Answer:
The farther star will appear 4 times fainter than the star that is near to the observer.
Explanation:
Since it is given that the luminosity of the 2 stars is same thus they radiate the same energy per unit time
Consider a spherical wave front of energy 'E' that leaves both the stars (Both radiate 'E' as they have same luminosity)
This Energy is spread over the whole surface area of sphere Thus when the wave front is at a distance 'r' the energy per unit surface area is given by

For the star that is twice away from the earth the distance is '2r' thus we will receive an energy given by
Hence we sense it as 4 times fainter than the nearer star.
The JWST is postioned about 1.5 million kilometers from the earth on the side facing away from the sun
G is the gravitational constant, which is approximately 6.6x10^-11 Nm/s^2. It has the same value regardless of the masses of both objects or the distance between them.
Answer:
v=-30 m/s
Explanation:
Using the equation:
v=u+at
v=?
u means initial velocity=0
a=-10 since it is going down we make the acceleration due to gravity negative
t=3
v=0+(-10)(3)
v=-30 m/s