Física Para Ingenieria y Ciencias Ohanian 3ed Solucionario PDF

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www.elsolucionario.net CHAPTER 1 SPACE, TIME, AND MASS Select odd-numbered solutions, marked with a dagger (†), appear in the Student Solutions Manual, available for purchase. Answers to all solutions below are underscored. 1-1. Assume a height of 5 ft 10 in. Then, 5′10″ = 70 in = 70 in × 2.54 cm/in...

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CHAPTER 1

SPACE, TIME, AND MASS

Select odd-numbered solutions, marked with a dagger (†), appear in the Student Solutions Manual, available for purchase. Answers to all solutions below are underscored. 1-1.

Assume a height of 5 ft 10 in. Then, 5′10″ = 70 in = 70 in × 2.54 cm/in = 178 cm.

1-2.

There are approximately 300 actual pages in this text (vol. 1). The thickness is 2.5 cm. Therefore each page = 2.5 cm/300 pg = 8.3 × 10−3 cm.

1-3.

100 yd × 0.914 m/yd = 91.4 m; 53 1/3 yd × 0.914 m/yd = 48.7 m 1 step 1000 m steps N = × = 1.7 × 103 0.60 m 1 km km 1 pica 17 1 pica L = 11 in × = 66 picas. W = in × = 51 picas 1 1 2 in in 6 6 −8 Virus: 2 × 10 m × 10/0.3048 ft/m × 12 in/ft = 8 × 10−7 in

1-4. 1-5. 1-6.

Similarly: Atom: 1 × 10−10 m × 39.4 in/m = 4 × 10−9 in Fe Nucleus: 8 × 10−15 m × 39.4 in/m = 3 × 10−13 in Proton: 2 × 10−15 m × 39.4 in/m = 8 × 10−14 in †1-7.

1-8.

Let’s convert 1/2 inch to mm using conversion factors, then use proportional reasoning to do the others until the number of significant figures becomes large. 1 in × 25.4 mm/in = 12.7 mm 2 1 12.7 mm in = = 6.35 mm 4 2 1 6.35 mm in = = 3.175 mm = 3.18 mm (to three significant figures) 8 2 1 3.175 mm in = = 1.5875 mm = 1.59 mm (to three significant figures) 16 2 The number of digits is becoming large, so let’s do direct conversions for the rest of the problems. 1 in × 25.4 mm/in = 0.794 mm 32 1 in × 25.4 mm/in = 0.397 mm 64 10−3 in 2.54 cm 10−2 m 10−6 µm 1 mil × × × × = 25.4 µm. mil in cm m 1000 µm 1 mil 1 mm × × = 39.4 mil mm 25.4 µm

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CHAPTER 1-9.

1

(a) Grapefruit diameter ≈ 0.1 m Ratio of grapefruit/sun = 0.1 m/(1.4 × 109 m) = 7 × 10−11 Earth diameter ≈ 13 × 106 m Comparative size of Earth = 13 × 106 m × (7 × 10−11) = 9 × 10−4 m ≈ 1 mm. Nearest star distance = 4 × 1016 m Comparative distance = 4 × 1016 m × (7 × 10−11) = 2.8 × 106 m (b) Head diameter ≈ 0.2 m Earth diameter ≈ 13 × 106 m Earth/head ratio ≈ 7 × 107 Size of atom = 10−10 m Comparative size of atom = 10−10 m × (7 × 107) = 7 × 10−3 m = 7 mm Size of red blood cell ≈ 7.5 × 10−6 m Comparative size of cell = 7.5 × 10−6 m × (7 × 107) ≈ 500 m = 1/ 2 km

1-10.

Distance to Q1208 + 1011 = 12.4 × 109 × 9.47 × 1015 = 1.17 × 1026 m Distance on the diagram (PRELUDE, p. 6) 1.17 × 1026 = = 7.8 × 105 m 20 1.5 × 10

1-11.

Size (diameter) of the sun = 2 × 6.46 × 108 = 1.4 × 109 m distance on the diagram (PRELUDE, p. 1.4 × 109 6) =  10−3 m = 1 mm 1.5 × 1012

1-12.

