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**What will the final velocity of 5.0 g bullet starting from rest if a net force of 45 N is applied over a distance of 0.80 m ?**

**2.**

**3.**i tried to solve it but i don't know of my anwer is correct

F=ma, F/m=a

45/.005 kg

a=9000m/s^-2

From this formula ( d=1/2at^2) we can say that

Time= 2d/a and then take under root

2×0.80/9000 (under root or square root)

=0.013 sec

initial velocity (u) =0

a=v-u/t

at=v-u

9000×0.013=v-0

117+0=v

answer is 117m/s^-1

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