A 7.9 g bullet leaves the muzzle of a rifle with ayvaci (na25565) – Chapter 4 Basics – balantic – (APPhysPd4Balant) 2 a speed of 443.4 m/s. What constant force is exerted on the bullet while it is traveling down the 0.7 m length of the barrel of the rifle? Answer in units of N.

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JemdetNasr JemdetNasr · Answer 1 · 2019-10-25T16:32:54+02:00

Answer:

[tex]1109.41[/tex] N

Explanation:

[tex]v_{o}[/tex] = initial velocity of the bullet = 0 m/s

[tex]v_{f}[/tex] = final velocity of the bullet as it leaves = 443.4 m/s

[tex]a[/tex] = acceleration of the bullet

[tex]d[/tex] = length of the barrel of the rifle = 0.7 m

the kinematics equation we can use must include the variables in the above list, hence

[tex]{v_{f}}^{2} = {v_{o}}^{2} + 2 a d[/tex]

[tex]{443.4}^{2} = {0}^{2} + 2 (0.7)a[/tex]

[tex]a = 140431.11[/tex] ms⁻²

[tex]m[/tex] = mass of the bullet = 7.9 g = 0.0079 kg

Force exerted on the bullet is given as

[tex]F = ma[/tex]

[tex]F = (0.0079)(140431.11)[/tex]

[tex]F = 1109.41[/tex] N