Question

A 24.7-g bullet is fired from a rifle. It takes 2.73 × 10-3 s for the bullet to travel the length of the barrel, and it exits the barrel with a speed of 784 m/s. Assuming that the acceleration of the bullet is constant, find the average net force exerted on the bullet.

Answer 1

Answer:

$F_a_v_g=7093333.33N*s$

Explanation:

The impulse or average force in classical mechanics is the variation in the linear momentum that a physical object experiences in a closed system. It is defined by the following equation:

$F_a_v_g=m*\frac{\Delta v}{\Delta t}=m*\frac{v_2-v_1}{t_2-t_1}$

Where:

$m=mass\hspace{3}of\hspace{3}the\hspace{3}object$

$v_2=final\hspace{3}velocity\hspace{3}of\hspace{3}the\hspace{3}object\hspace{3}at\hspace{3}the\hspace{3}end\hspace{3}of\hspace{3}the\hspace{3}time\hspace{3}interval$

$v_1=initial\hspace{3}velocity\hspace{3}of\hspace{3}the\hspace{3}object\hspace{3}when\hspace{3}the\hspace{3}time\hspace{3}interval\hspace{3}begins.$

$t_2=final\hspace{3}time$

$t_1=initial\hspace{3}time$

Asumming v1=0 and t1=0:

$F_a_v_g=m* \frac{v_2}{t_2} =(24.7)*\frac{784}{2.73*10^{*3} } =7093333.333N*s$