Hello!

Use the equation for momentum:

Plug in the given mass and velocity into the equation:


-17.555m/s
first I found the time it took for jacks stone to reach the bottom, using the formula vf = vi + at, vf and vi are final and initial velocities.
then i found the velocity at 6.6m using vf^2 = vi^2 + 2ad
and I found the time it took to get to 6.6m, so that I knew how long Jill waited to throw her stone, I used the formula d = t(vi+vf)/2, then i done total time - the time she waited, to get the time it took for there stones to hit the ground at the same time.
then to find the initial velocity of her throw I used the formula d = vit + (at^2)/2
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:
F = 2(50 N) - (50 N) = 50 N
Explanation:
The direction of F is the direction in which the two students are pushing.
Answer:
Time and velocity
Explanation:
The time taken for the velocity to double is very important to find the amount of acceleration the car acquires.
Acceleration is the rate of change of velocity with time.
Acceleration =
v is the initial velocity
u is the final velocity
t is the time taken
So, the velocity and time is needed to calculate the value of the acceleration the car undergoes.