185_Physics ProblemsTechnical Physics

185_Physics ProblemsTechnical Physics - Chapter 6 P6.69 At...

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Chapter 6 187 P6.69 At terminal velocity, the accelerating force of gravity is balanced by frictional drag: mg arv br v =+ 22 (a) mg v v + × −− 3 1 01 0 0 8 7 0 91 0 2 .. ej e j For water, mV == L N M O Q P ρπ 1000 4 3 10 5 3 kg m m 3 4 11 10 3 10 10 0 870 10 11 9 10 2 . ×=× + × e j vv Assuming v is small, ignore the second term on the right hand side: v = 00132 . m s . (b) mg v v + × 3 1 0 0 8 7 0 88 2 e j Here we cannot ignore the second term because the coefficients are of nearly equal magnitude. 4 11 10 3 10 10 0 870 10 3 10 3 10 4 0 870 4 11 20870 103 8 2 2 . . . . . + × = −± + = e j af a f v ms (c) mg v v + × 3 1 0 0 8 7 0 76 2 e j Assuming v > 1 m s, and ignoring the first term: 411 10 0870 10 56 2 ×= × v v = 687 . m s P6.70 v mg b bt m = F H G I K J F H G I K J L N M O Q P 1 exp where exp xe x a f = is the exponential function. At t →∞ , mg b T →= At t = 554 s 0500 1 900 .e x p . . b TT =− F H G I K J L N M M O Q P P s kg exp .
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