A bullet of mass 5 g travelling at a speed of 120m per sec penetrates deeply into a fixed target out is brought to rest ib0.01 sec calculate the distance penetrated by bullet and average force exerted on it
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Answered by
1
mass=5g= 0.003 kg
speed u = 120m/s
t=0.01 sec
v=0 m/s
average force = m(v-u) /t
= 48 N
acceleration =v-u/t
= -12000 m/s²
we know,
v²=u²+ 2as
thus
substituting values..
s =0.6m
speed u = 120m/s
t=0.01 sec
v=0 m/s
average force = m(v-u) /t
= 48 N
acceleration =v-u/t
= -12000 m/s²
we know,
v²=u²+ 2as
thus
substituting values..
s =0.6m
Answered by
1
mass = 5g = 0.005 kg
a = v - u/t
a= 0-120/0.01
a=-12000m/s2
1. f=ma
f= 0.005*-12000
f= -60 n
2. s=ut+1/2at2
s= 120(0.01)+1/2*-12000(0.01)(0.01)
1.2+1/2*-1.2
1.2-12/20
=1.2-0.6
=0.6m
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