It is wavelength i think so!!!!
Answer:
(a) v = 5.42m/s
(b) vo = 4.64m/s
(c) a = 2874.28m/s^2
(d) Δy = 5.11*10^-3m
Explanation:
(a) The velocity of the ball before it hits the floor is given by:
(1)
g: gravitational acceleration = 9.8m/s^2
h: height where the ball falls down = 1.50m

The speed of the ball is 5.42m/s
(b) To calculate the velocity of the ball, after it leaves the floor, you use the information of the maximum height reached by the ball after it leaves the floor.
You use the following formula:
(2)
vo: velocity of the ball where it starts its motion upward
You solve for vo and replace the values of the parameters:

The velocity of the ball is 4.64m/s
(c) The acceleration is given by:


The acceleration of the ball is 2874.28/s^2
(d) The compression of the ball is:

THe compression of the ball when it strikes the floor is 5.11*10^-3m
different because joules law talks about heat produce in an electric whiles ohm' law talks about potential difference
Answer:
fr = ½ m v₀²/x
Explanation:
This exercise the body must be on a ramp so that a component of the weight is counteracted by the friction force.
The best way to solve this exercise is to use the energy work theorem
W = ΔK
Where work is defined as the product of force by distance
W = fr x cos 180
The angle is because the friction force opposes the movement
Δk =
–K₀
ΔK = 0 - ½ m v₀²
We substitute
- fr x = - ½ m v₀²
fr = ½ m v₀²/x