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

Exam (elaborations) TEST BANK FOR University Phyiscs 12th Edition By Hugh D. Young, Roger A. Freedman, and Lewis Ford (Solution manual) UNITS, PHYSICAL QUANTITIES AND VECTORS 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 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 N =V /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 1 L =1000 cm3 and 1 cm =10 mm. EXECUTE: The volume of the kernel is: ( )( )( ) 3 kernel V = 10 mm 6 mm 3 mm =180 mm . The bottle’s volume is: ( ) ( 3 ) ( ) ( )3 ( )3 6 3 bottle V = 2.0 L ⎡⎣ 1000 cm 1.0 L ⎤⎦ ⎡⎣ 10 mm 1.0 cm ⎤⎦ = 2.0×10 mm . The number of kernels is then ( 6 3 ) ( 3 ) kernels bottle kernels N =V /V ≈ 2.0×10 mm 180 mm =11,000 kernels . EVALUATE: This estimate is highly dependent upon your estimate of the kernel dimensions. And since these dimensions vary amongst the different available types 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 106 . 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 cm3 to m3 to find the volume in m3 breathed in a year. SET UP: Assume 10 breaths/min . 1 y (365 d) 24 h 60 min 5.3 105 min 1 d 1 h = ⎛ ⎞⎛ ⎞ = × ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ . 102 cm =1 m so 106 cm3 =1 m3 . The volume of a sphere is 4 3 1 3 3 6 V = π r = π d , where r is the radius and d is the diameter. Don’t forget to account for four astronauts. EXECUTE: (a) The volume is 5 (4)(10 breaths/min)(500 10 6 m3 ) 5.3 10 min 1 104 m3 / yr 1 y − ⎛ × ⎞ × ⎜ ⎟ = × ⎝ ⎠ . (b) 6 1/ 3 6[1 104 m3 ] 1/ 3 d V 27 m π π ⎛ ⎞ ⎛ × ⎞ = ⎜ ⎟ = ⎜ ⎟ = ⎝ ⎠ ⎝ ⎠ EVALUATE: Our estimate assumes that each cm3 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 minutes 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 (10 per min) 60 min 24 h 365 days (80 y/lifetime) 4 108 1 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 beats 75 beats/min 60 min 24 h 365 days 80 yr 3 10 beats/lifespan 1 h 1 day yr lifespan N = ⎛⎜ ⎞⎟⎜⎛ ⎞⎟⎛⎜ ⎞⎟⎛⎜ ⎞⎟ = × ⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠ ( ) 9 3 7 blood 3 50 cm /beat 1 L 1 gal 3 10 beats 4 10 gal/lifespan 1000 cm 3.788 L lifespan V ⎛ ⎞⎛ ⎞⎛ × ⎞ = ⎜ ⎟⎜ ⎟⎜ ⎟ = × ⎝ ⎠⎝ ⎠⎝ ⎠ EVALUATE: This is a very large volume. Units, Physical Quantities and Vectors 1-5 1.25. IDENTIFY: Estimation problem SET UP: Estimate that the pile is 18 in.×18 in.×5 ft 8 in.. Use the density of gold to calculate the mass of gold in the pile and from this calculate the dollar value. EXECUTE: The volume of gold in the pile is V =18 in.×18 in.× 68 in. = 22,000 in.3. Convert to cm3 : V = 22,000 in.3(1000 cm3 / 61.02 in.3 ) = 3.6×105 cm3. The density of gold is 19.3 g/cm3, so the mass of this volume of gold is m = (19.3 g/cm3)(3.6×105 cm3 ) = 7×106 g. The monetary value of one gram is $10, so the gold has a value of ($10/ gram)(7×106 grams) = $7×107 , or about $100×106 (one hundred million dollars). EVALUATE: This is quite a large pile of gold, so such a large monetary value is reasonable. 1.26. IDENTIFY: Estimate the diameter of a drop and from that calculate the volume of a drop, in m3 . Convert m3 to L. SET UP: Estimate the diameter of a drop to be d = 2 mm. The volume of a spherical drop is 4 3 1 3 3 6 V = π r = π d . 