if the momentum is increased by 20%, what is the % change in its kinetic energy?
Answers
Answered by
0
Answer:
p=mv
ke=p^2/2m
p'=p+ 0.2p
=1.2p
ke'=p'^2/2m
=1.44p/2m
∆ke%=(ke'-ke) *100
i. e. 44%
Answered by
1
Answer:
Explanation:
%change in K. E is given by
1-{kE(initial)/kE(final))*100
We need to cal. Initial and final kE
Relation is p(square)/2m=kE
P(initial)= 1 (100%)ie 100/100
P(final)=1+0.2=1.2 (100+20)% ie 100+20)/100
KE props P(square)
K(initial) props (1)square
K(final) props 1.2(square)=1.44
From 1st formula
1-{1.44/(1)}=0.44*100
44%
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