It would be D I believe! Depending on the angle of the mirror and distance positioned!
I think it’s Energy is lost when machines don’t work right.
F = 2820.1 N
Explanation:
Let the (+)x-axis be up along the slope. The component of the weight of the crate along the slope is -mgsin15° (pointing down the slope). The force that keeps the crate from sliding is F. Therefore, we can write Newton's 2nd law along the x-axis as
Fnet = ma = 0 (a = 0 no sliding)
= F - mgsin15°
= 0
or
F = mgsin15°
= (120 kg)(9.8 m/s^2)sin15°
= 2820.1 N
Answer:
4.25 m/s
Explanation:
Force, F = 22 N
Time, t = 0.029 s
mass, m = 0.15 kg
initial velocity of the cue ball, u = 0
Let v be the final velocity of the cue ball.
Use newton's second law
Force = rate of change on momentum
F = m (v - u) / t
22 = 0.15 ( v - 0) / 0.029
v = 4.25 m/s
Thus, the velocity of cue ball after being struck is 4.25 m/s.