a man of 60 kg lifts a load of mass 60 kg to a height 60 m .His efficiency is
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Input power is the power produced by man to reach the top of the building along with the body
Input power = (M + m)gh/t
Output power is the power required to take the body to top of the building
Output power = mgh/t
Effeciency = (Output power / Input power) × 100
= [ (mgh/t) / ((M + m)gh/t)] × 100
= [m / (M + m)] × 100
= [60 kg / (60 kg + 60 kg)] × 100
= 50%
Input power = (M + m)gh/t
Output power is the power required to take the body to top of the building
Output power = mgh/t
Effeciency = (Output power / Input power) × 100
= [ (mgh/t) / ((M + m)gh/t)] × 100
= [m / (M + m)] × 100
= [60 kg / (60 kg + 60 kg)] × 100
= 50%
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Answer:
yes he is efficiency hjh
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