A boxed 12.0 kg computer monitor is dragged by friction 5.50 m up along the moving surface of a conveyor belt inclined at an ang
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Answer:
a ) 4.5 N.s
b) V =5 m/s
Explanation:
given,
mass of rifle(M) = 0.9 kg
mass of bullet(m) = 6 g = 0.006 kg
velocity of the bullet(v) = 750 m/s
a) momentum of bullet = m × v
= 750 × 0.006
= 4.5 N.s
b) recoil velocity
m × u + M × U = m × v + M × V
0 + 0 = 0.006 × 750 - 0.9 × V
V =
V =5 m/s