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Answer:
1.
2.
3.The results from part 1 and 2 agree when r = R.
Explanation:
The volume charge density is given as

We will investigate this question in two parts. First r < R, then r > R. We will show that at r = R, the solutions to both parts are equal to each other.
1. Since the cylinder is very long, Gauss’ Law can be applied.

The enclosed charge can be found by integrating the volume charge density over the inner cylinder enclosed by the imaginary Gaussian surface with radius ‘r’. The integration of E-field in the left-hand side of the Gauss’ Law is not needed, since E is constant at the chosen imaginary Gaussian surface, and the area integral is

where ‘h’ is the length of the imaginary Gaussian surface.

2. For r> R, the total charge of the enclosed cylinder is equal to the total charge of the cylinder. So,

3. At the boundary where r = R:

As can be seen from above, two E-field values are equal as predicted.
Answer:
250 J
Explanation:
Apply the formula:
K = 1/2.m.v²
K = 1/2.5.10²
K = 1/2.5.100
K = 5.50
K = 250 Joules
But, remember that if the speed is accelerating or not, if it is, then we need to know the point in time that the question is asking for the Kinetic Energy.
In this case, I think it is just a constant speed.
Yes, if we know the Earth's mass
Explanation:
The momentum of an object is a vector quantity given by the equation

where
m is the mass of the object
v is its velocity
In this case, we are asked if we can find the velocity of the Earth by starting from its momentum. Indeed, we can. In fact, we can rewrite the equation above as

Therefore, if we know the momentum of the Earth (p) and we know its mass as well (m), we can solve the formula to find the Earth's velocity.
Learn more about momentum:
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