental constants to find the value of the Planck length. [ ] 3 2 2 32 3 5 2 3 3 33 3 [ / ][ / ] [/] P Gh L MT ML T L L T M L L L c L T MT L L ⎡ ⎤ ⎡⎤ =→ = === ⎡ ⎤ ⎢ ⎥ ⎢⎥ ⎣ ⎦ ⎣ ⎦ ⎣⎦ A 11 3 2 34 2 35 3 8 3 (6.67 10 m /kg s )(6.63 10 kg m /s) 4 05 10 m (3.00 10 m/s) P Gh c − − × × − = = =. × × i i A Thus the order of magnitude is 35 10 m . − 40. The percentage accuracy is 5 7 2 m 100% 1 10 % . 2 10 m − × =× × The distance of 20,000,000 m needs to be distinguishable from 20,000,002 m, which means that 8 significant figures are needed in the distance measurements. 41. Multiply the number of chips per wafer by the number of wafers that can be made from a cylinder. We assume the number of chips per wafer is more accurate than 1 significant figure. chips 1 wafer 250 mm 5 chips 400 3.3 10 wafer 0.300 mm 1 cylinder cylinder ⎛ ⎞⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟ = × ⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠ 42. Assume that the alveoli are spherical and that the volume of a typical human lung is about 2 liters, which is 3 0.002 m . The diameter can be found from the volume of a sphere, 4 3 3 π r . Introduction, Measurement, Estimating 1-11 No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 3 4 4 3 3 3 3 1 3 3 3 8 33 3 4 8 ( /2) 6 6(2 10 ) (3 10 ) 2 10 m m 2 10 m 6 3 10 / d r d d d π π π π π − − − = = ⎡ ⎤ × × =× → = = × ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ × 43. We assume that there are 40 hours of work per week and that the typist works 50 weeks out of the year. 12 1 char 1 min 1 hour 1 week 1 year 4 (1.0 10 bytes) 4.629 10 years 1 byte 180 char 60 min 40 hour 50 weeks 46,000 years × × × × × × =× ≈ 44. The volume of water used by the people can be calculated as follows: 3 3 4 3 3 5 1200 L/day 365 days 1000 cm 1 km (4 10 people) 4.38 10 km /yr 4 people 1 yr 1 L 10 cm − ⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞ × = ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ × ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠ The depth of water is found by dividing the volume by the area. 3 3 5 5 2 4.38 10 km /yr km 10 cm 8.76 10 8.76 cm/yr 9 cm/yr 50 km yr 1 km V d A − − × ⎛ ⎞⎛ ⎞ == = × = ≈ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ 45. We approximate the jar as a cylinder with a uniform cross-sectional area. In counting the jelly beans in the top layer, we find about 25 jelly beans. Thus we estimate that one layer contains about 25 jelly beans. In counting vertically, we see that there are about 15 rows. Thus we estimate that there are 25 15 375 400 jelly beans ×= ≈ in the jar. 46. The volume of a sphere is given by 4 3 3 V r = π , so the radius is 1/3 3 . 4 V r π ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ For a 1-ton rock, the volume is calculated from the density, and then the diameter from the volume. 3 2000 lb 1 ft 3 (1 T) 10.8 ft 1 T 186 lb V ⎛ ⎞⎛ ⎞ = = ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠ 1/3 1/3 3 3 3(10.8 ft ) 2 2 2 2.74 ft 3 ft 4 4 V d r π π ⎡ ⎤ ⎛ ⎞ == = = ≈ ⎜ ⎟ ⎢ ⎥ ⎝ ⎠ ⎢ ⎥ ⎣ ⎦ 47. We do a “units conversion” from bytes to minutes, using the given CD reading rate. 6 6 8 bits 1 s 1 min (783.216 10 bytes) 74.592 min 75 min 1 byte 60 s 1.4 10 bits × ×× ×= ≈ × 48. A pencil has a diameter of about 0.7 cm. If held about 0.75 m from the eye, it can just block out the Moon. The ratio of pencil diameter to arm length is the same as the ratio of Moon diameter to Moon distance. From the diagram, we have the following ratios. [Show Less]
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