a car moving with a velocity of 72kmph stops engine just before ascending an inclined road if 25% of energy is wasted in overcoming friction the car rises to a height of(g=10m/s)
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hey mate your answer is here
KE just before approaching the inclined plane ..50m
25%of it ..50m/4
therefore PE=KE....................(1)
let the hieghtt be h
PE..mgh
using eq1
mgh=50m/4
h=50/4g
5/4=1.25m
KE just before approaching the inclined plane ..50m
25%of it ..50m/4
therefore PE=KE....................(1)
let the hieghtt be h
PE..mgh
using eq1
mgh=50m/4
h=50/4g
5/4=1.25m
isvarya42:
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I did not understand it
i dont think this is the right answer.
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