†1-13.

10−9 m = 6.33 × 10−13 m. According to Table 1-1, the diameter of an nm atom is about 1 × 10−10 m, so this is 6.33 × 10−3 times the diameter of an atom, or roughly 1/100 the diameter of an atom. 1 turn = 360°, so 5° × 1 turn/360° = 0.0139 turn. For an English thread, 1in 0.0254 m 106 µm 0.0139 turn × × × = 4.41 µm. For a metric thread, 80 turns in m ∆l = 10−6 × 633 nm ×

0.5 mm 10−3 m 106 µm × × = 6.94 µm. turn mm m 1 nmi = 1852 m; Circumference of Earth = 4.00 × 107 m Circumference = (4.00 × 107 m)/1852 m/nmi = 21, 600 nmi 0.0139 turn ×

1-14.

Also 360° × 60 min/deg = 21, 600 min, so 1nmi ⇒ 1min

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CHAPTER †1-15.

For one of the triangles, (R + 1.75 m)2 = R2 + (4700 m)2. Expand this to get R 2 + 2(1.75 m) R + (1.75 m) 2 = R 2 + (4700 m) 2 . We expect R to be much larger than 1.75 m, so we can ignore (1.75 m)2 relative to all the other terms. The R2 terms cancel, leaving (3.50 m)R = (4700 m)2, which gives R = 6.3 × 106 m.

1-16.

9400

R R + 1.75 m

22 yr, 5 mo, 23 days = (8035 + 153 + 23) = 8211 days (This excludes leap years and assumes average 30.5-day month.) 1 day = 1 day × 24 h/day × 60 min/h × 60 s/min = 86,400 s 8211 days = 8211 days × 86,400 s/day = 7.1 × 108 s

1-17. 1-18. 1-19.

1-20. †1-21.

1-22. 1-23. 1-24. 1-25. 1-26.

1 yr = 365.25 days. Therefore, 4.5 × 109 yr = 4.5 × 109 yr × 365.25 day/yr × 86,400 s/day = 1.4 × 1017 s 60 s 1 calculation = 3.6 × 1012 calculations/h. × min 10−9 s 3600 s 60 s 2 h 9 min 21s = 2 h × + 9 min × + 21s = 7761 s h min 3600 s 60 s 2 h 24 min 51s = 2 h × + 24 min × + 51s = 8692 s h min 365.25 solar days/year 24 h 1 sidereal day × × = 23.934 h/sidereal day. Using 1 h = 60 366.25 sidereal days/year solar day N =1h ×

60 min h

×

min to convert the 0.934 h to minutes gives 1 sidereal day = 23 h 56 min. s N = 4 ticks × 3.2 × 107 × 10 years = 1.2 × 109 ticks year 1h 1day 106 s × × = 11.6 days 3600 s 24h 7 days 24 h 3600 s 1week × × = 168 h. 168 h × = 6.048 × 105 s hr week day beats 1 min 3.2 × 107 s = 3.8 × 107 beats/year ticks × × min 60 s yr (a) June 24−25: (20 − 4) s/day = 16 s/24 h = 0.67 s/h N = 71

June 25−26: (34 − 20) s/day = 14 s/24 h = 0.58 s/h June 26−27: (51 − 34) s/day = 17 s/24 h = 0.71 s/h (b) Average rate = (51 − 4)/3 day = 47 s/(24 × 3)h = 0.65 s/h

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1

CHAPTER

1

(c) 10h30m on June 30 is 70.5 h after noon June 27. By average loss, watch should have lost 70.5 h × 0.65 s/h = 46 s. Combined with loss of 51 s on June 27 gives total loss of 97 s. Therefore the correct WWV time is 10h 31m 37s . With the largest rate, loss is 70.5 h × 0.71 s/h = 50 s. Combined loss is 51 + 50 = 101 s. Then estimated WWV time is 10h 31m 41s . The wristwatch can be trusted to †1-27.

