The number of protons determines the element. Many elements also have many isotopes. Such as carbon-14 Carbon-12 and carbon-10.
Answer:
The magnitude of gravitational force between two masses is
.
Explanation:
Given that,
Mass of first lead ball, 
Mass of the other lead ball, 
The center of a large ball is separated by 0.057 m from the center of a small ball, r = 0.057 m
We need to find the magnitude of the gravitational force between the masses. It is given by the formula of the gravitational force. It is given by :

So, the magnitude of gravitational force between two masses is
. Hence, this is the required solution.
The area of the circle with radius r is
A = πr²
The rate of change of area with respect to time is

The rate of change of the radius is given as

Therefore

When r = 10 ft, obtain

Answer: - 40π ft²/s (or - 127.5 ft²/s)
Answer:
m³/(kg⋅s²)
Explanation:
Hello.
In this case, since the involved formula is:

By writing a dimensional analysis with the proper algebra handling, we obtain:
![N[=]G*\frac{kg*kg}{m^2}\\ \\kg*\frac{m}{s^2}[=]G *\frac{kg*kg}{m^2}\\\\G[=]\frac{kg*m*m^2}{kg^2*s^2}\\ \\G[=]\frac{m^3}{kg*s^2}](https://tex.z-dn.net/?f=N%5B%3D%5DG%2A%5Cfrac%7Bkg%2Akg%7D%7Bm%5E2%7D%5C%5C%20%5C%5Ckg%2A%5Cfrac%7Bm%7D%7Bs%5E2%7D%5B%3D%5DG%20%2A%5Cfrac%7Bkg%2Akg%7D%7Bm%5E2%7D%5C%5C%5C%5CG%5B%3D%5D%5Cfrac%7Bkg%2Am%2Am%5E2%7D%7Bkg%5E2%2As%5E2%7D%5C%5C%20%5C%5CG%5B%3D%5D%5Cfrac%7Bm%5E3%7D%7Bkg%2As%5E2%7D)
Thus, answer is:
m³/(kg⋅s²)
Note that the [=] is used to indicate the units of G.
Best regards
I think the answer should be the last one. Magnets attract magnets with unlike poles and repel magnets with like poles