Answer:
The magnitude will be "2212 N".
Explanation:
The given values are:
Initial speed, u = 250 m/s
Final speed, v = 0
Mass, m = 16.0 = 0.016 kg
Distance, d = 22.6 cm
On converting cm into m, we get
d = 0.226 m
As we know,
⇒  [tex]v^2-u^2=-2ad[/tex]
⇒  [tex]a=\frac{v^2-u^2}{-2d}[/tex]
On putting the estimated values, we get
⇒   [tex]=\frac{(0)^2-(250)^2}{-2\times 0.226}[/tex]
⇒   [tex]=\frac{0-62500}{-0.452}[/tex]
⇒   [tex]=13,8276 \ m/s^2[/tex]
Now,
Force, [tex]F=ma[/tex]
On putting values, we get
⇒       [tex]=0.016\times 138276[/tex]
⇒       [tex]=2212 \ N[/tex] (magnitude)
The magnitude of the stopping force on the bullet is 2,212.4 N in opposite direction to the bullet.
The given parameters;
The acceleration of the bullet is calculated as follows;
v² = u² - 2ad
where;
0 = 250² - a(2 x 0.226)
0.452a = 62500
[tex]a = \frac{62500}{0.452} \\\\a = 138,274.34 \ m/s^2\\\\[/tex]
The magnitude of the stopping force on the bullet is calculated as follows;
F = ma
F = 0.016 x 138,274.34
F = 2,212.4 N
Thus, the magnitude of the stopping force on the bullet is 2,212.4 N in opposite direction to the bullet.
Learn more here: https://brainly.com/question/19887955
