Answer:

Explanation:
Acceleration is given by

where
is the change in velocity
is the time interval in which the change in velocity occurs
To find the acceleration at 1 second, we can take the data at t = 1 s and t = 2. We find:


So, the acceleration is

Answer A is incorrect
A crest is just one point. It is not the distance between 2 crests.
B is incorrect
A trough is just 1 point. It is not the distance between 2 troughs.
C is incorrect.
the amplitude measures the height of a crest from the middle of the wave to the crest (or trough).
D is the correct answer. That is the distance between 2 crests or 2 troughs
The amount of movement, linear momentum, momentum or momentum is a physical quantity derived from a vector type that describes the movement of a body in any mechanical theory. In classical mechanics, the amount of movement is defined as the product of body mass and its velocity at a given time.
p= mv
Where,
m = mass
v = Velocity
Our values are given as,


Replacing we have that,


Therefore the momentum is 
Maybe you can split up the questions. I will try to answer your first question.
1. In an elastic collision, momentum is conserved. The momentum before the collision is equal to the momentum after the collision. This is a consequence of Newton's 3rd law. (Action = Reaction)
2. Momentum: p = m₁v₁ + m₂v₂
m₁ mass of ball A
v₁ velocity of ball A
m₂ mass of ball B
v₂ velocity of ball B
Momentum before the collision:
p = 2*9 + 3*(-6) = 18 - 18 = 0
Momentum after the collision:
p = 2*(-9) + 3*6 = -18 + 18 = 0
3: mv + m(-v) = m(-v) + m(v)
the velocities would reverse.
4.This question is not factual since the energy of an elastic collision must also be conserved. The final velocities should be: v₁ = -1 m/s and v₂ = 5 m/s. That said assuming the given velocities were correct:
before collision
p = 10*3 + 5*(-3) = 30 - 15 = 15
after collision:
p = 10*(-2) + 5 * v₂ = 15
v₂ = 7
5.You figure out.
Answer:
The answer is a for Plato users.
Explanation:
Since the angle of the refracted ray moves away from the normal, it must be traveling in a faster medium.