A pump is used to lift 500 kg of water from a depth of 80 m in 10s (take g=10m/s^2).Calculate:
(i) work done by the pump (ii) the power at which pump works
(iii) the power rating of the pump if it’s efficiency is 80%.
Answers
Answered by
2
Explanation:
work done =mgh =400000J
power= mgh/t= 400000/10=40000
Answered by
26
a) W=FS
W=mg×S
500×10×80
4,00,000J
b) P=W/t
=400000/10=40000W=40kW
c) efficiency=useful power/power input
40%=40/Power input
Power input = 40 × 100/40
Power input = 40kW
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