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
Answer:
5.740 m
Explanation:
PE = mgh
900.0 J = (16.00 kg) (9.8 m/s²) h
h = 5.740 m
Combine all of the x's on one side of the equation and then finish the problem!
Answer:
The pressure in the water is 
Explanation:
Given that,
Depth = 101 m
Let P be the pressure at the bottom of water at a depth
We need to calculate the pressure in water
Using formula of pressure

Where,
= atmospheric pressure
= pressure in water
Put the value of
in to the formula

Put the value into the formula


Hence, The pressure in the water is 