Answer: The small spherical planet called "Glob" has a mass of 7.88×1018 kg and a radius of 6.32×104 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×103 m, above the surface of the planet, before it falls back down.
1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.
2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.
Explanation: To find the answer, we need to know about the different equations of planetary motion.
<h3>How to find the initial speed of the rock as it left the astronaut's hand?</h3>
- We have the expression for the initial velocity as,

- Thus, to find v, we have to find the acceleration due to gravity of glob. For this, we have,

- Now, the velocity will become,

<h3>How to find the speed of the satellite?</h3>
- As we know that, by equating both centripetal force and the gravitational force, we get the equation of speed of a satellite as,

Thus, we can conclude that,
1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.
2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.
Learn more about the equations of planetary motion here:
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250 mph? i think that’s correct lol, if it is mark brainliest please;)
<span>Answer:
Total kinetic energy at the bottom = 0.5(1+0.4) mv^2 = mgh
V^2 = 7*9.8/0.7
V = 9.9m/s
ω = V/r = 9.9/1.7 = 5.8rad/s
Answer c. 5.8 rad/s</span>
Explanation:
It is given that,
= -40 mi/h,
= -40 mi/h
The negative sign indicates that x and y are decreasing.
We have to find
. Equation for the given variables according to the Pythagoras theorem is as follows.

Now, we will differentiate each side w.r.t 't' as follows.

or, 
So, when x = 4 mi, and y = 3 mi then z = 5 mi.
As, 
= 
= 
= 52
Thus, we can conclude that the cars are approaching at a rate of 52 mi/h.