103 cm3 =1 L . EXECUTE: 1 3 3 3 6 V = π (0.2 cm) = 4×10− cm . The number of drops in 1.0 L is 3 5 3 3 1000 cm 2 10 4 10− cm = × × EVALUATE: Since V ∼ d3 , if our estimate of the diameter of a drop is off by a factor of 2 then our estimate of the number of drops is off by a factor of 8. 1.27. IDENTIFY: Estimate the number of students and the average number of pizzas eaten by each student in a school year. SET UP: Assume a school of thousand students, each of whom averages ten pizzas a year (perhaps an underestimate) EXECUTE: They eat a total of 104 pizzas. EVALUATE: The same answer applies to a school of 250 students averaging 40 pizzas a year each. 1.28. IDENTIFY: The number of bills is the distance to the moon divided by the thickness of one bill. SET UP: Estimate the thickness of a dollar bills by measuring a short stack, say ten, and dividing the measurement by the total number of bills. I obtain a thickness of roughly 1 mm. From Appendix F, the distance from the earth to the moon is 3.8×108 m. EXECUTE: 8 3 12 12 bills 3.8 10 m 10 mm 3.8 10 bills 4 10 bills 0.1 mm/bill 1 m N ⎛ × ⎞⎛ ⎞ = ⎜ ⎟⎜ ⎟ = × ≈ × ⎝ ⎠⎝ ⎠ EVALUATE: This answer represents 4 trillion dollars! The cost of a single space shuttle mission in 2005 is significantly less – roughly 1 billion dollars. 1.29. IDENTIFY: The cost would equal the number of dollar bills required; the surface area of the U.S. divided by the surface area of a single dollar bill. SET UP: By drawing a rectangle on a map of the U.S., the approximate area is 2600 mi by 1300 mi or 3,380,000 mi2 . This estimate is within 10 percent of the actual area, 3,794,083 mi2 . The population is roughly 8 3.0 10 × while the area of a dollar bill, as measured with a ruler, is approximately 18 6 in. by 5 8 2 in. EXECUTE: ( 2 )[( ) ( )]2 ( ) ( ) 2 16 2 U.S. A = 3,380,000 mi 5280 ft / 1 mi ⎡⎣ 12 in. 1 ft ⎤⎦ =1.4×10 in. ( )( ) 2 bill A = 6.125 in. 2.625 in. =16.1 in. ( 16 2 ) ( 2 ) 14 bills U.S. bill Total cost = N = A A = 1.4×10 in. 16.1 in. / bill = 9×10 bills Cost per person = (9×1014 dollars) /(3.0×108 persons) = 3×106dollars/person EVALUATE: The actual cost would be somewhat larger, because the land isn’t flat. 1.30. IDENTIFY: The displacements must be added as vectors and the magnitude of the sum depends on the relative orientation of the two displacements. SET UP: The sum with the largest magnitude is when the two displacements are parallel and the sum with the smallest magnitude is when the two displacements are antiparallel. EXECUTE: The orientations of the displacements that give the desired sum are shown in Figure 1.30. EVALUATE: The orientations of the two displacements can be chosen such that the sum has any value between 0.6 m and 4.2 m. Figure 1.30 1-6 Chapter 1 1.31. IDENTIFY: Draw each subsequent displacement tail to head with the previous displacement. The resultant displacement is the single vector that points from the starting point to the stopping point. SET UP: Call the three displacements A " , B " , and C " . The resultant displacement R " is given by R= A+ B+C " " " " . EXECUTE: The vector addition diagram is given in Figure 1.31. Careful measurement gives that R " is 7.8 km, 38# north of east . EVALUATE: The magnitude of the resultant displacement, 7.8 km, is less than the sum of the magnitudes of the individual displacements, 2.6 km+ 4.0 km + 3.1 km . Figure 1.31 1.32. IDENTIFY: Draw the vector addition diagram, so scale. SET UP: The two vectors