PHSY218_CHP01_31-40

# PHSY218_CHP01_31-40 - 1.31: 7.8 km, 38 north of east 1.32:...

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1.31: east of north 38 km, 8 . 7 1.32: a) 11.1 m @ o 6 . 77 b) 28.5 m @ o 202 c) 11.1 m @ o 258 d) 28.5 m @ o 22 1.33: west. of south 41 m, 144

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1.34: 1.35: ( 29 ( 29 m. 6 . 9 37.0 cos m 0 . 12 m, 2 . 7 37.0 sin m 0 . 12 ; = = = = y x A A A ( 29 ( 29 ( 29 ( 29 m. 2 . 5 60.0 sin m 0 . 6 m, 0 . 3 60.0 cos m 0 . 6 ; m. 6 . 9 40.0 sin m 0 . 15 m, 5 . 11 40.0 cos m 0 . 15 ; - = - = - = - = - = - = = = y x y x C C B B C B 1.36: 500 . 0 m 2.00 m 00 . 1 tan (a) - = - = = X y A A θ ( 29 ( 29 ( 29 ( 29 207 6 . 26 180 500 . 0 tan 500 . 0 m 2.00 m 00 . 1 tan (d) 153 6 . 26 180 500 . 0 tan 500 . 0 m 2.00 m 00 . 1 tan ) ( 6 . 26 500 . 0 tan 500 . 0 m 2.00 m 00 . 1 tan (b) 333 6 . 26 360 500 . 0 tan 1 1 1 1 = + = = = - - = = = - = - = - = - = = = = = = = = - = - = - - - - θ A A θ θ A A θ c θ A A θ θ x y x y x y 1.37: Take the + x -direction to be forward and the + y -direction to be upward. Then the second force has components N 433 4 . 32 cos 2 2 = = F F x and
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## This homework help was uploaded on 03/25/2008 for the course PHYS 218 taught by Professor Safonov during the Spring '06 term at Texas A&M.

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PHSY218_CHP01_31-40 - 1.31: 7.8 km, 38 north of east 1.32:...

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