I may be wrong, but I think you're trying to say that Planet-A is
<em>3 times as far from the sun</em> as Planet-C is.
If that's the real question, then the answer is that the period of Orbit-A
is about<em> 5.2</em> times as long as the period of Orbit-C .
Orbital period ≈ (proportional to) (the orbital distance) ^ 3/2 power.
This was empirically demonstrated about 350 years ago by Johannes
and his brilliant Kepple, and derived about 100 years later by Newton
from his formula for the forces of gravity.
The total quantity of electrons that have flowed through a circuit is a
quantity of charge, measured in Coulombs, or in Ampere-seconds.
The <em><u>rate</u></em> of flow of electrons, or more accurately the rate of flow of
the charge on them, is electrical current. Its unit is the Ampere.
1 Ampere is 1 Coulomb of charge per second.
Answer:
The solution and the explanation are in the Explanation section.
Explanation:
According to the diagram that is in the attached image, the EFFORT force at point A and the load is at O point. The torque due to weight is:
TA = W * (a * cosθ)
The torque due to effort at C point is:
TC = E * (b * cosθ)
The net torque is equal to 0, we have:
Tnet = 0
W * (a * cosθ) - E * (b * cosθ) = 0

From the figure, you can observe that a/b < 1, thus E < W
Answer:
Recall that the electric field outside a uniformly charged solid sphere is exactly the same as if the charge were all at a point in the centre of the sphere:

lnside the sphere, the electric field also acts like a point charge, but only for the proportion of the charge further inside than the point r:

To find the potential, we integrate the electric field on a path from infinity (where of course, we take the direct path so that we can write the it as a 1 D integral):

=![\frac{q}{4\pi e_{0} } [\frac{1}{R} -\frac{r^{2}-R^{2} }{2R^{3} } ]](https://tex.z-dn.net/?f=%5Cfrac%7Bq%7D%7B4%5Cpi%20e_%7B0%7D%20%7D%20%5B%5Cfrac%7B1%7D%7BR%7D%20-%5Cfrac%7Br%5E%7B2%7D-R%5E%7B2%7D%20%20%7D%7B2R%5E%7B3%7D%20%7D%20%5D)
∴NOTE: Graph is attached