Consider a car moving on a straight road with a speed of 100m/s.the distance at which the car can be stopped is when coefficient of kinetic friction is equal to.5
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nu=a/g
a=0.5x9.8=4.9ms^-2
v=0,u=100,a=-4.9(Retardation)
v x v - u x u= 2as
S= -(100 X 100)/ 2 X -4.9
S=1020.40m
=1.02040km.
a=0.5x9.8=4.9ms^-2
v=0,u=100,a=-4.9(Retardation)
v x v - u x u= 2as
S= -(100 X 100)/ 2 X -4.9
S=1020.40m
=1.02040km.
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the answer is 1.02040 km
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