A 15.7-g bullet is fired from a rifle. It takes 3.00 × 10-3 s for the bullet to travel the length of the barrel, and it exits the barrel with a speed of 779 m/s. Assuming that the acceleration of the bullet is constant, find the average net force exerted on the bullet.

F=ma=m•v/t=.

To find the average net force exerted on the bullet, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by the acceleration of the object (F = m*a).

First, let's find the acceleration of the bullet. We know that the initial velocity of the bullet (u) is 0 m/s, the final velocity (v) is 779 m/s, and the time taken (t) is 3.00 x 10^-3 s.

Using the formula:

v = u + a*t

where:
v = final velocity
u = initial velocity
a = acceleration
t = time

We can rearrange the formula to solve for acceleration:

a = (v - u) / t

Substituting the given values:

a = (779 m/s - 0 m/s) / (3.00 x 10^-3 s)
= 772666.67 m/s^2

Next, we can calculate the mass of the bullet. We are given that the bullet weighs 15.7 g. To convert this to kilograms:

mass = 15.7 g * (1 kg / 1000 g)
= 0.0157 kg

Now, we can find the average net force exerted on the bullet using Newton's second law:

F = m * a
= 0.0157 kg * 772666.67 m/s^2
= 12115.67 N

Therefore, the average net force exerted on the bullet is approximately 12115.67 Newtons.