CHAPTER 1: Introduction, Measurement, Estimating Answers to Questions 1. (a) Fundamental standards should be accessible, invariable, indestructible, and reproducible. A particular person’s foot would not be very accessible, since the person could not be at more than one place at a time. The standard would be somewhat invariable if the person were an adult, but even then, due to swelling or injury, the length of the standard foot could change. The standard would not be indestructible – the foot would not last forever. The standard could be reproducible – tracings or plaster casts could be made as secondary standards. (b) If any person’s foot were to be used as a standard, “standard” would vary significantly depending on the person whose foot happened to be used most recently for a measurement. The standard would be very accessible, because wherever a measurement was needed, it would be very easy to find someone with feet. The standard would be extremely variable – perhaps by a factor of 2. That also renders the standard as not reproducible, because there could be many reproductions that were quite different from each other. The standard would be almost indestructible in that there is essentially a limitless supply of feet to be used. 2. There are various ways to alter the signs. The number of meters could be expressed in one significant figure, as “900 m (3000 ft)”. Or, the number of feet could be expressed with the same precision as the number of meters, as “914 m (2999 ft)”. The signs could also be moved to different locations, where the number of meters was more exact. For example, if a sign was placed where the elevation was really 1000 m to the nearest meter, then the sign could read “1000 m (3280 ft)”. 3. Including more digits in an answer does not necessarily increase its accuracy. The accuracy of an answer is determined by the accuracy of the physical measurement on which the answer is based. If you draw a circle, measure its diameter to be 168 mm and its circumference to be 527 mm, their quotient, representing , is 3.136904762. The last seven digits are meaningless – they imply a greater accuracy than is possible with the measurements. 4. The problem is that the precision of the two measurements are quite different. It would be more appropriate to give the metric distance as 11 km, so that the numbers are given to about the same precision (nearest mile or nearest km). 5. A measurement must be measured against a scale, and the units provide that scale. Units must be specified or the answer is meaningless – the answer could mean a variety of quantities, and could be interpreted in a variety of ways. Some units are understood, such as when you ask someone how old they are. You assume their answer is in years. But if you ask someone how long it will be until they are done with their task, and they answer “five”, does that mean five minutes or five hours or five days? If you are in an international airport, and you ask the price of some object, what does the answer “ten” mean? Ten dollars, or ten pounds, or ten marks, or ten euros? 6. If the jar is rectangular, for example, you could count the number of marbles along each dimension, and then multiply those three numbers together for an estimate of the total number of marbles. If the jar is cylindrical, you could count the marbles in one cross section, and then multiply by the number of layers of marbles. Another approach would be to estimate the volume of one marble. If we assume that the marbles are stacked such that their centers are all on vertical and horizontal lines, then each marble would require a cube of edge 2R, or a volume of 8R3, where R is the radius of a marble. The number of marbles would then be the volume of the container divided by 8R3. © 2005 Pearson Education, Inc., Upper Saddle River, NJ. 