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488_Physics ProblemsTechnical Physics

488_Physics ProblemsTechnical Physics - 490 Wave Motion T...

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490 Wave Motion P16.55 (a) v T = = × = µ 80 0 5 00 10 179 3 . . N kg 2.00 m m s e j (b) From Equation 16.21, P = 1 2 2 2 µ ω v A and ω π λ = F H G I K J 2 v P P P = F H G I K J = = F H I K = × = × 1 2 2 2 2 0 040 0 179 0 160 1 77 10 17 7 2 2 2 2 3 2 2 5 00 10 2 3 2 4 3 µ π λ π µ λ π vA v A v . . . . . kg 2.00 m m m s m W kW b g b g a f P16.56 v T = µ and in this case T mg = ; therefore, m v g = µ 2 . Now v f = λ implies v k = ω so that m g k = F H G I K J = L N M O Q P = µ ω π π 2 1 1 2 0 250 9 80 18 0 750 14 7 . . . . kg m m s s m kg 2 . *P16.57 Let M = mass of block, m = mass of string. For the block, F ma = implies T mv r m r b = = 2 2 ω . The speed of a wave on the string is then v T M r r M m t r v m M t m M m r = = = = = = = = = µ ω ω ω θ ω 2 1 0 003 2 0 084 3 . . kg 0.450 kg
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