1-28.

about ±4 s on June 30. 1 day = 24 hr = 86,400 s. The earth rotates 360º per day, which corresponds to a rotation rate of 360° 60 min × = 0.250 min/s. A timing error of 1 s will result in an angular error of 0.250 86, 400 s degree min. According to Problem 1-14, 1 min = 1852 m, so the corresponding error in position is 0.250 min/s × 1852 m = 463 m = 0.463 km. 140 lb-mass = 140 lb-mass × 0.454 kg/lb-mass = 64 kg 140 lb-mass = 140 lb-mass × 0.454 kg/lb-mass × 1/14.6 slug/kg = 4.35slug 140 lb-mass = 140 lb-mass × 0.454 kg/lb-mass × 1 amu/(1.67 × 10−27 kg) = 3.8 × 10 28 amu

†1-29.

m planets = (0.33 + 4.9 + 5.98 + 0.64 + 1900 + 553 + 87.3 + 10.8 + 0.66) × 1024 kg = 2.56 × 1027 kg (to three significant figures). msun = 1.99 × 1030 kg, so the total mass is mtotal = msun + m planets = 1.99 × 1030 kg + 2.56 × 1027 kg = 1.99 × 1030 kg (to three significant figures). The fraction of the total mass included in the planets is m planets 2.56 × 1027 × 100% = × 100% = 0.134%. The fraction of the mass in the sun is 100% − mtotal 1.99 × 1030

1-30.

0.134% = 99.9%. largest length 1 × 1026 m 4 × 1017 s 40 longest time = = 5 × 10 . = = 4 × 1041. smallest length 2 × 10−15 m shortest time 1 × 10−24 s largest mass 1 × 1053 kg = = 1 × 1083. The first two ratios are within an order of magnitude of −31 smallest mass 9 × 10 kg

†1-31.

1-32.

1-33.

1-34.

each other, and the third is roughly equal to the square of the other two. (Note that the first two ratios will keep increasing because the universe is expanding and aging.) From the periodic table in the Appendix, we see that the uranium nucleus contains about 238 nucleons, each with the mass of a proton. Table 1.7 gives 1.7 × 10−27 kg for the mass of a proton. Then the total mass of the electrons is (92)( 9.1 × 10−31 kg) = 8.4 × 10−29 kg, and the total mass of the nucleus is (238)( 1.7 × 10−27 kg) = 4.0 × 10−25 kg. To two significant figures, the total mass of the atom is 4.0 × 10−25 kg. The fraction of the total mass in the electrons is 8.4 × 10−29/4.0 × 10−25 = 2.1 × 10−4 = 0.021%. The fraction of mass in the nucleus is 99.98%. Let m represent the mass of material and M represent the mass of one mole. Then atoms ⎛ 0.1 × 10−6 g ⎞ ⎛m⎞ 14 N = N A ⎜ ⎟ = 6.02204 × 1023 ×⎜ ⎟ = 3 × 10 atoms. mol ⎝M ⎠ ⎝ 197 g/mol ⎠ 1 lb avoirdupois = 0.453 59 kg = 435.59 g (given in the text) 0.453 59 kg 1 lb troy = 0.822 86 lb avoidupois × = 0.373 24 kg = 373.24 g lb avoidupois ⎞ atoms ⎛ 10−21 kg ⎛m⎞ 4 N = N A ⎜ ⎟ = 6.02204 × 1023 ×⎜ ⎟ = 1.1 × 10 atoms mol 0.055 85 kg/mol ⎝M ⎠ ⎝ ⎠

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CHAPTER 1-35.

1

(a) Since the density of water is 1 g/cm3, 250 cm3 of water has a mass of 250 g, which is 250 g/(18 g/mol) = 14 mol. Therefore, the number of molecules is 14 mol × 6.02 × 1023 molecules/mol = 8.4 × 1024 molecules. (b) Mass of sea-water (density = 1030 kg/m3) is 1.3 × 1018 m3 × 1030 kg/m3 = 1.3 × 1021 kg = 1.3 × 1024 g Number of molecules of sea-water is 1.3 × 1024 g × 1 mol/18 g × 6.02 × 1023 molecules/mol = 4.3 × 1046 molecules. (c) Ratio of molecules of water from cup to molecules from sea is 8.4 × 1024/4.3 × 1046 = 2.0 × 10−22 . The probability of a single molecule drawn from the ocean being originally from the cup is therefore 2.0 × 10−22. Drawing a cup full then should result in 2.0 × 10−22 × 8.4 × 1024 = 1680 molecules.

