Efficiency- How much energy is wasted and how much useful energy is produced
Money- How expensive to install them
Location- She has to consider the location where she places them so that there is enough sunlight and light energy for the solar panels, obviously she should have solar panels if the location receives a sufficient amount of sunlight
Answer:
The radius of the new planet is ~2.04 * 10⁶ m, or 2,041,752 m.
Explanation:
We can use Newton's Law of Universal Gravitation:
Let's look at Newton's 2nd Law:
We can set these equations equal to each other:
The mass of the second mass (astronaut) cancels out. We are left with:
We are solving for the radius of the new planet, so we can rearrange the equation:
Substitute in our known values given in the problem (<u><em>G = 6.67 * 10⁻¹¹ </em></u><em> ; </em><u><em>M = 7.5 * 10²³</em></u><em> ; </em><u><em>a = 12</em></u>).
The radius of the new planet is ~2.04 * 10⁶ m.
In a way it’s true because you can get a ticket for getting caught littering
210 Pb ---> -ie + 210 B:
84 8.3
Answer:
Same magnitude of the 10 nc charge cause the electric field is external.
Explanation:
To do a better explanation, let's go and suppose we have an electric field of, 1300 N/C with a 10 nC charge.
As the system we are talking about is really big, and the charge is small, we can assume always if the charge is sitting right in the same point where the electric field is, then, the electric field would not suffer any kind of alteration in it's value. Therefore, no matter what value of the charge is sitting here, the electric field is independent of the charge, so it would not feel any alteration. However, the force that the charge is feeling would be stronger than in the first case.
F = qE
If charge is doubled, then the force would be bigger in the second case than in the first case, but electric field remain the same value.
