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A 5.30-g lead bullet traveling at 390 m/s is stopped by a large tree. If half the kinetic energy of the bullet is transformed into internal energy and remains with the bullet while the other half is transmitted to the tree, what is the increase in temperature of the bullet?

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Answered by Elena
KE=mv²/2
½KE=mcΔT
ΔT= KE/2mc= mv²/4mc=
= v²/4c=390²/4•130=292.5℃
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