1-36.

The mass of H2 and He, respectively, is 0.70 × 1.99 × 1030 kg = 1.39 × 1030 kg and 0.30 × 1.99 × 1030 kg = 0.60 × 1030 kg. The atomic mass of hydrogen and helium, respectively, is 1 g/mol and 4 g/mol. Therefore, the number of hydrogen atoms is NH = 1.39 × 1033 g × 1 mol/g × 6.02 × 1023 molecules/mol = 8.38 × 1056 molecules. The number of helium atoms is NHe = 0.60 × 1033 g × 1 mol/4 g × 6.02 × 1023 molecules/mol = 9.03 × 1055 atoms. Total number = 9.28 × 1056

†1-37.

1-38.

Molecular mass of N2 = 28 g/mol Molecular mass of O2 = 32 g/mol Molecular mass of Ar = 40 g/mol Therefore, 1000 g of air will contain: 755 g N2 = 755 g/(28 g/mol) = 27.0 mol 232 g O2 = (232/32) mol = 7.25 mol 13 g Ar = (13/40) mol = 0.325 mol The percentage by number of molecules of these substances is: N2: 27.0/(27.0 + 7.25 + 0.325) × 100% = 27.0/34.575 × 100% = 78.1% O2: (7.25/34.575) × 100% = 21% Ar: (0.325/34.575) × 100% = 0.9% Therefore, the “molecular mass” of air is (0.781 × 28) + (0.21 × 32) + (0.009 × 40) = 28.95 g/mol. The mass of elements is: Oxygen: 0.65 × 73 kg = 47.4 kg; atomic mass = 16 Carbon: 0.185 × 73 kg = 13.5 kg; atomic mass = 12 Hydrogen: 0.095 × 73 kg = 6.94 kg; atomic mass = 1.008 Nitrogen: 0.033 × 73 kg = 2.41 kg; atomic mass = 14 Calcium: 0.015 × 73 kg = 1.09 kg; atomic mass = 40.08 Phosphorous: 0.01 × 73 kg = 0.73 kg; atomic mass = 31

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CHAPTER

†1-39.

1-40. 1-41. 1-42.

1-43.

1

The number of atoms of these substances is: O: 47.4 × 103 g × 1/16 mol/g × 6.02 × 1023/mol C: 13.5 × 103 g × 1/12 mol/g × 6.02 × 1023/mol H: 6.94 × 103 g × 1 mol/g × 6.02 × 1023/mol N: 2.41 × 103 g × 1/14 mol/g × 6.02 × 1023/mol Ca: 1.09 × 103 g × 1/40 mol/g × 6.02 × 1023/mol P: 0.73 × 103 g × 1/31 mol/g × 6.02 × 1023/mol Total number of atoms: 0.53° = 0.53/360 × 2π rad = 9.25 × 10−3 rad d = 9.25 × 10−3 rad × 1.5 × 1011 m = 1.4 × 109 m r = 6.9 × 108 m

= 1.78 × 1027 = 6.77 × 1026 = 4.17 × 1027 = 1.04 × 1026 = 1.64 × 1025 = 1.42 × 1025 = 6.76 × 1027

