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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Answer:
The statements are missing!
Answer:
The Answer is below!!
Explanation:
The larger the area of the parachute, the more air needs to be pushed out of the way, and so the slower it descends. the independent variable is the shape of the parachutes' canopies. The dependent variable is the drop speeds of the parachutes. How long does it take for the parachutes to reach the ground? Measure this using a stopwatch.
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A falling object (directly downward) is slowed down by air resistance. In turn it would take longer to fall.
Answer: option 1 : the electric potential will decrease with an increase in y
Explanation: The electric potential (V) is related to distance (in this case y) by the formulae below
V = kq/y
Where k = 1/4πε0
Where V = electric potential,
k = electric constant = 9×10^9,
y = distance of potential relative to a reference point, ε0 = permittivity of free space
q = magnitude of electronic charge = 1.609×10^-19 c
From the formulae, we can see that q and k are constants, only potential (V) and distance (y) are variables.
We have that
V = k/y
We see the potential(V) is inversely proportional to distance (y).
This implies that an increase in distance results to a decreasing potential and a decrease in distance results to an increase in potential.
This fact makes option 1 the correct answer