Answer:
Explanation:
Given:
- gravitational field strength of moon at a distance R from its center,

- Distance of the satellite from the center of the moon,

<u>Now as we know that the value of gravity of any heavenly body is at height h is given as:</u>

∴The gravitational field strength will become one-fourth of what it is at the surface of the moon.
Answer:
1. 0 vh g
Explanation:
As air resistance is negligible, horizontal speaking, nothing is affecting the velocity. So the horizontal velocity of the projectile stays the same, vh.
As for vertical velocity, since there's always a constant gravitational acceleration acting downward, namely g, the vertical speed will decrease until it reaches the top where it is 0, it then starting to increase in magnitude, downward, due to its gravitational acceleration g.
Answer:

Explanation:
The rotation rate of the man is:



The resultant rotation rate of the system is computed from the Principle of Angular Momentum Conservation:
![(90\,kg)\cdot (5\,m)^{2}\cdot (0.16\,\frac{rad}{s} ) = [(90\,kg)\cdot (5\,m)^{2}+20000\,kg\cdot m^{2}]\cdot \omega](https://tex.z-dn.net/?f=%2890%5C%2Ckg%29%5Ccdot%20%285%5C%2Cm%29%5E%7B2%7D%5Ccdot%20%280.16%5C%2C%5Cfrac%7Brad%7D%7Bs%7D%20%29%20%3D%20%5B%2890%5C%2Ckg%29%5Ccdot%20%285%5C%2Cm%29%5E%7B2%7D%2B20000%5C%2Ckg%5Ccdot%20m%5E%7B2%7D%5D%5Ccdot%20%5Comega)
The final angular speed is:
