479_Physics ProblemsTechnical Physics

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

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Chapter 16 481 P16.23 v T == × = µ 1350 500 10 520 3 kg m s kg m ms 2 . P16.24 (a) ωπ π = 2 2 500 3 140 f af rad s , k v = ω 3140 196 16 0 . r adm yx t 200 10 160 4 .s i n . m ej bg (b) v T × 196 410 10 3 kg m . T = 158 N P16.25 TM g = is the tension; v T Mg MgL m L t m L = = is the wave speed. Then, MgL m L t = 2 2 and g Lm Mt × × = 2 3 3 2 1 60 4 00 10 300 10 164 .. . . m k g kg 3.61 s 2 P16.26 v T = Tv A v r v T T = = µρ ρ 22 2 2 4 2 2 8 920 7 50 10 200 631 kg m m m s N 3 . P16.27 Since μ is constant, T v T v 2 2 2 1 1 2 and T v v T 2 2 1 2 1 2 30 0 20 0 600 135 = F H G I K J = F H G I K J = . . N N . P16.28 The period of the pendulum is T L g = 2 Let F represent the tension in the string (to avoid confusion with the period) when the pendulum is
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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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