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430_Physics ProblemsTechnical Physics

# 430_Physics ProblemsTechnical Physics - 432 P14.64 Fluid...

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432 Fluid Mechanics P14.64 Looking at the top scale and the iron block: TBF g 1 += , Fe where BV g m g == F H G I K J ρρ ρ 00 Fe Fe Fe is the buoyant force exerted on the iron block by the oil. Thus, TF Bm g m g g 10 =− F H G I K J , Fe Fe Fe Fe or Tm g 1 0 1 F H G I K J Fe Fe is the reading on the top scale. Now, consider the bottom scale, which exerts an upward force of T 2 on the beaker–oil–iron combination. F y = 0: TTF F F gg g 12 0 + −−= ,, , beaker oil Fe F F T mmmg m g g b 21 0 0 1 =+ + = + + F H G I K J , beaker oil Fe Fe Fe Fe bg or m m g b 20 0 + F H G I K J L N M M O Q P P Fe Fe is the reading on the bottom scale. P14.65 Cu g V = 3083 . Zn Cu Zn Cu g Zn xV x V xx x x a f a f a f +− = F H G I K J = F H G I K J F H G I K J = = 5 1 7 30831 2517 1 7133 8960 1 09004 90 04% . . .. . . . . . %. P14.66 (a) From Fm a = Bm gm gm a m m a −− ==+ shell He total shell He (1) Where g = water and mV He He = Also, Vr d 4 36 3 3 π Putting these into equation (1) above,
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