A very elastic rubber ball is dropped from a certain height and hits the floor with a (2pts) downward speed v. Since it is so el
1 answer:
Answer:
3.True. The magnitude of momentum is the same
Explanation:
Let's propose the solution of the problem
The initial moment is
p₀ = m v
The final moment
= m (-v)
p₀ = -
Now we can review the claims
1. False. We see that the moment module is the same, but its direction changes
2. False. The impulse is a vector
3.True. The magnitude of momentum is the same
You might be interested in
Answer:
The motion is over-damped when λ^2 - w^2 > 0 or when > 0.86
The motion is critically when λ^2 - w^2 = 0 or when = 0.86
The motion is under-damped when λ^2 - w^2 < 0 or when < 0.86
Explanation:
Using the newton second law
k is the spring constante
b positive damping constant
m mass attached
x(t) is the displacement from the equilibrium position
Converting units of weights in units of mass (equation of motion)
From hook's law we can calculate the spring constant k
If we put m and k into the DE, we get
Denoting the constants
2λ = =
λ = b/0.215
λ^2 - w^2 =
This way,
The motion is over-damped when λ^2 - w^2 > 0 or when > 0.86
The motion is critically when λ^2 - w^2 = 0 or when = 0.86
The motion is under-damped when λ^2 - w^2 < 0 or when < 0.86