Question

A bullet is fired from a gun .The force on the bullet is given by F=600-2X105t,where F is in newton and t is in second.The force on the bullet becomes zero as soon as it leaves the barrel.the time for which the bullet was in the barrel of gun is

Answer 1

F = 600 - 2 * 10⁵t

0 = 600 - 2 * 10⁵ t

t = 600 / (2 * 10⁵)

t = 3 * 10⁻³ s

t = 3 milliseconds

Time taken by bullet to leave the barrel is 3 milliseconds

F.dt = (600 - 2* 10⁵t).dt

J = [600t - (2 * 10⁵ t²/2)]

J = [600*(3*10^-3) - (2 * 10^5 * (3 * 10^-3)^2 / 2)]

J = 0.9 Ns

Impulse imparted to bullet is 0.9 Ns

hope it helps ☺☺☺☺

Answer 2

Answer:

When F=0

then 600-2*10^5t=0

therefore, t= 600/2*10^5  = 3*10^-3

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