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.
Sorry, I don't know but I think the correct answer is the first option.
(A)
Explanation:
We can see that the resistors are connected in parallel so all of them have the same voltage of 100 V. We also know that
Since resistor Y dissipates 100 W of power, we can solve for the current as
