Or, G = [M1 L1 T-2] × [L]2 × [M]-2 = [M-1 L3 T-2]. Therefore, the gravitational constant is dimensionally represented as M-1 L3 T-2.
Correct me if I’m wrong but I think it’s C
Answer:
If you add the 2 velocity vectors you get the velocity of the third vector which is the (change in) velocity of the cue ball after striking the cushion
V * 2 * cos 45 = 4 * cos 45 = 2.83 m/s
For this problem, we would be using the formula: Vf^2 = Vi^2 + 2ad
where:
Vf = 400m/s
Vi = 300m/s
a = ?
d = 4.0km
= 4000m
400^2 = 300^2 + 2a4000
a = [ 160000 - 90000 ] / 8000
a = 8.75m/s^2
rounding it off to 2 significant figures, will give us 8.8 m/s^2.
Answer:
Because of the way light bounces off of objects.
