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Answer to Question #80695 in Physics for Dirk McPherson
Question #80695
If 40 MJ of work are done in lifting 5Mg (or 5 t) of coal is 120s, what is the power required?
Expert's answer
power=(work done)/time
=(〖40×10〗^6 J)/(120 s)=3.33×〖10〗^5 W=333 kW
=(〖40×10〗^6 J)/(120 s)=3.33×〖10〗^5 W=333 kW
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