a) work done is equal to change in the potential energy
work done = mgΔh=65×10×(3700−2300)=650×1400=910kJmg\Delta h=65\times10\times(3700-2300)=650\times1400=910kJmgΔh=65×10×(3700−2300)=650×1400=910kJ
b) power = Work donetime=910×1035×60×60=50.55W\frac{Work\ done }{time}=\frac{910\times10^3}{5\times60\times60}=50.55 WtimeWork done=5×60×60910×103=50.55W
1 HP =735 W
50.55 W= 0.068 HP
c) efficiency = outputinput×100\frac{output }{input}\times100inputoutput×100
15=50.55input×100input=337W15=\frac{50.55}{input}\times100\\input=337 W15=input50.55×100input=337W
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