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. 1 Chapter 1 Introduction, Measurement, Estimating 7. The result should be written as 8.32 cm. The factor of 2 used to convert radius to diameter is exact – it has no uncertainty, and so does not change the number of significant figures. 8. sin30.0o 0.500 9. Since the size of large eggs can vary by 10%, the random large egg used in a recipe has a size with an uncertainty of about 5%. Thus the amount of the other ingredients can also vary by about 5% and not adversely affect the recipe. 10. In estimating the number of car mechanics, the assumptions and estimates needed are: the population of the city the number of cars per person in the city the number of cars that a mechanic can repair in a day the number of days that a mechanic works in a year the number of times that a car is taken to a mechanic, per year We estimate that there is 1 car for every 2 people, that a mechanic can repair 3 cars per day, that a mechanic works 250 days a year, and that a car needs to be repaired twice per year. (a) For San Francisco, we estimate the population at one million people. The number of mechanics is found by the following calculation. repairs 2 1 car year 1 yr 1 mechanic 1 106people 1300 mechanics 2 people 1 car 250 workdays repairs 3 workday (b) For Upland, Indiana, the population is about 4000. The number of mechanics is found by a similar calculation, and would be 5 mechanics. There are actually two repair shops in Upland, employing a total of 6 mechanics. Solutions to Problems 1. (a) 14 billion years 1.4 1010years (b) 1.4 1010y 3.156 107s 1 y 4.4 1017s 2. (a) 214 3 significant figures (b) 81.60 4 significant figures (c) 7.03 3 significant figures (d) 0.03 1 significant figure (e) 0.0086 2 significant figures (f) 3236 4 significant figures (g) 8700 2 significant figures © 2005 Pearson Education, Inc., Upper Saddle River, NJ. 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. 2 Giancoli Physics: Principles with Applications, 6th Edition 3. (a) 1.156 1.156 100 (b) 21.8 2.18 101 (c) 0.0068 6.8 10 3 (d) 27.635 2.7635 101 (e) 0.219 2.19 10 1 (f) 444 4.44 102 4. (a) 8.69 104 86,900 (b) 9.1 103 9,100 (c) 8.8 10 1 0.88 (d) 4.76 102 476 (e) 3.62 10 5 0.0000362 5. The uncertainty is taken to be 0.01 m. 0.01 m % uncertainty 100% 1% 1.57 m 0.25 m 6. % uncertainty 100% 6.6% 3.76 m 0.2 s 7. (a) % uncertainty 100% 4% 5 s 0.2 s (b) % uncertainty 100% 0.4% 50 s 0.2 s (c) % uncertainty 100% 0.07% 300 s 8. To add values with significant figures, adjust all values to be added so that their exponents are all the same. 9.2 103s 8.3 104s 0.008 106s 9.2 103s 83 103s 8 103s 9.2 83 8 103s 100 103s 1.00 105s When adding, keep the least accurate value, and so keep to the “ones” place in the parentheses. 9. 2.079 102m 0.082 10 1 1.7 m . When multiplying, the result should have as many digits as the number with the least number of significant digits used in the calculation. 10. 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 0.1 104cm. © 2005 Pearson Education, Inc., Upper Saddle River, NJ. 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. 3 Chapter 1 Introduction, Measurement, Estimating A r2 3.8 104cm 2 4.5 109cm2 specified specified A r2 3.7 104cm 2 4.30 109cm2 min min A r2 3.9 104cm 2 4.78 109cm2 max max A 1 A A 1 4.78 109cm2 4.30 109cm2 0.24 109cm2 2 max min 2 Thus the area should be quoted as A 4.5 0.2 109cm2 11. To find the approximate uncertainty in the volume, calculate the volume for the specified radius, the minimum radius, and the maximum radius. Subtract the extreme volumes. The uncertainty in the volume is then half this variation in volume. V 4 r3 4 2.86 m 3 9.80 101m3 specified 3 specified 3 V 4 r3 