The radius of the cylinder is equal to half the diameter:

The volume of the cylinder is given by:

where h is the heigth of the cylinder. Converting into meters,

And the density of the material will be given by the ratio between the mass and the volume:

Answer:
Explanation:
Given
Car speed decreases at a constant rate from 64 mi/h to 30 mi/h
in 3 sec


we know acceleration is given by 


negative indicates that it is stopping the car
Distance traveled



s=63.038 m
Answer:
acceleration, a = 9.8 m/s²
Explanation:
'A ball is dropped from the top of a building' indicates that the initial velocity of the ball is zero.
u = 0 m/s
After 2 seconds, velocity of the ball is 19.6 m/s.
t = 2s, v = 19.6 m/s
Using
v = u + at
19.6 = 0 + 2a
a = 9.8 m/s²
Answer:
block velocity v = 0.09186 = 9.18 10⁻² m/s and speed bollet v₀ = 11.5 m / s
Explanation:
We will solve this problem using the concepts of the moment, let's try a system formed by the two bodies, the bullet and the block; In this system all scaffolds during the crash are internal, consequently, the moment is preserved.
Let's write the moment in two moments before the crash and after the crash, let's call the mass of the bullet (m) and the mass of the Block (M)
Before the crash
p₀ = m v₀ + 0
After the crash
= (m + M) v
p₀ = 
m v₀ = (m + M) v (1)
Now let's lock after the two bodies are joined, in this case the mechanical energy is conserved, write it in two moments after the crash and when you have the maximum compression of the spring
Initial
Em₀ = K = ½ m v2
Final
E
= Ke = ½ k x2
Emo = E
½ m v² = ½ k x²
v² = k/m x²
Let's look for the spring constant (k), with Hook's law
F = -k x
k = -F / x
k = - 0.75 / -0.25
k = 3 N / m
Let's calculate the speed
v = √(k/m) x
v = √ (3/8.00) 0.15
v = 0.09186 = 9.18 10⁻² m/s
This is the spped of the block plus bullet rsystem right after the crash
We substitute calculate in equation (1)
m v₀ = (m + M) v
v₀ = v (m + M) / m
v₀ = 0.09186 (0.008 + 0.992) /0.008
v₀ = 11.5 m / s