Answer:A brick is released with no initial speed from the roof of a building and strikes the ground in 2.50 s, encountering no ..
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Answer:

Explanation:
The original equation is:

We notice that:
- we have 1 atom of Fe on the left, and 2 atoms of Fe on the right
- we have 2 atoms of O on the left, and 3 atoms of O on the right
Therefore, the equation is not balanced.
In order to balance it, we can add:
- a coefficient 3 in front of 
- a coefficient 2 in front of 
So we have:

Now the oxygen is balanced, but the iron it not balanced yet, since we have 1 Fe on the left and 4 on the right. Therefore, we should add a coefficient 4 on the Fe on the left:

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)
Answer:
1902.75 kg
Explanation:
From Law of conservation of momentum,
m₁u₁ + m₂u₂ = V (m₁ + m₂).................... Equation 1
make m₂ the subject of the equation,
m₂ = (m₁V - m₁u₁)/(u₂-V)..................... Equation 2
Where m₁ = mass of the truck, m₂ = mass of the car, u₁ initial velocity of the truck, u₂ = initial velocity of the car V = common velocity
Given: m₁ = 2537 kg, u₁ = 14, V= 8 m/s, u₂ = 0 m/s ( as the car was at rest waiting at a traffic light)
Substituting into equation 2.
m₂ =[2537(8) - 2537(14)]/(0-8)
m₂ = (20296-35518)/-8
m₂ = -15222/-8
m₂ = 1902.75 kg.
Thus the mass of the car = 1902.75 kg