m⎞ days h s ⎛ 1 ly = ⎜ 3.00 × 108 ⎟ × 365.25 × 24 × 3600 = 9.47 × 1015 m s ⎠ year day h ⎝ 15 1 ly = 9.47 × 10 m 2.2 × 106 ly = 2.2 × 106 ly × 9.47 × 1015 m/ly = 2.1 × 1022 m 1 light-s (ls) = 3.00 × 108 m/s × 1 s = 3.00 × 108 m 1 light-min (lm) = 3.00 × 108 m/s × 60 s = 1.80 × 1010 m E-S distance = 1.50 × 1011 m × 1/(1.80 × 1010) lm/m = 8.3 lm E-M distance = 3.84 × 108 m × 1/(3.00 × 108) ls/m = 1.28 ls 1 2π 1 degree = 1 second of arc = × rad = 4.85 × 10−6 rad 3600 360 3600 1 AU = 4.85 × 10−6 (a) pc 1 pc = 1 × 1/(4.85 × 10−6)AU = 2.06 × 105 AU (b) 1 pc = 2.06 × 105 AU × 1.496 × 1011 m/AU = 3.08 × 1016 m 1 ly = 9.47 × 1015 m 1 pc = 3.08 × 1016 m × 1 ly/(9.47 × 1015 m) = 3.25 ly (c) 1 ly = 3.00 × 108 m/s × 1 yr = 3.00 × 108 m/s × 3.156 × 107 s = 9.47 × 1015 m 2

1-44.

⎛ 1 ft ⎞ 2 1m × ⎜ ⎟ = 10.76 ft . Note that the conversion factor must be squared. ⎝ 0.3048 m ⎠ 2

3

†1-45. 1-46. †1-47.

⎛ 1 ft ⎞ 3 1m3 × ⎜ ⎟ = 35.31 ft . Note that the conversion factor must be cubed. 0.3048 m ⎝ ⎠ 1 m2 Use the result from 1−44: A = (78 ft)(27 ft) × z = 196 m2. 10.76 ft 2 His height is measured to a precision of 0.1 inch, so we want to see how many significant digits this implies. 8 feet = 96 inches (exactly), so to the nearest 0.1 inch his height can be expressed as 107.1 inches, which contains four significant figures. Converting 11.1 inches to feet gives his height in feet: 8 ft + 11.1 in = 8.925 ft, which also contains four significant figures. Thus his m height in meters should be specified to four figures: 8.925 ft × 0.3048 = 2.720 m. ft

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CHAPTER

1

2

1-48.

⎛ 1 m2 ⎞ ⎛ 3 ft ⎞ = 4.46 × 103 m2, using the result from 1−44. A = 100 yd × 53.33 yd × ⎜ × ⎜ ⎟ 2 ⎟ yd 10.76 ft ⎝ ⎠ ⎝ ⎠ 3

†1-49.

⎛ 1 kg ⎞ ⎛ 100 cm ⎞ 3 3 8.9 g/cm3 × ⎜ ⎟×⎜ ⎟ = 8.9 × 10 kg/m . Note that the cm to m conversion factor 1000 g m ⎝ ⎠ ⎝ ⎠ must be cubed! 1 ft = 0.3048 m, 1 lb = 0.454 kg. 3 ⎛ 1 lb ⎞ ⎛ 0.3048 m ⎞ 2 3 3 8.9 × 103 kg/m3 × ⎜ × ⎟ ⎜ ⎟ = 555 lb/ft , or 5.6 × 10 lb/ft to two significant 0.454 kg ft ⎝ ⎠ ⎝ ⎠ figures. Again, note that the m to ft conversion factor must be cubed. 1 ft = 12 in, so 3

1-50.

1-51. 1-52.

†1-53.

1-54. †1-55.

1-56.

†1-57.

1-58.

⎛ 1 ft ⎞ 3 555 lb/ft 3 × ⎜ ⎟ = 0.32 lb/in . ⎝ 12 in ⎠ Assume a mass of 73 kg and a density of 1000 kg/m3. mass V= = 73 kg × 1/1000 m3/kg = 0.073 m3 density cm3 3600 s 10−6 m3 = 7.9 m3/day. × 24 h × × 3 s h cm 5.2 × 103 cm3 1 liter = 103 cm3, so 5.2 liter = 5.2 × 103 cm3. t = = 57 s. 92 cm3 /s 10−4 m 2 1 cm 2 × cm 2 = 108 transistors. If they’re stacked, N = the number of transistors per N = m2 10−12 transistor layer (108) × number of layers. If the cube is 1 cm high and each layer is 10−7 m, or 10−5 cm, thick, then the cube holds 105 layers and the cube can hold 108 × 105 = 1013 transistors! 3 g 1 lb liter 3 cm −3 kg × 10 × 10 × × 3.875 1 gal = 3.785 liter. Density = 1.00 = 8.34 3 cm g liter 0.4536 kg gal 92

