[benson]Sol-Ch01 - C H A P T E R 1 Exercises 1(a 55 mi/h X...

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Unformatted text preview: C H A P T E R 1 Exercises 1. (a) 55 mi/h X 5280 ft/mi x l h/36OO s = 80.7 ft/s; (b) 24.6 m/s 2. 13440 furlongs/fortnight 3. 10—3 kg/(lo'2 m)3 = 103 kg/m3. 4. 3.156 x 107 s 5. (a) 9.47 x 1012 km; (b) 7.2 AU/h 6. (a) 1.0073 u; (b) 1.67493 x 10‘27 kg 7. 0.514 m/s 8. (a) 0.984 ft/ns; (b) 1.86 x 105 mi/s 9. 134 in2 10. 30 mi/gal X 1 gal/3.79 L x 1.6 km/mi = 12.6 km/L or 7.85 L/lOO km 11. (a) 5, (b) 3, (c) 4, (d) 2 to 4 12. (a) 6.5 x 10‘9 s; (b) 1.28 x 10‘5 m; (c) 2 x 1010 w; (d) 3 x 10‘4 A; (e) 1.5 x 10"12 A 13. (a) A = nr2 = 55.4 m2, (b) A = 4nr2 = 2.7 m2, (c) v = 4nr3/3 = 52.17 m3. 14. (a) 2.5 x 10‘1; (b) 5.00 x 10’3; (c) 7.6300 x 10‘4 15. 3.33 x 103. 16. (a) 48.0; (b) 403.2 17. (a) 1.495 x 1011 m, (b) 5.893 x 10'7 m, (c) 2 x 10"10 m, (d) 4 x 10‘15 m. 18. (a) 15.69; (b) 25.9 19. (a) 91.440 m, (b) 0.40469 hectares 2 20. (a) 6.2 x 103; (b) 2.73 x 101; (c) 6.00000 x 102 21. (a) 2%, (b) 4%, (c) 6%. 22. 243 i 4.5 cm2 23. (a) 4nR2 = 5 x 1014 m2, (b) 4nR3/3 = 3 x 1020 m3 (c) RS/RE = 100, so (RS/RE)3 = 1 x 106 24. 2 x 105 25. 14.5 min. error in one day 26. 1670 km/h 27. For a 2—h movie at 30 frames/second, find 2 x 105 frames. 28. 3 x 10'5 m 29. (a) (2 km/d)(400 d/y)(70 y) = 5 x 104 km. (b) At 2 kg/d we find 5 x 104 kg. 30. 106 31. 109 m3 32. 104 grains 33. 0.1 m3 34. m‘1L3T'2 35. (a) Correct, (b) wrong, (c) correct LT’4 36. [A] = LT‘Z, [B] 37. (a) (2.68 m, 2.25 m), (b) (—1.15 m, -l.38 m), (c) (-1.80 m, 1.26 m), (d) (1.99 m, —1.67 m) 38. (a) 5.00 m, 53 1°; (b) 3.61 m, 124°; (c) 2.92 m, 329°; (d) 2.24 m, 206° 39. [w] = T'l, [k] = M T'2 40. V = 4nr3/3, so r = 9.14 cm. 3 41. (a) v = nr2h, h = 14.5 cm. (b) Area = 2nrh + 2nr2 = 330 cm2. 42. 1 y = 3.15576 x 107 s and n = 3.14159. Thus error is 0.449%. 43. Using 3.786 L per gal. and 1 mi = 1.609 km, find 25.9 m.p.g. 44. For each person Ax = 1.6 m. Circumference = 4 x 107 m, so we need 2.5 X 10’ people. 45. (a) 4.96 x 102; (b) 2.6 x 104 46. Area = 480 ft2 = 44.595 m2, so cost is $32.97. 47. (4 x 10'3 m3)/20 m2 = 0.2 mm. 48. (a) kg.m2/s2; (b) E = (10'3 kg)c2 = 9 x 1013 kg.m2/52. 49. 0 = s/R in radians, or sin(0.5") = 0.5 AU/l parsec, give 1 parsec = 2.063 X 105 AU. 50. 64.206 cmz. 51. 1 " = 2.54 cm exactly, so g = 32.1740 ft/s2 52. 1 m2 = 10.764 ftz. Cost is 21.05 per m2. Problems 1. With a 0.5 m diameter, the circumference is C e 1.5 m. If the tread lasts for 7 x 107 m, the number of revolutions is 7 x 107 m/1.5 m z 4.5 x 107 rev. The tread depth is about 1 cm, so the loss per revolution is 0.01 m/(4.5 x 107) z 2 x 10"10 m. This about roughly the size of an atom. 2. a a v2/r 3. T = kx my = Mx T‘ZX My. The unit of k is N/m, so [k] = MT'2 Equating the exponents: 1 = -2x; 0 = x + y. Thus x = -l/2 and y = + 1/2 and so T = C(m/k)1/2 4. x a at2 ...
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This note was uploaded on 03/09/2010 for the course PHYISCS 0122 taught by Professor Xiyanh during the Spring '10 term at University of Alaska Southeast.

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[benson]Sol-Ch01 - C H A P T E R 1 Exercises 1(a 55 mi/h X...

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