159_Physics ProblemsTechnical Physics

# 159_Physics ProblemsTechnical Physics - Chapter 6 e je j 2...

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Chapter 6 161 P6.3 (a) F mv r == ×× × 2 31 6 2 10 8 9 1 10 2 2 01 0 0530 10 832 10 .. . . kg m s m N inward ej e j (b) a v r × × 2 6 2 10 22 220 10 913 10 . . . ms m ms inwa rd 2 P6.4 Neglecting relativistic effects. Fm a mv r c 2 F × × −− 21 6 6 11 0 2998 10 0480 622 10 27 7 2 12 . . . . kg m N a f P6.5 (a) static friction (b) ma f n mg ±± ± ± ii j j =++ − Fn m g y ==− 0 thus nm g = and v r fnm g r = = 2 µµ . Then µ = v rg 2 2 50 0 30 0 980 00850 . . . cm s cm cm s 2 bg af . P6.6 (a) a yy = , mg mv r moon down down = 2 vg r × + × = × moon 2 m m 1 5 2 1 71 0 1 0 0 1 6 51 0 63 3 . e j (b) v r T = 2 π , T = × × = 81 0 165 10 684 10 190 6 3 3 . . m s h P6.7 g = since a y = 0 The force causing the centripetal acceleration is the frictional force
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## This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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