Let F be the magnitude of the force. The impulse of this force while the ball is in contact with the wall is
Ft = F (0.0210 s)
and this impulse is equal to the change in the ball's momentum,
m ∆v = (1.30 kg) (6.50 m/s - (-10.5 m/s)) = (1.30 kg) (17.0 m/s)
Solve for F :
F (0.0210 s) = (1.30 kg) (17.0 m/s)
F = (1.30 kg) (17.0 m/s) / (0.0210 s)
F ≈ 1050 N
Answer: C
Explanation:
In collision, whether elastic or inelastic collisions, momentum is always conserved. That is, the momentum before collision will be equal to the momentum after collision.
Change in momentum of the system will be momentum after collision minus total momentum before collision.
Since momentum is a vector quantity, the direction will also be considered.
Momentum = MV - mU
Let
M = 800 kg is going north
at V = 20 m/s and the other car
m= 800 kg is going south
at U = 10m/s.
Substitute all the parameters into the formula
Momentum = (800 × 20) - (800 × 10)
= 8000 kgm/s
The final momentum after collision will also be equal to 8000 kgm/s
Change in momentum = 8000 - 8000
Change in momentum = 0
Answer:
Explanation:
In an experimental research, the control group is the group that serves as the neutral group that is not given any form of treatment and serves as the group in which the experimental groups are firstly compared to. Thus, <u>the control group in the question described is the Third group</u>.
While experimental groups are the groups that receive treatments required to make an inference from the experiment. From this description, <u>it can be deduced that the First and the Second group are the experimental groups.</u>
Anwser
The strength of a force is usually expressed by its magnitude. We have also to specify the direction in which a force acts.
