# Soluciones fisica i

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A.46

The object does not hit the Earth; its minimum radius is 1.33RE as shown in the diagram below. Its period is 1.09 � 104 s. A circular orbit would require speed 5.60 km/s. 51. (a) 4.43 m/s (b) The siphon can be no higher than 10.3 m. 53. 12.6 m/s 55. 1.91 m 59. 0.604 m
y

49. 103 m/s

57. (a)

23. 520 m/s25. 1.64 m/s2

27. 13.5 N

29. 586 m/s

31. 0.329 s

63. 17.3 N and 31.7 N 65. 90.04% 67. 758 Pa 69. 4.43 m/s 71. (a) 1.25 cm 73. (a) 3.307 g (b) 3.271 g (c) 3.48 � 10�4 N 75. (c) 1.70 m2
x a h Li L

33. (a) s and kg · m/s2 (b) time interval (period) and force (tension)

37. 55.1 Hz

39. (a) 62.5 m/s (b) 7.85 m (c) 7.96 Hz (d) 21.1 W (b) 13.8 m/s

41. √2 �0

CHAPTER 15
1. (a)The motion repeats precisely. (b) 1.82 s (c) No, the force is not in the form of Hooke’s law. 3. (a) 1.50 Hz, 0.667 s (b) 4.00 m (c) � rad (d) 2.83 m 5. (b) 18.8 cm/s, 0.333 s (d) 12.0 cm 7. (a) 2.40 s (b) 0.417 Hz (c) 2.62 rad/s 9. 40.9 N/m 11. (a) 40.0 cm/s, 160 cm/s2 (b) 32.0 cm/s, � 96.0 cm/s2 (c) 0.232 s 13. 0.628 m/s 15. (a) 0.542 kg (b) 1.81 s (c) 1.20 m/s2 17. 2.23 m/s 19. (a) 28.0 mJ (b)1.02 m/s (c) 12.2 mJ (d) 15.8 mJ 21. (a) E increases by a factor of 4. (b) vmax is doubled. (c) a max is doubled. (d) Period is unchanged. 23. 2.60 cm and � 2.60 cm 25. (b) 0.628 s 27. (a) 35.7 m (b) 29.1 s 29. �100 s 31. Assuming simple harmonic motion, (a) 0.820 m/s (b) 2.57 rad/s2 (c) 0.641 N. More precisely, (a) 0.817 m/s (b) 2.54 rad/s2 (c) 0.634 N 35. 0.944 kg � m2 39. (a) 5.00 � 10�7 kg �m2 (b) 3.16 � 10�4 N � m/rad 41. 1.00 � 10�3 s�1 43. (a) 7.00 Hz (b) 2.00% (c) 10.6 s 45. (a) 1.00 s (b) 5.09 cm 47. 318 N 49. 1.74 Hz 51. (a) 2Mg ; Mg(1� y/L) (b) T � (4�/3)(2L/g)1/2; 2.68 s (b) 2.95 m/s (c) 4.34 kPa 53. 6.62 cm 55. 9.19 � 1013 Hz (c) 178 cm /s2, 0.500 s (b) dT � dt 2�a 2g 1/2[L i

43. (a) A � 40 (b) A � 7.00, B � 0, C � 3.00. One can take the dot product of the given equationwith each one of ˆ, ˆ, i j ˆ and k. (c) A � 0, B � 7.00 mm, C � 3.00/m, D � 4.00/s, E � 2.00. Consider the average value of both sides of the given equation to ﬁnd A. Then consider the maximum value of both sides to ﬁnd B. One can evaluate the partial derivative of both sides of the given equation with respect to x and separately with respect to t to obtain equations yielding C and D upon chosensubstitutions for x and t. Then substitute x � 0 and t � 0 to obtain E.

�(dM/dt) � (dM/dt)t/2�a 2]1/2
(c) T � 2�g�1/2 [Li � (dM/dt)t/2�a2]1/2 59. f � (2�L)�1 (gL � kh2/M)1/2 61. (b) 1.23 Hz 63. (a) 3.00 s (b) 14.3 J (c) 25.5°

47. � 1 min

ˆ 49. (a) (3.33i ) m/s (d) 11.0 m/s

(b) � 5.48 cm (c) 0.667 m, 5.00 Hz

51. 0.456 m/s

53. (a) 39.2 N (b) 0.892 m (c) 83.6 m/s

CHAPTER 14
1.0.111 kg 3. 6.24 MPa 5. 5.27 � 1018 kg 7. 1.62 m 9. 7.74 � 10�3 m2 11. 271 kN horizontally backward
1 13. P0 � 2�d √g 2 � a 2

55. (a) 179 m/s (b) 17.7 kW

65. If the cyclist goes over them at one certain speed, the washboard bumps can excite a resonance vibration of the bike, so large in amplitude as to make the rider lose control. �101 m

61. (a) (0.707)2(L/g)1/2

(b)L/4

63. 3.86 � 10�4

65. (b) 31.6 m/s

73. For �max � 5.00° there is precise agreement. For �max � 100° there are large differences, and the period is 23% greater than small-angle period. 75. (b) after 42.1 min

67. (a)

��3 A02e �2bx 2k

(b)

��3 2 A (c) e�2bx 2k 0

69. (a) �0 � (�L � �0)x/L

15. 0.722 mm 17. 10.5 m; no; some alcohol and water evaporate 19. 98.6 kPa 21. (a)1.57 Pa, 1.55 � 10�2 atm, 11.8 mm Hg (b) The ﬂuid level in the tap should rise. (c) Blockage of ﬂow of the cerebrospinal ﬂuid 23. 0.258 N 25. (a) 9.80 N (b) 6.17 N 27. (a) 1.017 9 � 103 N down, 1.029 7 � 103 N up (b) 86.2 N 29. (a) 7.00 cm (b) 2.80 kg 33. 1 430 m3 35. 1 250 kg/m3 and 500 kg/m3 37. 1.28 � 104 m2 39. (a) 17.7 m/s (b) 1.73 mm 41. 31.6 m/s 43. 0.247 cm 45. (a) 1 atm � 15.0 MPa 47. 2.51...