4 2.77 m 3 8.903 101m3 min 3 min 3 V 4 r3 4 2.95 m 3 10.754 101m3 max 3 max 3 V 1 V V 1 10.754 101m3 8.903 101m3 0.926 101m3 2 max min 2 V 0.923 101m3 The percent uncertainty is 100 0.09444 9% V 9.80 101m3 specified 12. (a) 286.6 mm 286.6 10 3m 0.286 6 m (b) 85 V 85 10 6V 0.000 085 V (c) 760 mg 760 10 6kg 0.000 760 kg (if last zero is significant) (d) 60.0 ps 60.0 10 12s 0.000 000 000 0600 s (e) 22.5 fm 22.5 10 15m 0.000 000 000 000 022 5 m (f) 2.50 gigavolts 2.5 109volts 2,500,000,000 volts 13. (a) 1 106volts 1 megavolt 1 Mvolt (b) 2 10 6meters 2 micrometers 2 m (c) 6 103days 6 kilodays 6 kdays (d) 18 102bucks 18 hectobucks 18 hbucks (e) 8 10 9pieces 8 nanopieces 8 npieces 14. (a) Assuming a height of 5 feet 10 inches, then 5'10" 70 in 1 m 39.37 in 1.8 m (b) Assuming a weight of 165 lbs, then 165 lbs 0.456 kg 1 lb 75.2 kg Technically, pounds and mass measure two separate properties. To make this conversion, we have to assume that we are at a location where the acceleration due to gravity is 9.8 m/s2. © 2005 Pearson Education, Inc., Upper Saddle River, NJ. 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. 4 Giancoli Physics: Principles with Applications, 6th Edition 15. (a) 93 million miles 93 106miles 1610 m 1 mile 1.5 1011m (b) 1.5 1011m 150 109m 150 gigameters or 1.5 1011m 0.15 1012m 0.15 terameters 16. (a) 1 ft2 1 ft2 1 yd 3 ft 2 0.111 yd2 (b) 1 m2 1 m2 3.28 ft 1 m 2 10.8 ft2 17. Use the speed of the airplane to convert the travel distance into a time. 1 h 3600 s 1.00 km 3.8 s 950 km 1 h 18. (a) 1.0 10 10m 1.0 10 10m 39.37 in 1 m 3.9 10 9in 1 m 1 atom (b) 1.0 cm 1.0 108atoms 100 cm 1.0 10 10m 19. To add values with significant figures, adjust all values to be added so that their units are all the same. 1.80 m 142.5 cm 5.34 105 m 1.80 m 1.425 m 0.534 m 3.759 m 3.76 m When adding, 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. 0.621 mi 20. (a) 1k h 0.621mi h 1 km 3.28 ft (b) 1m s 3.28ft s 1 m 1000 m 1 h (c) 1km h 0.278m s 1 km 3600 s 21. One mile is 1.61 103m. It is 110 m longer than a 1500-m race. The percentage difference is 110 m 100% 7.3% 1500 m 22. (a) 1.00 ly 2.998 108m s 3.156 107s 9.46 1015m 9.462 1015m 1 AU (b) 1.00 ly 6.31 104AU 1.00 ly 1.50 1011m 1 AU 3600 s (c) 2.998 108m s 7.20AU h 1.50 1011m 1 hr © 2005 Pearson Education, Inc., Upper Saddle River, NJ. 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. 5 Chapter 1 Introduction, Measurement, Estimating 23. The surface area of a sphere is found by A 4 r2 4 d 2 2 d2. (a) A D2 3.48 106m 2 3.80 1013m2 Moon Moon A D2 D 2 R 2 6.38 106m 2 (b) Earth Earth Earth Earth 13.4 A D2 D R 1.74 106m Moon Moon Moon Moon 24. (a) 2800 2.8 103 1 103 103 (b) 86.30 102 8.630 103 10 103 104 (c) 0.0076 7.6 10 3 10 10 3 10 2 (d) 15.0 108 1.5 109 1 109 109 25. The textbook is approximately 20 cm deep and 4 cm wide. With books on both sides of a shelf, with a little extra space, the shelf would need to be about 50 cm deep. If the aisle is 1.5 meter wide, then about 1/4 of the floor space is covered by shelving. The number of books on a single shelf level is 1 book then 1 3500m2 8.75 104books. With 8 shelves of books, the total number 4 0.25 m 0.04 m of books stored is as follows. books 8.75 104 8 shelves 7 10 5 books . shelf level 26. The distance across the United States is about 3000 miles. 3000 mi 1 km 0.621 mi 1 hr 10 km 500 hr Of course, it would take more time on the clock for the runner to run across the U.S. The runner could obviously not run for 500 hours non-stop. If they could run for 5 hours a day, then it would take about 100 days for them to cross the country. 27. 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 5,500 m2. The mower has a cutting width of 0.5 meters. Thus the distance to be walked is Area 5500m2 d 11000 m 11 km width 0.5 m At a speed of 1 km/hr, then it will take about 11 h to mow the field. 