When the force on some area is doubled and the area doesn't change,
then the pressure on that area has doubled.
Explanation:
Given that,
Mass of a freight car, 
Speed of a freight car, 
Mass of a scrap metal, 
(a) Let us assume that the final velocity of the loaded freight car is V. The momentum of the system will remain conserved as follows :

So, the final velocity of the loaded freight car is 0.182 m/s.
(b) Lost on kinetic energy = final kinetic energy - initial kinetic energy
![\Delta K=\dfrac{1}{2}[(m_1+m_2)V^2-m_1u_1^2)]\\\\=\dfrac{1}{2}\times [(30,000+110,000 )0.182^2-30000(0.85)^2]\\\\=-8518.82\ J](https://tex.z-dn.net/?f=%5CDelta%20K%3D%5Cdfrac%7B1%7D%7B2%7D%5B%28m_1%2Bm_2%29V%5E2-m_1u_1%5E2%29%5D%5C%5C%5C%5C%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%20%5B%2830%2C000%2B110%2C000%20%290.182%5E2-30000%280.85%29%5E2%5D%5C%5C%5C%5C%3D-8518.82%5C%20J)
Lost in kinetic energy is 8518.82. Negative sign shows loss.
Answer:
When she stops at a fast pace the energy and wind will take the cup forward and it will most likeley brake
Explanation:
I'm not entirely sure this is what you were looking for but I hope this helped!
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The use of force to move an object is called work. This only applies if the object moves.
This state of motionlessness occurs because all of the kinetic energy in the car is absorbed by the spring in the form of elastic potential energy. The mathematical representation is:
1/2 mv² = 1/2 kx²
25m = kx², where m is the mass of the cart, k is the spring constant and x is the spring's extension.