428_Physics ProblemsTechnical Physics

# 428_Physics ProblemsTechnical Physics - 430 P14.56 Fluid...

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430 Fluid Mechanics P14.56 Assume v inside 0 P PP ++= + + =− = × + × = 001 1 2 1 000 30 0 1 000 9 80 0 500 1 4 50 10 4 90 10 455 2 53 atm atm kPa gauge bg a f a fa f .. . P14.57 The “balanced” condition is one in which the apparent weight of the body equals the apparent weight of the weights. This condition can be written as: FBFB gg −= ′ − where B and B are the buoyant forces on the body and weights respectively. The buoyant force experienced by an object of volume V in air equals: Buoyant force Volume of object air = ρ g FIG. P14.57 so we have BV g = air and ′= F H G I K J B F g g g air . Therefore, FF V F g g g =′ + − F H G I K J air . P14.58 The cross–sectional area above water is 246 0 600 0 200 0 566 0 330 0600 113 113 0330 0709 709 2 2 . . . . . rad 2 cm cm cm cm cm gcm k gm 2 all 2 water under wood all wood 33 π ρρ a f a fa f af == = = A gA A g 0.400 cm 0.80 cm FIG. P14.58
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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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