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OpenStudy (anonymous):
A water pump is run by an electric motor with a power rating of 750W. It is used to pump water from a reservoir up to a height of 37 m and into a water tower at a rate of 1.48 kg per second. What is the useful energy output
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OpenStudy (anonymous):
power of motor =750W=750joules/s potential energy generated by 1.48kg of water at 37 m =mgh=1.48*9.8*37=536.648joules 536.648joules of PE is generated in 1 second by the pump . useful energy output=(536.648/750)*100=71.55% sorry if I went wrong.
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