171_Physics ProblemsTechnical Physics

# 171_Physics ProblemsTechnical Physics - Chapter 6 P6.39 af...

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Chapter 6 173 P6.39 (a) vt ve i ct a f = vv e i c 20 0 5 00 20 0 .. . s a f == , v i = 10 0 . ms . So 500 100 20 0 . = e c and −= F H G I K J 20 0 1 2 .l n c c =− = × −− ln . . 1 2 21 20 0 347 10 ch s (b) At t = 40 0 . s c = 100 0250 250 40 0 . . . ms bg af (c) e i ct = s dv dt cv e cv i ct = P6.40 Fm a = −−= =− + =+ = + = + zz kmv m dv dt kdt dv v kd t v d v kt v kt vk t v v v t t v v v v 2 2 0 2 1 0 0 0 0 0 0 0 0 0 1 11 1 1 a f *P6.41 (a) From Problem 40, v dx dt v t dx v dt t k vkd t t x k t x k t x k t xt t x t + = + = + + z 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 0 1 1 1 ln ln ln ln (b) We have ln 1 0 += t kx 1 0 t e kx so v v t v e v kx kx = + = 0 0 0 0 1 *P6.42 We write kmv D Av 22 1 2 ρ so k DA m e e kx × = −× 2 0305120 42 10 20145 53 10
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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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