A " and B " are specified in the figure that accompanies the problem. EXECUTE: (a) The diagram for C = A+ B " " " is given in Figure 1.32a. Measuring the length and angle of C " gives C = 9.0 m and an angle of θ = 34° . (b) The diagram for D = A− B " " " is given in Figure 1.32b. Measuring the length and angle of D " gives D = 22 mand an angle of θ = 250° . (c) −A− B = −(A+ B) " " , so −A− B " " has a magnitude of 9.0 m (the same as A+ B " " ) and an angle with the +x axis of 214° (opposite to the direction of A+ B " " ). (d) B − A = −(A− B) " " " " , so − B A " " has a magnitude of 22 m and an angle with the +x axis of 70° (opposite to the direction of A− B " " ). EVALUATE: The vector −A " is equal in magnitude and opposite in direction to the vector A " . Figure 1.32 1.33. IDENTIFY: Since she returns to the starting point, the vectors sum of the four displacements must be zero. SET UP: Call the three given displacements A " , B " , and C " , and call the fourth displacement D " . A+ B + C + D = 0 " " " " . EXECUTE: The vector addition diagram is sketched in Figure 1.33. Careful measurement gives that D " is144 m, 41# south of west. Units, Physical Quantities and Vectors 1-7 EVALUATE: D " is equal in magnitude and opposite in direction to the sum A+ B+C " " " . Figure 1.33 1.34. IDENTIFY and SET UP: Use a ruler and protractor to draw the vectors described. Then draw the corresponding horizontal and vertical components. EXECUTE: (a) Figure 1.34 gives components 4.7 m, 8.1 m. (b) Figure 1.34 gives components −15.6 km,15.6 km . (c) Figure 1.34 gives components 3.82 cm, − 5.07 cm . EVALUATE: The signs of the components depend on the quadrant in which the vector lies. Figure 1.34 1.35. IDENTIFY: For each vector V " , use that cos x V =V θ and sin y V =V θ , when θ is the angle V " makes with the +x axis, measured counterclockwise from the axis. SET UP: For A " , 270.0 θ = ° . For B " , θ = 60.0° . For C " , 205.0 θ = ° . For D " , θ =143.0° . EXECUTE: 0 x A = , 8.00 m y A = − . 7.50 m x B = , 13.0 m y B = . 10.9 m x C = − , 5.07 m y C = − . 7.99 m x D = − , 6.02 m y D = . EVALUATE: The signs of the components correspond to the quadrant in which the vector lies. 1.36. IDENTIFY: tan y x A A θ = , for θ measured counterclockwise from the +x -axis. SET UP: A sketch of x A , y A and A " tells us the quadrant in which A " lies. EXECUTE: (a) tan 1.00 m 0.500 2.00 m y X A θ A − = = = − . θ = tan−1 (−0.500) = 360° − 26.6° = 333° . (b) tan 1.00 m 0.500 2.00 m y x A θ A = = = . θ = tan−1 (0.500) = 26.6° . (c) tan 1.00 m 0.500 2.00 m y x A θ A = = =− − . θ = tan−1 (−0.500) =180° − 26.6° =153° . (d) tan 1.00 m 0.500 2.00 m y x A θ A − = = = − . θ = tan−1 (0.500) =180° + 26.6° = 207° EVALUATE: The angles 26.6° and 207° have the same tangent. Our sketch tells us which is the correct value of θ . 1.37. IDENTIFY: Find the vector sum of the two forces. SET UP: Use components to add the two forces. Take the +x-direction to be forward and the + y-direction to be upward. 1-8 Chapter 1 EXECUTE: The second force has components F2x = F2 cos32.4° = 433 N and 2 2sin 32.4 275 N. y F = F ° = The first force has components 1 725 N x F = and 1 0. y F = 1 2 1158 N x x x F = F + F = and 1 2 275 N y y y F = F + F = The resultant force is 1190 N in the direction 13.4° above the forward direction. EVALUATE: Since the two forces are not in the same direction the magnitude of their vector sum is less than the sum of their magnitudes. 