Assuming 50% of heat is absorbed by the bullet the increase in temperature in there is
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Solution :
Since specific heat of lead is given in Joules, hence use
W=Q
instead of
W=JQ
. Þ
1
2
×(
1
2
m
v
2
)=m.c.Δθ
Þ
Δθ=
v
2
4c
=
(300)
2
4×150
=150
∘
C
.
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