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A force of `5N` changes the velocity of a body from `10 ms^(-1)` to `20 ms^(-1)` in `5 sec`. How much force is required to bring about the same change in `2 sec`? |
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Answer» From `F = (dp)/(dt)`, we have `F_(1)=(dp)/(dt_(1))` and `F_(2)=(dp)/(dt_(2))` `:. (F_(2))/F_(1)=(dt_(1))/(dt_(2))=5/2 F_(2)=5/2F_(1)=5/2xx5=12.5N` |
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