1.38. IDENTIFY: Find the vector sum of the three given displacements. SET UP: Use coordinates for which +x is east and + y is north. The driver’s vector displacements are: A = 2.6 km, 0° of north; B = 4.0 km, 0° of east; C = 3.1 km, 45° north of east $ $ $ . EXECUTE: 0 4.0 km (3.1 km)cos(45 ) 6.2 km x x x x R = A + B + C = + + # = ; y y y y R = A + B + C = 2.6 km + 0 + (3.1 km)(sin45# ) = 4.8 km ; 2 2 7.8 km x y R = R + R = ; θ = tan−1⎡⎣(4.8 km) (6.2 km)⎤⎦ = 38# ; R = 7.8 km, 38# north of east. $ This result is confirmed by the sketch in Figure 1.38. EVALUATE: Both x R and y R are positive and R " is in the first quadrant. Figure 1.38 1.39. IDENTIFY: If C = A+ B " " " , then x x x C = A + B and y y y C = A + B . Use x C and y C to find the magnitude and direction of C " . SET UP: From Figure 1.34 in the textbook, 0 x A = , 8.00 m y A = − and sin30.0 7.50 m xB = +B ° = , cos30.0 13.0 m yB = +B ° = . EXECUTE: (a) C = A+ B " " " so 7.50 m x x x C = A + B = and 5.00 m y y y C = A + B = + . C = 9.01 m . tan 5.00 m 7.50 m y x C C θ = = and θ = 33.7° . (b) B+ A= A+ B " " " " , so B+ A " " has magnitude 9.01 m and direction specified by 33.7° . (c) D = A− B " " " so 7.50 m x x x D = A − B = − and 21.0 m y y y D = A − B = − . D = 22.3 m . tan 21.0 m 7.50 m y x D D φ − = = − and 70.3 φ = ° . D " is in the 3rd quadrant and the angle θ counterclockwise from the +x axis is 180° + 70.3° = 250.3° . (d) B − A = −(A− B) " " " " , so − B A " " has magnitude 22.3 m and direction specified by θ = 70.3° . EVALUATE: These results agree with those calculated from a scale drawing in Problem 1.32. 1.40. IDENTIFY: Use Equations (1.7) and (1.8) to calculate the magnitude and direction of each of the given vectors. SET UP: A sketch of x A , y A and A " tells us the quadrant in which A " lies. EXECUTE: (a) (−8.60 cm)2 + (5.20 cm)2 =10.0 cm, arctan 5.20 148.8 8.60 ⎛ ⎞ = ° ⎜ − ⎟ ⎝ ⎠ (which is 180° − 31.2° ). (b) (−9.7 m)2 + (−2.45 m)2 =10.0 m, arctan 2. . 9.7 ⎛ − ⎞ ⎜ ⎟ = ° + ° = ° ⎝ − ⎠ (c) (7.75 km)2 + (−2.70 km)2 = 8.21 km, arctan 2.7 340.8 7.75 ⎛ − ⎞ ⎜ ⎟ = ° ⎝ ⎠ (which is 360° −19.2° ). EVALUATE: In each case the angle is measured counterclockwise from the +x axis. Our results for θ agree with our sketches. Units, Physical Quantities and Vectors 1-9 1.41. IDENTIFY: Vector addition problem. We are given the magnitude and direction of three vectors and are asked to find their sum. SET UP: A = 3.25 km B = 4.75 km C =1.50 km Figure 1.41a Select a coordinate system where x + is east and y + is north. Let , A " B " and C " be the three displacements of the professor. Then the resultant displacement R " is given by R = A+ B + C. " " " " By the method of components, x x x x R = A + B + C and . y y y y R = A + B + C Find the x and y components of each vector; add them to find the components of the resultant. Then the magnitude and direction of the resultant can be found from its x and y components that we have calculated. As always it is essential to draw a sketch. EXECUTE: 0, x A = 3.25 km y A = + 4.75 km, x B = − 0 y B = 0, x C = 1.50 km y C = − x x x x R = A + B +C 0 4.75 km 0 4.75 km x R = − + =− y y y y R = A + B +C 3.25 km 0 1.50 km 1.75 km y R= +− = Figure 1.41b