Water is pumped steadily out of a flooded basement at a speed of 5.30 m/s through a uniform hose of radius 9.70 mm. The hose passes out through a window 2.90 m above the water line. How much power is supplied by the pump?
Answers
Answer:
Here is the answer
Explanation:
The cross-sectional area of the hose is
A = πr2
Here, radius of the hose is r.
Insert the values of in the equation R = vA gives
R = v (πr2)
Using the above values in the equation Rm = ρR
gives
Rm = ρv (πr2)
Substitute the values of K, U and Rm in the equation P = (K+U) Rm gives
P = [(1/2) v2 + gh] [ρv (πr2)]
To obtain the power supplied by the pump, substitute 5.30 m/s for v, 9.8 m/s2 for g, 2.90 m for h, 1000 kg/m3 for ρ and 9.70 mm for r in the above equation gives
P = [(1/2) v2 + gh] [ρv (πr2)]
= [(1/2) (5.30 m/s)2 + (9.8 m/s)2 (2.90 m)] [(1000 kg/m3) (5.30 m/s) (3.14) (9.70 mm)2 (10-3 m/1 mm)2]
= (14.045 m2/s2 + 28.42 m2/s2) (1.56584578 kg/s)
= (66.494 kg.m2/s) [1 W/(1kg.m2/s)]
= 66.494 W
Rounding off to three significant figures, the power supplied by the pump is 66.5 W.
Answer:
This is the answerer 81
Explanation:
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