434_Physics ProblemsTechnical Physics

434_Physics - 436 P14.72 Fluid Mechanics(a The flow rate Av as given may be expressed as follows 25.0 liters = 0.833 liters s = 833 cm3 s 30.0 s

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436 Fluid Mechanics P14.72 (a) The flow rate, Av , as given may be expressed as follows: 25 0 0833 833 . . liters 30.0 s liters s cm s 3 == . The area of the faucet tap is π cm 2 , so we can find the velocity as v A flow rate cm s cm cm s m s 3 2 833 265 2 65 .. (b) We choose point 1 to be in the entrance pipe and point 2 to be at the faucet tap. Av 11 22 = gives v 1 0295 = . m s. Bernoulli’s equation is: PP v v g y y 12 2 2 1 2 21 1 2 −= − + ρρ ej bg and gives 3 3 1 2 10 2 65 0 295 10 9 80 2 00 + kg m m s m s kg m m s m 33 2 b g e j af . . or P gauge Pa =−= × 4 231 10 . . P14.73 (a) Since the upward buoyant force is balanced by the weight of the sphere, mg Vg R g 1 3 4 3 F H G I K J . In this problem, ρ = 078945 . g c m 3 at 20.0°C, and R = 100 . cm so we find: mR 1 3 3 4 3 4 3 3307 = F H G I K J = L N M O Q P = ρπ . gcm cm
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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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