
Assuming vertical acceleration of
, the speed after x seconds of falling is 
For the answer to the question above, let us first start with relaxation time. it is the absence of an external electric field, the free electrons in a metallic substance will move in random directions so that the resultant velocity of free electrons in any direction is equal to zero. While the Collision time it is<span> the mean </span>time<span> required for the direction of motion of an individual type particle to deviate through approximately as a consequence of </span>collisions<span> with particles of type.</span>
Answer:
The kinetic energy is: 50[J]
Explanation:
The ball is having a potential energy of 100 [J], therefore
PE = [J]
The elevation is 10 [m], and at this point the ball is having only potential energy, the kinetic energy is zero.
![E_{p} =m*g*h\\where:\\g= gravity[m/s^{2} ]\\m = mass [kg]\\m= \frac{E_{p} }{g*h}\\ m= \frac{100}{9.81*10}\\\\m= 1.01[kg]\\\\](https://tex.z-dn.net/?f=E_%7Bp%7D%20%3Dm%2Ag%2Ah%5C%5Cwhere%3A%5C%5Cg%3D%20gravity%5Bm%2Fs%5E%7B2%7D%20%5D%5C%5Cm%20%3D%20mass%20%5Bkg%5D%5C%5Cm%3D%20%5Cfrac%7BE_%7Bp%7D%20%7D%7Bg%2Ah%7D%5C%5C%20m%3D%20%5Cfrac%7B100%7D%7B9.81%2A10%7D%5C%5C%5C%5Cm%3D%201.01%5Bkg%5D%5C%5C%5C%5C)
In the moment when the ball starts to fall, it will lose potential energy and the potential energy will be transforme in kinetic energy.
When the elevation is 5 [m], we have a potential energy of
![P_{e} =m*g*h\\P_{e} =1.01*9.81*5\\\\P_{e} = 50 [J]\\](https://tex.z-dn.net/?f=P_%7Be%7D%20%3Dm%2Ag%2Ah%5C%5CP_%7Be%7D%20%3D1.01%2A9.81%2A5%5C%5C%5C%5CP_%7Be%7D%20%3D%2050%20%5BJ%5D%5C%5C)
This energy is equal to the kinetic energy, therefore
Ke= 50 [J]
Given mass= 1kg
Weight on earth = mg(gravity of earth) = 9.8N
weight on moon = mg(gravity of moon)= 1.62N
weight on outer space mg(gravity outer space = 0) = 0N
Answer:
The momentum would be doubled
Explanation:
The magnitude of the momentum of the freight train is given by:

where
m is the mass of the train
v is its speed
In this problem, we have that the speed of the train is unchanged, while the mass of the train is doubled:

therefore, the new momentum is

so, the momentum has also doubled.