A bullet of mass 5, g,travelling with a speed | vedclass

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(B) The initial kinetic energy of the bullet is $K = \frac{1}{2} m v^2$.Given $m = 5\, g$ and $v = 210\, m/s = 21000\, cm/s$.$K = \frac{1}{2} 	imes 5

Vedclass · Accepted Answer

$87.5$

Vedclass · Answer

(B) The initial kinetic energy of the bullet is $K = \frac{1}{2} m v^2$.Given $m = 5\, g$ and $v = 210\, m/s = 21000\, cm/s$.$K = \frac{1}{2} 	imes 5 	imes (21000)^2 = 1.1025 	imes 10^9\, ergs$.One half of this energy is converted into heat in the bullet:$Q = \frac{1}{2} K = \frac{1}{2} 	imes 1.1025 	imes 10^9 = 5.5125 	imes 10^8\, ergs$.We know $Q = m s \Delta T$,where $s = 0.030\, cal/(g \cdot ^{\circ}C) = 0.030 	imes 4.2 	imes 10^7\, ergs/(g \cdot ^{\circ}C) = 1.26 	imes 10^6\, ergs/(g \cdot ^{\circ}C)$.Substituting the values:$5.5125 	imes 10^8 = 5 	imes (1.26 	imes 10^6) 	imes \Delta T$.$\Delta T = \frac{5.5125 	imes 10^8}{6.3 	imes 10^6} = \frac{551.25}{6.3} = 87.5^{\circ}C$.

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