A marksman holds a rifle with a mass of 2.57 kg loosely, so it can recoil freely. He fires a bullet with a mass of 3.65 g horizontally with a velocity relative to the ground of 318.4 m/s. What is the final kinetic energy in J of the rifle

KE = 0.262 J

Explanation:

given data

rifle with  mass  Mr = 2.57 kg

bullet with mass Mb = 3.65 g

velocity relative to the ground Vb = 318.4 m/s

solution

we get here final KE of rifle that is express as

KE = 0.5 × Mr × (Vr)²    ..............1

here Vr is

Mb × Vb = Mr × Vr         .........2

put here value

3.65 ×  × 318.4 = 2.57 × Vr

Vr = 0.452 m/s

now put value in eq 1 we get

KE = 0.5 × 2.57 × (0.452)²

KE = 0.262 J

the answer is ampere

the ampere is used to measure electricity. an ampere is a unit of electric current equal. a unit of measure for the flow of current in a circuit. one ampere is the amount of current flow provided when one volt of electrical pressure is applied against one ohm of resistance.