Solucionario de fisica

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. 1-2. Assume a height of 5 ft 10 in. Then, 5′10″ = 70 in = 70 in × 2.54 cm/in = 178 cm. There are approximately 300 actual pages in this text (vol. 1). The thickness is 2.5 cm. Thereforeeach page = 2.5 cm/300 pg = 8.3 × 10−3 cm.

1-3. 1-4.
1-5.
1-6.

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
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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1-9.

1

(a) Grapefruit diameter ≈ 0.1 mRatio 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 Size (diameter) of the sun = 2 × 6.46 × 108 = 1.4 × 109 m distance on the diagram (PRELUDE, p. 1.4 × 1096) = 10−3 m = 1 mm 1.5 × 1012 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 × Also 360° × 60 min/deg = 21, 600 min, so 1nmi ⇒ 1min

1-11.

1-12.

†1-13.

1-14.

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†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 = R2 + (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

9400

R R + 1.75 m

1-16.

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 × 60s/min = 86,400 s 8211 days = 8211 days × 86,400 s/day = 7.1 × 108 s

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

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...