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Fluid II-solutions

Fluid II-solutions - le(txl65 Fluid II mohanty(81440 This...

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le (txl65) – Fluid II – mohanty – (81440) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points A fountain sends a stream of water 23 . 8 m up into the air. If the base of the stream is 8 . 9 cm in diame- ter, what power is required to send the water to this height? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 31339 . 3 W. Explanation: Let : h 1 = 23 . 8 m , d = 8 . 9 cm = 0 . 089 m , r = 4 . 45 cm = 0 . 0445 m , and g = 9 . 8 m / s 2 . Using conservation of energy, 1 2 m v 2 = m g h v = radicalbig 2 g h = radicalBig 2 (9 . 8 m / s 2 ) (23 . 8 m) = 21 . 5981 m / s is the velocity of the stream at its base. The flow rate at the base is π r 2 v = π (0 . 0445 m) 2 (21 . 5981 m / s) = 0 . 134365 m 3 / s . Using the density of water as 1000 kg / m 3 , we see that 134 . 365 kg of water emerges from the fountain each second. m t = ρ π r 2 v = (1000 kg / m 3 ) π (0 . 0445 m) 2 × (21 . 5981 m / s) = 134 . 365 kg . Thus, each second, sufficient work is done to raise 134 . 365 kg to a height of 23 . 8 m or P = W t = m g h t = ρ π r 2 v g h t = (1000 kg / m 3 ) π (0 . 0445 m) 2 × (21 . 5981 m / s) (9 . 8 m / s 2 ) 23 . 8 m 1 s = 31339 . 3 W . 002 10.0points A styrofoam slab has a thickness of 10 . 3 cm and a density of 319 kg / m 3 . What is the area of the slab if it floats just awash (top of slab is even with the water sur- face) in fresh water when a 65 . 1 kg swimmer is aboard? The density of water is 1000 kg / m 3 . Correct answer: 0 . 928104 m 2 . Explanation: Let : d = 10 . 3 cm , ρ s = 319 kg / m 3 , and m = 65 . 1 kg . Since the slab floats awash, the buoyant force is B = ρ w V g = ρ w d A g , and is equal in magnitude to the weight of the slab and the swimmer: ρ w d A g = ρ s d A g + m g A = m d ( ρ w - ρ s ) = 65 . 1 kg (10 . 3 cm) (1000 kg / m 3 - 319 kg / m 3 ) = 0 . 928104 m 2 .
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