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479_Physics ProblemsTechnical Physics

479_Physics ProblemsTechnical Physics - Chapter 16 P16.23...

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Chapter 16 481 P16.23 v T = = × = µ 1 350 5 00 10 520 3 kg m s kg m m s 2 . P16.24 (a) ω π π = = = 2 2 500 3 140 f a f rad s , k v = = = ω 3 140 196 16 0 . rad m y x t = × 2 00 10 16 0 3 140 4 . sin . m e j b g (b) v T = = × 196 4 10 10 3 m s kg m . T = 158 N P16.25 T Mg = is the tension; v T Mg MgL m L t m L = = = = µ is the wave speed. Then, MgL m L t = 2 2 and g Lm Mt = = × × = 2 3 3 2 1 60 4 00 10 3 00 10 1 64 . . . . m kg kg 3.61 s m s 2 e j e j P16.26 v T = µ T v Av r v T T = = = = × = µ ρ ρπ π 2 2 2 2 4 2 2 8 920 7 50 10 200 631 kg m m m s N 3 e j a f e j b g . P16.27 Since μ is constant, µ = = T v T v 2 2 2 1 1 2 and T v v T 2 2 1 2 1 2 30 0 20 0 6 00 13 5 = F H G I K J = F H G I K J = . . . . m s m s N N a f . P16.28 The period of the pendulum is T L g =
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