Asked by Tim
A lawnmower engine with an efficiency of 0.29 rejects 5500 J of heat every second. What is the magnitude of the work that the engine does in one second?
Answers
Answered by
bobpursley
.29=(input-waste)/input
solving for input..
input(1-.29)=waste
but work=.29 input
work/.29 (.71)=5500
work=.29*5500/.71
check that
solving for input..
input(1-.29)=waste
but work=.29 input
work/.29 (.71)=5500
work=.29*5500/.71
check that
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