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434_Physics ProblemsTechnical Physics

# 434_Physics ProblemsTechnical Physics - 436 P14.72 Fluid...

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436 Fluid Mechanics P14.72 (a) The flow rate, Av , as given may be expressed as follows: 25 0 0 833 833 . . 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 π . . (b) We choose point 1 to be in the entrance pipe and point 2 to be at the faucet tap. A v A v 1 1 2 2 = gives v 1 0 295 = . m s . Bernoulli’s equation is: P P v v g y y 1 2 2 2 1 2 2 1 1 2 = + ρ ρ e j b g and gives P P 1 2 3 2 2 3 1 2 10 2 65 0 295 10 9 80 2 00 = + kg m m s m s kg m m s m 3 3 2 e j b g b g e je j a f . . . . or P P P gauge Pa = = × 1 2 4 2 31 10 . . P14.73 (a) Since the upward buoyant force is balanced by the weight of the sphere, m g Vg R g 1 3 4 3 = = F H G I K J ρ ρ π . In this problem, ρ = 0 789 45 . g cm 3 at 20.0°C, and R = 1 00 . cm so we find: m R 1 3 3 4 3 0 789 45 4 3 1 00 3 307 = F H G I K J = L
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