Answer:
Option A. 180000 Kgm/s.
Explanation:
From the question given above, the following data were obtained:
For Train Car A:
Mass of train car A = 45000 Kg
Velocity of train car A = 4 m/s
Momentum of train car A =?
For Train Car B:
Mass of train car B = 45000 Kg
Velocity of train car B = 0 m/s
Momentum is simply defined as the product of mass and velocity. Mathematically, it can be expressed as:
Momentum = mass × velocity
With the above formula, the momentum of train car A before collision can be obtained as follow:
Mass of train car A = 45000 Kg
Velocity of train car A = 4 m/s
Momentum of train car A =?
Momentum = mass × velocity
Momentum = 45000 × 4
Momentum of train car A = 180000 Kgm/s
An output for is the force a person applies to a simple machine.
Answer:
a) 0 < r < R: E = 0, R < r < 2R: E = KQ/r^2, r > 2R: E = 2KQ/r^2
b) See the picture
Explanation:
We can use Gauss's law to find the electric field in all the regions:
EA = qen/e0 where qen is the enclosed charge
Remember that the electric field everywhere outside a sphere is:
E(r) = q/(4*pi*eo*r^2) = Kq/r^2
a)
- For 0 < r < R: There is not enclosed charge because all of it remains on the outer layer of the conducting sphere, therefore E = 0 EA = 0/e0 = 0 E = 0
- For R < r < 2R: Here the enclosed charge is equal Q E = Q/(4*pi*eo*r^2) = KQ/r^2
- For r > 2R: Here the enclosed charge is equal 2Q E = Q/(4*pi*eo*r^2) + Q/(4*pi*eo*r^2) = 2Q/(4*pi*eo*r^2) = 2KQ/r^2
b) At the beginning there is no electric field this is why you see a line in zero, In R the electric field is maximum and then it starts to decrease exponentially with the distance and finally in 2R the field increase a little due to the second sphere to then continue decreasing exponentially with the distance
Using kinematics, approximately 10.1 seconds