If momentum is increased by 20% then K.E increases by? plz explain plzzzzz
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Hi Mate!!!
K.E = p² / 2m
{( K.E )f -( K.E )I/ K.E)I. } ×100 =. ( pf² - pi² / pi² ) ×100
Pf= 100+20. pi = 100.
Change in K.E = ( 120)² - (100)
K.E = p² / 2m
{( K.E )f -( K.E )I/ K.E)I. } ×100 =. ( pf² - pi² / pi² ) ×100
Pf= 100+20. pi = 100.
Change in K.E = ( 120)² - (100)
bashirda:
correct answer is 44%
Answered by
1
may be it will help u
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