Homework%233

Homework%233 - 1. We apply Eq. 4-35 to solve for speed v...

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1 . We apply Eq. 4-35 to solve for speed v and Eq. 4-34 to find acceleration a . (a) Since the radius of Earth is 6.37 × 10 6 m, the radius of the satellite orbit is r = (6.37 × 10 6 + 640 × 10 3 ) m = 7.01 × 10 6 m. Therefore, the speed of the satellite is v r T == × 2 2 7 01 10 98 0 60 749 10 6 3 π . ./ m i n . m min s m/s. c h bg b g (b) The magnitude of the acceleration is a v r × × = 2 3 2 6 701 10 800 . . .. m/s m 2 c h
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2. (a) The circumference is c = 2 π r = 2 (0.15 m) = 0.94 m. (b) With T = (60 s)/1200 = 0.050 s, the speed is v = c / T = (0.94 m)/(0.050 s) = 19 m/s. This is equivalent to using Eq. 4-35. (c) The magnitude of the acceleration is a = v 2 / r = (19 m/s) 2 /(0.15 m) = 2.4 × 10 3 m/s 2 . (d) The period of revolution is (1200 rev/min) –1 = 8.3 × 10 –4 min which becomes, in SI units, T = 0.050 s = 50 ms.
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5 . We note that ma = (–16 N) i ^ + (12 N) j ^ . With the other forces as specified in the problem, then Newton’s second law gives the third force as F 3 = F 1 F 2 =(–34 N) i ^ (12 N) j ^ .
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6 . To solve the problem, we note that acceleration is the second time derivative of the
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Homework%233 - 1. We apply Eq. 4-35 to solve for speed v...

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