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. Then approximate the lifetime as 70 years. 70 y 365 d 1 y 2 L 1 d 5 104L 29. Consider the body to be a cylinder, about 170 cm tall, and about 12 cm in cross-sectional radius (a 30-inch waist). The volume of a cylinder is given by the area of the cross section times the height. V r2h 12 cm 2 170 cm 9 104cm3 8 104cm3 © 2005 Pearson Education, Inc., Upper Saddle River, NJ. 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. 6 Giancoli Physics: Principles with Applications, 6th Edition 30. Estimate one side of a house to be about 40 feet long, and about 10 feet high. Then the wall area of that particular wall is 400 ft2. There would perhaps be 4 windows in that wall, each about 3 ft wide and 4 feet tall, so 12 ft2 per window, or about 50 ft2 of window per wall. Thus the percentage of wall 50 ft2 area that is window area is 100 12.5%. Thus a rough estimate would be 10% 15% of 400 ft2 the house’s outside wall area. 31. Assume that the tires last for 5 years, and so there is a tread wearing of 0.2 cm/year. Assume the average tire has a radius of 40 cm, and a width of 10 cm. Thus the volume of rubber that is becoming pollution each year from one tire is the surface area of the tire, times the thickness per year that is wearing. Also assume that there are 150,000,000 automobiles in the country – approximately one automobile for every two people. So the mass wear per year is given by Mass Surface area Thickness wear density of rubber # of tires year tire year 2 0.4 m 0.1 m 0.002 m y 1200kg m3 600,000,000 tires 4 108kg y 32. For the equation v At3 Bt, the units of At3 must be the same as the units of v. So the units of A must be the same as the units of v t3 , which would be distance time4 . Also, the units of Bt must be the same as the units of v. So the units of B must be the same as the units of v t, which would be distance time2 . 33. (a) The quantity vt2 has units of m s s2 m s, which do not match with the units of meters for x. The quantity 2at has units m s2 s m s, which also do not match with the units of meters for x. Thus this equation cannot be correct . (b) The quantity v t has units of m s s m, and 1at2 has units of m s2 s2 m. Thus, 0 2 since each term has units of meters, this equation can be correct . (c) The quantity v t has units of m s s m, and 2at2 has units of m s2 s2 m. Thus, 0 since each term has units of meters, this equation can be correct . 2 m 34. The percentage accuracy is 100% 1 10 5% . The distance of 20,000,000 m needs to 2 107m be distinguishable from 20,000,002 m, which means that 8 significant figures are needed in the distance measurements. 35. Multiply the number of chips per wafer times the number of wafers that can be made fro a cylinder. chips 1 wafer 300 mm chips 100 50,000 wafer 0.60 mm 1 cylinder cylinder © 2005 Pearson Education, Inc., Upper Saddle River, NJ. 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 Chapter 1 Introduction, Measurement, Estimating 3.156 107s 36. (a) # of seconds in 1.00 y: 1.00 y 1.00 y 3.16 107s 1 y 3.156 107s 1 109ns (b) # of nanoseconds in 1.00 y: 1.00 y 1.00 y 3.16 1016ns 1 y 1 s 1 y (c) # of years in 1.00 s: 1.00 s 1.00 s 3.17 10 8y 3.156 107s 37. Assume that the alveoli are spherical, and that the volume of a typical human lung is about 2 liters, which is .002 m3. The diameter can be found from the volume of a sphere, 4 r3. 3 d3 4 r3 4 d 2 3 3 3 6 . 1/3 d3 6 2 10 3 3 108 2 10 3m 3 d m3 2 10 4m 6 3 108 104m2 3.28 ft 2 1 acre 38. 