430_Physics ProblemsTechnical Physics

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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,
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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 .

Ask a homework question - tutors are online