# physics

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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.)

• physics -

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

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