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Sagot :
The average useful power output of robot is 0.0001 W
The time taken by robot to lift bricks is 630000000 sec
Given:
work done by robot = 5.5 j
time taken to do the work = 12.0 hours
weight of bricks = 3500 kg
length of bricks = 1.8 m
To Find:
The average useful power output of robot is
The time taken by robot to lift bricks is
Solution: Useful power output means the electric or mechanical energy made available for use, exclusive of any such energy used in the power production process.
P = W/t
P = 5.5/12*3600
P = 0.0001 W
So, average useful power output of robot is 0.0001 W
t = W/P
W = mgh So t = mgh/P
t = 3500x10x1.8/0.0001
t = 630000000 sec
So, time taken by robot to lift bricks is 630000000 sec
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