The reason an astronaut in an earth satellite feels weightless is that the astronaut is falling.
Option a
<u>Explanation: </u>
The other options except Option is not applicable since the gravitational force is a long range force, in which the satellite revolves very close to the surface of the Earth where the gravity is felt.The zero weight experienced by the astronaut in a satellite is due to the earth pulling along with satellite. Due to gravitational force of the Earth,the astronaut falls freely .
But why not the satellite comes down due to gravity when its launched in space. The fact is that the satellite is launched with velocity of tangent direction and it is very high. The centripetal force balances the gravity.
I'm not sure what you're asking but the earth has the ability to infinitely continue to spin or the earth completes 365.25 rotations during a full cycle.
So simply it to 120m/m for 120 minutes. So then you multiply 120x120 and that equals 14,400
Answer:
h' = 603.08 m
Explanation:
First, we will calculate the initial velocity of the pellet on the surface of Earth by using third equation of motion:
2gh = Vf² - Vi²
where,
g = acceleration due to gravity on the surface of earth = - 9.8 m/s² (negative sign due to upward motion)
h = height of pellet = 100 m
Vf = final velocity of pellet = 0 m/s (since, pellet will momentarily stop at highest point)
Vi = Initial Velocity of Pellet = ?
Therefore,
(2)(-9.8 m/s²)(100 m) = (0 m/s)² - Vi²
Vi = √(1960 m²/s²)
Vi = 44.27 m/s
Now, we use this equation at the surface of moon with same initial velocity:
2g'h' = Vf² - Vi²
where,
g' = acceleration due to gravity on the surface of moon = 1.625 m/s²
h' = maximum height gained by pellet on moon = ?
Therefore,
2(1.625 m/s²)h' = (44.27 m/s)² - (0 m/s)²
h' = (1960 m²/s²)/(3.25 m/s²)
<u>h' = 603.08 m</u>
Answer:
changing the direction in which a force is exerted
