The particle will accelerate 5m/s every second until it reaches a maximum of whatever your graph/diagram goes to, I'm in physical science and this is somewhat similar to what I am doing now but I'm not sure if that was what your looking for.
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
Because even though the moon is smaller, therefore a weaker gravitational pull, the moon is much closer to the earth than the sun, thus having a greater gravitational pull
They go in the boxes in this order:
density
2.meter
3.matter
4.hypothesis
5.control
6.kilogram
Answer:
The change in velocity is 15.83 [m/s]
Explanation:
Using the Newton's second law we have:
ΣF = m*a
The force in the graph is 185 N, therefore:
![185=0.369*a\\Where\\a=acceleration made it by the force [m/s^2]](https://tex.z-dn.net/?f=185%3D0.369%2Aa%5C%5CWhere%5C%5Ca%3Dacceleration%20made%20it%20by%20the%20force%20%5Bm%2Fs%5E2%5D)
![a=501.35[m/s^2]](https://tex.z-dn.net/?f=a%3D501.35%5Bm%2Fs%5E2%5D)
Now using the following kinematic equation:
![V^{2}=Vi^{2} + 2*a*(x-xi) \\where\\V=final velocity [m/s]\\Vi= initial velocity [m/s] = 0 the hockey disk is in rest when receives the hit.\\ x = Final position [m] = 0.4 m\\xi = initial position [m] = 0.15m\\](https://tex.z-dn.net/?f=V%5E%7B2%7D%3DVi%5E%7B2%7D%20%2B%202%2Aa%2A%28x-xi%29%20%5C%5Cwhere%5C%5CV%3Dfinal%20velocity%20%5Bm%2Fs%5D%5C%5CVi%3D%20initial%20velocity%20%5Bm%2Fs%5D%20%3D%200%20the%20hockey%20disk%20is%20in%20rest%20when%20receives%20the%20hit.%5C%5C%20x%20%3D%20Final%20position%20%5Bm%5D%20%3D%200.4%20m%5C%5Cxi%20%3D%20initial%20position%20%5Bm%5D%20%3D%200.15m%5C%5C)
Now replacing the values:
![V^{2}=0 + 2*501.35*(0.4-0.15)\\ \\V= 15.83[m/s]](https://tex.z-dn.net/?f=V%5E%7B2%7D%3D0%20%2B%202%2A501.35%2A%280.4-0.15%29%5C%5C%20%5C%5CV%3D%2015.83%5Bm%2Fs%5D)