Asked by stephanie
                What is the average useful power output of a person who does 5.50×106 J of useful work in 6.50 h? 
Working at this rate, how long will it take this person to lift 1850 kg of bricks 1.20 m to a platform? (Work done to lift his body can be omitted because it is not considered useful output here.)
            
            
        Working at this rate, how long will it take this person to lift 1850 kg of bricks 1.20 m to a platform? (Work done to lift his body can be omitted because it is not considered useful output here.)
Answers
                    Answered by
            Damon
            
    assume you mean 5.50 * 10^6 Joules
6.50 hr * 3600 s/hr = time in seconds
power = Joules / time in seconds
( which is Watts )
part B:
power * time = 1850 * 9.81 * 1.20
    
6.50 hr * 3600 s/hr = time in seconds
power = Joules / time in seconds
( which is Watts )
part B:
power * time = 1850 * 9.81 * 1.20
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