The answer is (A) hope it helps
Explanation:
If the center of the load is directly above the vertebrae, there is no torque in the system. This is a good thing so that the vertebrae are not put out of alignment over time. (Of course, this still doesn't prevent compression of the vertebrae over time, which is a possibility.)
<span>We can use Coulomb's law to find the force F acting on the proton that is released.
F = k x Q1 x Q2 / r^2
k = 9 x 10^9
Q1 is the charge on one proton which is 1.6 x 10^{-19} C
Q2 is the same charge on the other proton
r is the distance between the protons
F = (9x10^9) x (1.6 x 10^{-19} C) x (1.6 x 10^{-19} C) / (10^{-3})^2
F = 2.304 x 10^{-22} N
We can use the force to find the acceleration.
F = ma
a = F / m
a = (2.304 x 10^{-22} N) / (1.67 x 10^{-27} kg)
a = 1.38 x 10^5 m/s^2
The initial acceleration of the proton is 1.38 x 10^5 m/s^2</span>
The acceleration due to gravity is g/4
The acceleration above the earth surface is given by the relation
g^'=gr^2/〖(h+r)〗^2
Since the satellite orbits the earth in a orbit of radius equal to earth radius, therefore
g^'=(gr^2)/〖(r+r)〗^2 =g/4
Thus the acceleration due to gravity on the satellite is g/4.
