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The momentumof a body increased doubled .What is the percentage of increment in its kinetic energy ? |
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Answer» Solution :`K = 1/mv^(2) = (p^(2))/(2M) "" …(1)` `K = (p^(2))/(2m) ""….(2)` `p. =2p ` ` :. K = (4p^(2))/(2m)` ` :. K . = 4K ` Increase in KINETIC energy= `(DeltaK)/K xx100` `= (3K)/K XX 100` = 300 % |
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