Answer:
r = 4.24x10⁴ km.
Explanation:
To find the radius of such an orbit we need to use Kepler's third law:

<em>where T₁: is the orbital period of the geosynchronous Earth satellite = 1 d, T₂: is the orbital period of the moon = 0.07481 y, r₁: is the radius of such an orbit and r₂: is the orbital radius of the moon = 3.84x10⁵ km. </em>
From equation (1), r₁ is:
Therefore, the radius of such an orbit is 4.24x10⁴ km.
I hope it helps you!
There’s no picture so how r we supposed to answer it
It should be 0.25kg because you converter from g to kg and since 1g<1kg so you move the decimal to the left
The lens equation gives d relation between focal length, object distance n image distance.
1/f = 1/v + 1/u
seldon
The correct answer should be bounced off since the light ray hit the surface and reflected towards a new location. It therefore bounced off.