A lead bulled strikes at target with a velocity of 480 m$s^{- 1}$. If the bullet falls dead, the rise in temperature of bullet ($c$ = 0.03), assuming that heat produced is equally shared between the bullet and target is

Question

Dr. Abubacker · Accepted Answer

(b) $mc$ ($\mathrm{\Delta}T) = \frac{\frac{1}{2}\left( \frac{1}{2}m\ v^{2} ight)}{J}$
$\mathrm{\Delta}T = \frac{v^{2}}{4Jc} = \ \frac{480 	imes 480}{4 	imes 4.2 	imes (0.03 	imes 10^{3})} = 457℃$

Q. A lead bulled strikes at target with a velocity of 480 m$s^{- 1}$. If the bullet falls dead, the rise in temperature of bullet ($c$ = 0.03), assuming that heat produced is equally shared between the bullet and target is

Explanation

Reviewed by Dr. Abubacker

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