lb/gal (a) (3.6 × 104) × (2.049 × 10−2) = (3.6)(2.049) × 104−2 = 7.4 × 102 (b) (2.581 × 102) − (7.264 × 101) = (2.581 − 0.7264) × 102 = 1.855 × 102 7.9832 × 10−2 (c) 0.079832 ÷ 9.43 = = 0.847 × 10−2−0 = 8.47 × 10−3 0 9.43 × 10 3 3m 3(2.0 × 1030 kg) m m 3 g −6 m = = = = 1.5 g/cm3 × 10 × 10 3 8 3 3 4 π π V 4 R 4 (7.0 × 10 m) kg cm 3 πR 3 3 3m 3(2.0 × 1030 kg) 1 metric ton m m −6 m = = = Density = × × 10 3 4 V 4π (20 × 103 m)3 103 kg cm3 π R 3 4π R 3 = 6.0 × 107 metric tons/cm3 Oceans of the earth have 1.3 × 1018 m3 of water. The mass of the oceans is 1.3 × 1018 m3 × 1030 kg/m3 = 1.3 × 1021 kg. Mass of the earth is 5.98 × 1024 kg, so that the percentage of the mass of the earth that is water is: (1.3 × 1021 kg/5.98 × 1024 kg) × 100% = 0.02%.

Density =

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CHAPTER

1

†1-59.

From Table 1.10, 1 liter = 10−3 m3. According to data given in the “Conversion of Units” section of the chapter, the density of water is 1000 kg/m3, so 10−3 m3 1 min = 5.00 × 10−3 m3 /s, and 300 liters/min × × liter 60 s 5.00 × 10−3 m3 /s × 1000 kg/m3 = 5.00 kg/s.

1-60.

1 in = 1 in × 2.54 cm/in × 1/100 m/cm = 0.0254 m. Therefore volume on 1 m2 is V = 0.0254 m × 1 m2 = 0.0254 m3 /m 2 .

†1-61.

Mass of this much water is 0.0254 m3/m2 × 1000 kg/m3 = 25.4 kg/m2. 3m m m This can be solved using proportional reasoning. Density = , which means = = 3 4 π 4 V R 3 πR 3 1/ 3

⎛ mlead mcopper mlead R R = , from which we get = ⎜ lead copper 3 3 ⎜m Rlead Rcopper ⎝ copper

⎞ ⎟⎟ ⎠

1/ 3

= (4.8 × 10

−15

⎛ 3.5 ⎞ m) ⎜ ⎟ ⎝ 1.06 ⎠

1/ 3

1/ 3 ⎛ moxygen ⎞ ⎛ 0.27 ⎞ −15 = 7.1 × 10 m. Likewise we get Roxygen = Rcopper ⎜ ⎟⎟ = (4.8 × 10 m) ⎜ ⎟ ⎜m ⎝ 1.06 ⎠ ⎝ copper ⎠ = 3.0 × 10−15 m. Note that it was not necessary to include the factor (× 10−25) when expressing the masses because it cancels in the ratio. mass 4 where V = πr3. Density is calculated as volume 3 Planet Mass (kg) Vol (m3) Density (kg/m3) 23 20 Pluto 6.6 × 10 ? 1.13 × 10 5800 ? 23 19 Mercury 3.3 × 10 5.94 × 10 5600 1.09 × 1021 5500 Earth 5.98 × 1024 24 20 9.51 × 10 5100 Venus 4.9 × 10 Mars 6.40 × 1023 1.62 × 1020 4000 26 22 4.58 × 10 2250 Neptune 1.03 × 10 5.65 × 1022 1550 Uranus 8.73 × 1025 27 24 Jupiter 1.90 × 10 1.52 × 10 1250 9.23 × 1023 600 Saturn 5.53 × 1026 One dimension of each roof segment must be divided by cos 45°, or multiplied by 2. Then total roof area = floor area × 2 = 250 m2 × 2 = 354 m2. Use the radius of the earth from Table 1-1. ∆s 15 m ∆θ = = = 2.3 × 10−6 radian. In ∆θ ∆s = 15 m R 6.4 × 106 m

−15

1-62.

