Question

If the speed of a vehicle is reduced by 10%, by how much the KE changes?

Answer 1

Answer:

Firstly,

let mass=m

velocity=v

KE=1/2 (mv^2)

latter,

mass=m

velocity=[v-{v×(10/100)}]

=9v/10

Now,

KE=1/2×m×(9v/10)^2

Change in KE={1/2mv^2}-{1/2×m×(9v/10)^2}

=1/2m(19v^2/100)

%change in KE=([1/2m(19/100)v^2]/[1/2mv^2])×100

=19%

THANK YOU

Answer 2

Answer:

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