Answer:
The magnitude of the electrostatic force is 120.85 N
Explanation:
We can use Coulomb's law to find the electrostatic force between the down quarks.
In scalar form, Coulomb's law states that for charges
and
separated by a distance d, the magnitude of the electrostatic force F between them is:
![F = k \frac{|q_1q_2|}{d^2}](https://tex.z-dn.net/?f=F%20%3D%20k%20%5Cfrac%7B%7Cq_1q_2%7C%7D%7Bd%5E2%7D)
where
is Coulomb's constant.
Taking the values:
![d = 4.6 \ 10^{-15} m](https://tex.z-dn.net/?f=d%20%3D%204.6%20%5C%2010%5E%7B-15%7D%20m)
![q_1 = q_2 = - \frac{e}{3} = - \frac{1.6 \ 10^{-19} \ C}{3}](https://tex.z-dn.net/?f=q_1%20%3D%20q_2%20%3D%20-%20%5Cfrac%7Be%7D%7B3%7D%20%3D%20-%20%5Cfrac%7B1.6%20%5C%2010%5E%7B-19%7D%20%5C%20C%7D%7B3%7D)
and knowing the value of the Coulomb's constant:
![k = 8.99 \ 10 ^{9} \frac{N m^2}{C^2}](https://tex.z-dn.net/?f=k%20%3D%208.99%20%5C%2010%20%5E%7B9%7D%20%5Cfrac%7BN%20m%5E2%7D%7BC%5E2%7D)
Taking all this in consideration:
![F = 8.99 \ 10 ^{9} \frac{N m^2}{C^2} \frac{ (- \frac{1.6 \ 10^{-19} \ C}{3} ) ^2}{(4.6 \ 10^{-15} m)^2}](https://tex.z-dn.net/?f=F%20%3D%208.99%20%5C%2010%20%5E%7B9%7D%20%5Cfrac%7BN%20m%5E2%7D%7BC%5E2%7D%20%5Cfrac%7B%20%28-%20%5Cfrac%7B1.6%20%5C%2010%5E%7B-19%7D%20%5C%20C%7D%7B3%7D%20%29%20%5E2%7D%7B%284.6%20%5C%2010%5E%7B-15%7D%20m%29%5E2%7D)
![F = 120.85 \ N](https://tex.z-dn.net/?f=F%20%3D%20120.85%20%20%5C%20N)
<span>12-50t=70t, t= 0.1h = 6 minutes.</span>
The answer is C. elastic potential energy
Answer:
The height of the object is 5007.4 miles.
Explanation:
Given that,
Weight of object = 200 lb
We need to calculate the value of ![Gmm_{e}](https://tex.z-dn.net/?f=Gmm_%7Be%7D)
Using formula of gravitational force
![F=\dfrac{Gmm_{e}}{r^2}](https://tex.z-dn.net/?f=F%3D%5Cdfrac%7BGmm_%7Be%7D%7D%7Br%5E2%7D)
Put the value into the formula
![200=\dfrac{Gmm_{e}}{(3958.756)^2}](https://tex.z-dn.net/?f=200%3D%5Cdfrac%7BGmm_%7Be%7D%7D%7B%283958.756%29%5E2%7D)
![200\times(3958.756)^2=Gmm_{e}](https://tex.z-dn.net/?f=200%5Ctimes%283958.756%29%5E2%3DGmm_%7Be%7D)
![Gmm_{e}=3.134\times10^{9}](https://tex.z-dn.net/?f=Gmm_%7Be%7D%3D3.134%5Ctimes10%5E%7B9%7D)
We need to calculate the height of the object
Using formula of gravitational force
![F=\dfrac{Gmm_{e}}{r^2}](https://tex.z-dn.net/?f=F%3D%5Cdfrac%7BGmm_%7Be%7D%7D%7Br%5E2%7D)
Put the value into the formula
![125=\dfrac{200\times(3958.756)^2}{r^2}](https://tex.z-dn.net/?f=125%3D%5Cdfrac%7B200%5Ctimes%283958.756%29%5E2%7D%7Br%5E2%7D)
![r^2=\dfrac{200\times(3958.756)^2}{125}](https://tex.z-dn.net/?f=r%5E2%3D%5Cdfrac%7B200%5Ctimes%283958.756%29%5E2%7D%7B125%7D)
![r^2=25074798.5](https://tex.z-dn.net/?f=r%5E2%3D25074798.5)
![r=\sqrt{25074798.5}](https://tex.z-dn.net/?f=r%3D%5Csqrt%7B25074798.5%7D)
![r=5007.4\ miles](https://tex.z-dn.net/?f=r%3D5007.4%5C%20miles)
Hence. The height of the object is 5007.4 miles.