1-63. 1-64.

R

degrees, this is (2.3 × 10−6 radian) × (180°/π radian) = 1.3 × 10−4 degree. †1-65.

The slope is the tangent of the required angle. For a slope of 1:5, θ = tan −1

1 = 11°. For a slope of 5

1 = 5.7°. For a slope angle of 0.1°, tan θ = 1.7 × 10−3, so the slope is 10 1:( 1.7 × 10−3)−1 = 1: 5.7 × 102, so the rise is about 1 atom per 570 atoms. 1:10, θ = tan −1

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CHAPTER

1-66.

†1-67.

1-68.

1-69.

1-70.

h , where D = 75 m. Both D and θ are D specified to two significant figures, so the value calculated for h must be specified to two figures. Thus h = D tan θ = (75 m)( tan 78°) = 3.5 × 102 m.

1

tan θ =

h

θ D

In this calculation, assume that the calendar year is exactly 365 days long and a circle contains exactly 360°, so these are not interpreted as numbers with only three significant figures. The 365 × 360° = 359.76°. In four angle the earth moves through in one calendar year is θ = 365.24 years, including one leap year with exactly 366 days, the total number of days is 3 × 365 + 366 = 1461 × 360° = 1440.0°, 1461 days. The angle the earth moves through in four years is θ = 365.24 which is exactly four complete circles to five significant figures. (The angle is actually a little larger than 1440.0°, which is why there are some four-year intervals that do not include an extra day.) The sun will set at Marchena n-minutes after 8 PM where minutes in a day × distance between islands 24 × 60 × 60 × 103 = = 2.15 min n= circumference of the earth 4 × 107 Therefore, the sun sets at Marchena at 2.15 min after 8 PM. The diameter, d of the tree trunk is related to its length, L by d = AL3/2, where A is a constant 7.6 = A(81)3/2 if L = 90 m, then 3/ 2 ⎛ 90 ⎞ 3/2 d = A(90) = ⎜ ⎟ 7.6 = 8.9 m. ⎝ 81 ⎠ The mass, m of the tree trunk is related to d and L by 2 ⎛d⎞ m = ρπ ⎜ ⎟ L, where ρ is the mass density. ⎝2⎠ π 6100 = ρ (7.6) 2 81 4 for L = 90 m, 2 π ⎛ 8.9 ⎞ 90 = 9295 tons m = ρ (8.9) 2 90 = 6100 ⎜ ⎟ 4 ⎝ 7.6 ⎠ 81 Distance from pole to equator = 1/4 circumference. Therefore 1 π π × 6.37 × 106 m = 1.00 × 107 m. d = (2πr ) = r = 4 2 2 Straight line distance is, using the Pythagorean theorem: d=

r 2 + r 2 = 2r 2 =

2 × 6.37 × 106 m = 9.0 × 106 m.

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CHAPTER

†1-71.

1-72.

1-73.

1

M 235 g/mol = 3.902 × 10−22 g = 3.902 × 10−25 kg. In atomic mass units, = NA 23 atoms 6.02204 × 10 mol 3.902 × 10−25 kg = 235.0 u. this is matom = kg 1.66054 × 10−27 u The molar mass M of water (H2O) is 18.0 g/mol. The mass m of 1 liter (1000 cm3) of water is m 1000 g molecules 1000 g. N = × NA = × 6.02204 × 1023 = 3.35 × 1025 molecules. Each M 18.0 g/mol mol matom =

molecule contains one oxygen atom and two hydrogen atoms, so there are 3.35 × 1025 oxygen atoms and 6.70 × 1025 hydrogen atoms. Molecular mass of N2 is 2 × 14 g/mol = 28 g/mol. ...