# Lesson 4 - u s F s = (1450 kg)(9.8 kg/N)(1.1) F s = 1.6 x...

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Lesson 4 13) G = 6.67 x 10 -11 (N * m 2 ) / kg 2 m e = 6.38 x 10 24 kg m m = 7.35 x 10 22 kg r = 3.84 x 10 8 m x = ? (Gm m ) / (r – x) 2 = (Gm e ) / r 2 x = r – (m m x r 2 ) / (m e ) x = 3.84 x 10 8 m – (7.35 x 10 22 kg x 1.47 x 10 17 m) / (6.38 x 10 24 kg) x = 3.4 x 10 8 m The distance would be 3.4 x 10 8 m. 14) f = 1000 / s r = 8.4 x 10 -2 m a c = 4(pi) 2 rf 2 a c = 3.3 x 10 6 m/s 2 g = a c / g g = (3.4 x 10 5 )g The acceleration in terms of g is (3.4 x 10 5 )g. 15) v = 225 km/h = 62.5 m/s m = 1450 kg = 15 o v 2 = grtan r = v 2 /g tan15 o r = (62.5 m/s) 2 / [(9.8 m/s 2 )(tan15 o )] r = 1487.6 m The radius for the curvature is 1.5 x 10 3 m.

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b) a c = v 2 / r a c = (62.5 m/s) 2 / (1487.6 m) a c = 2.6 m/s 2 The centripetal acceleration of the car is 2.6 m/s 2 . c) F s = F n
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Unformatted text preview: u s F s = (1450 kg)(9.8 kg/N)(1.1) F s = 1.6 x 10 4 N The magnitude of the static friction is 1.6 x 10 4 N. d) u s = F s / F n u s = (1.6 x 10 4 N) / (1450 kg)(9.8 kg/N) u s = 1.1 The coefficient would be 1.1. 16)a) t = 5.93 x 10 7 s r = 2.28 x 10 11 m v = [2(pi)(r)] / T v = [2(3.1416)(2.28 10 11 m)] / (5.93 x 10 7 s) v = 2.4 x 10 4 m/s The orbital speed of mars is 2.4 x 10 4 m/s. b) v 2 = Gm 1 / r m 1 = v 2 r / G m 1 = [(2.4 x 10 4 m/s) 2 (2.28 x 10 11 m)] / (6.67 x 10-11 (N * m 2 ) / kg 2 ) m 1 = 1.97 x 10 30 kg The mass of the sun is 1.97 x 10 30 kg....
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## This note was uploaded on 12/13/2009 for the course ENGG eg 404 taught by Professor Weisenhaupt during the Spring '09 term at University of the Bío-Bío.

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Lesson 4 - u s F s = (1450 kg)(9.8 kg/N)(1.1) F s = 1.6 x...

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