Answer:
Part a)
![\frac{F_e}{F_g} = 2.74 \times 10^{12}](https://tex.z-dn.net/?f=%5Cfrac%7BF_e%7D%7BF_g%7D%20%3D%202.74%20%5Ctimes%2010%5E%7B12%7D)
Part b)
![q = 3.37 \times 10^{-4} C](https://tex.z-dn.net/?f=q%20%3D%203.37%20%5Ctimes%2010%5E%7B-4%7D%20C)
Explanation:
As we know that electric force on electric charge is given as
![F = qE](https://tex.z-dn.net/?f=F%20%3D%20qE)
here we have
![q = 1.6 \times 10^{-19}C](https://tex.z-dn.net/?f=q%20%3D%201.6%20%5Ctimes%2010%5E%7B-19%7DC)
E = 153 N/C
now force is given as
![F = (1.6 \times 10^{-19})(153) = 2.45 \times 10^{-17} N](https://tex.z-dn.net/?f=F%20%3D%20%281.6%20%5Ctimes%2010%5E%7B-19%7D%29%28153%29%20%3D%202.45%20%5Ctimes%2010%5E%7B-17%7D%20N)
Gravitational force on electric charge near surface of earth is given as
![F_g = mg](https://tex.z-dn.net/?f=F_g%20%3D%20mg)
![F_g = (9.1 \times 10^{-31})(9.81) = 8.93 \times 10^[-30} N](https://tex.z-dn.net/?f=F_g%20%3D%20%289.1%20%5Ctimes%2010%5E%7B-31%7D%29%289.81%29%20%3D%208.93%20%5Ctimes%2010%5E%5B-30%7D%20N)
now the ratio of two forces is given as
![\frac{F_e}{F_g} = \frac{2.45 \times 10^{-17}}{8.93 \times 10^{-30}}](https://tex.z-dn.net/?f=%5Cfrac%7BF_e%7D%7BF_g%7D%20%3D%20%5Cfrac%7B2.45%20%5Ctimes%2010%5E%7B-17%7D%7D%7B8.93%20%5Ctimes%2010%5E%7B-30%7D%7D)
![\frac{F_e}{F_g} = 2.74 \times 10^{12}](https://tex.z-dn.net/?f=%5Cfrac%7BF_e%7D%7BF_g%7D%20%3D%202.74%20%5Ctimes%2010%5E%7B12%7D)
Part b)
Now the ball is balanced by the electric force and the force of gravity on it
so here we have
![F_g = qE](https://tex.z-dn.net/?f=F_g%20%3D%20qE)
![mg = qE](https://tex.z-dn.net/?f=mg%20%3D%20qE)
![(5.25 \times 10^{-3})(9.81) = q(153)](https://tex.z-dn.net/?f=%285.25%20%5Ctimes%2010%5E%7B-3%7D%29%289.81%29%20%3D%20q%28153%29)
here we have
![q = 3.37 \times 10^{-4} C](https://tex.z-dn.net/?f=q%20%3D%203.37%20%5Ctimes%2010%5E%7B-4%7D%20C)
According to the Jefferson lab, "The scientific definition of work is: using a force to move an object a distance (when both the force and the motion of the object are in the same direction.)"
The distance of the galaxy is 32.86 Mpc.
Using the hubble law, v = H₀D where v = apparent velocity of galaxy = 2300 km/s, H = hubble constant = 70 km/s/Mpc and D = distance of galaxy.
Since we require the distance of the galaxy, we make D subject of the formula in the equation. So, we have
D = v/H₀
Substituting the values of the variables into the equation, we have
D = 2300 km/s ÷ 70 km/s/Mpc
D = 32.86 Mpc
So, the distance of the galaxy is 32.86 Mpc
Learn more about hubble law here:
brainly.com/question/18484687