1 hectare 1 hectare 2.69 acres 1 hectare 1 m 4 104ft2 10 15kg 1 proton or neutron 39. (a) 1012protons or neutrons 1 bacterium 10 27kg 10 17kg 1 proton or neutron (b) 1010protons or neutrons 1 DNA molecule 10 27kg 102kg 1 proton or neutron (c) 1029protons or neutrons 1 human 10 27kg 1041kg 1 proton or neutron (d) 1068protons or neutrons 1 galaxy 10 27kg 40. There are about 300,000,000 people in the United States. Assume that half of them have cars, that they each drive 12,000 miles per year, and their cars get 20 miles per gallon of gasoline. 1 automobile 12,000 mi 1 gallon 3 108people 1 1011gallons y 2 people 1 y 20 mi 41. Approximate the gumball machine as a rectangular box with a square cross-sectional area. In counting gumballs across the bottom, there are about 10 in a row. Thus we estimate that one layer contains about 100 gumballs. In counting vertically, we see that there are bout 15 rows. Thus we estimate that there are about 1500 gumballs in the machine. 42. The volume of water used by the people can be calculated as follows: 1200 L day 365 day 1000 cm3 1 km 3 4 104people 4.4 10 3km3 y 4 people 1 y 1 L 10 5 cm © 2005 Pearson Education, Inc., Upper Saddle River, NJ. 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. 8 Giancoli Physics: Principles with Applications, 6th Edition The depth of water is found by dividing the volume by the area. V 4.4 10 3km3 y km 105cm d 8.76 10 5 8.76cm y 9cm y A 50 km2 y 1 km 43. The volume of a sphere is given by V 4 r3. For our 1-ton rock, we can calculate the volume to be 3 2000 lb 1ft3 V 1 T 10.8 ft3. 1 T 186 lb Then the radius is found by 1/3 3V 1/3 3 10.8 ft3 d 2r 2 2 2.74 ft 3 ft 4 4 44. To calculate the mass of water, we need to find the volume of water, and then convert the volume to mass. 105cm 2 10 3kg 1 ton 4 101 km2 1.0 cm 4 105ton 1 km 1 cm3 103kg To find the number of gallons, convert the volume to gallons. 105cm 2 1 L 1 gal 4 101 km2 1.0 cm 1 108gal 1 km 1 103cm3 3.78 L 45. 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. Pencil Moon Pencil Distance Moon Distance Pencil diameter Moon diameter Pencil distance Moon distance pencil diameter 7 10 3m Moon diameter Moon distance 3.8 105km 3500 km pencil distance 0.75 m 46. The person walks 4km h, 10 hours each day. The radius of the Earth is about 6380 km, and the distance around the world at the equator is the circumference, 2 R . We assume that the person Earth can “walk on water”, and so ignore the existence of the oceans. 1 h 1 d 2 6380 km 1 10 3 d 4 km 10 h © 2005 Pearson Education, Inc., Upper Saddle River, NJ. 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. 9 Chapter 1 Introduction, Measurement, Estimating 47. A cubit is about a half of a meter, by measuring several people’s forearms. Thus the dimensions of Noah’s ark would be 150 m long , 25 m wide, 15 m high . The volume of the ark is found by multiplying the three dimensions. V 150 m 25 m 15 m 5.625 104m3 6 104m3 48. The volume of the oil will be the area times the thickness. The area is r2 d 2 2, and so 1 m 3 1000cm3 V 100 cm V d 2 2 t d 2 2 3 10 3 m . t 2 10 10m 49. Consider the diagram shown. L is the distance she walks upstream, which is about 120 yards. Find the distance across the river from the diagram. d tan60 o d Ltan60o 120 yd tan60o 210 yd d L 3 ft 0.305 m 60o 210 yd 190 m 1 yd 1 ft L 8 s 1 y 50. 100% 3 10 5% 1 y 3.156 107s 51. The volume of a sphere is found by V 4 r3. 3 V 4 R3 4 1.74 106m 3 2.21 1019m3 Moon 3 Moon 3 V 4 R3 R 3 6.38 106m 3 Earth 3 Earth Earth 49.3. V 4 R3 R 1.74 106m Moon 3 Moon Moon Thus it would take about 49.3 Moons to create a volume equal to that of the Earth. o o 10 10m 1 nm 52. (a) 1.0A 1.0A 0.10 nm o 10 9m 1A o o 10 10m 1 fm (b) 1.0A 1.0A 1.0 105fm o 10 15m 1A o 1A o (c) 1.0 m 1.0 m 1.0 1010A 10 10m o 9.46 1015m 1A o (d) 1.0 ly 1.0 ly 9.5 1025A 1 ly 10 10m © 2005 Pearson Education, Inc., Upper Saddle River, NJ. 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. 10