A bullet is fired from a gun. The force on th | vedclass

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Question

When force on the bullet is zero, then $0=600-2 	imes 10^{5} t$

or $\quad t=\frac{600}{2 	imes 10^{5}}=3 	imes 10^{-3} \mathrm{s}$

Now, $\qua

Vedclass · Accepted Answer

$0.9\ kg / s$

Vedclass · Answer

When force on the bullet is zero, then $0=600-2 	imes 10^{5} t$

or $\quad t=\frac{600}{2 	imes 10^{5}}=3 	imes 10^{-3} \mathrm{s}$

Now, $\quad I=\int_{0}^{t} F d t=\int_{0}^{3 	imes 10^{-3}}\left(600-2 	imes 10^{5} tight) d t$

$=\left[600 t-\frac{2 	imes 10^{5} 	imes t^{2}}{2}ight]_{0}^{3 	imes 10^{-3}}$

$=\left[600 	imes 3 	imes 10^{-3}-10^{5} 	imes 9 	imes 10^{-6}ight]$

$=\left\lceil 18 	imes 10^{-1}-0.9ightceil=\lceil 1.8-0.9ceil= 0.9 N-s$

A bullet is fired from a gun. The force on the bullet is given by:

$F = 600 - 2 \times 10^5\ t$

Where $F$ is in newton and $t$ in second. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet?

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A bullet is fired from a gun. The force on the bullet is given by: $F = 600 - 2 \times 10^5\ t$ Where $F$ is in newton and $t$ in second. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet?