Answer:
The gravity on this planet is stronger than that of earth.
Explanation:
First we need to find the acceleration due to gravity value of this planet to compare its gravity force with that of the earth. Hence, we will use second equation of motion:
h = Vi t + (0.5)gt²
where,
h = height or depth of crater = 100 m
Vi = Initial Velocity of rock = 0 m/s
t = time = 4 s
g = acceleration due to gravity on this planet = ?
Therefore,
100 m = (0 m/s)(4 s) + (0.5)(g)(4 s)²
g = (200 m)/(16 s²)
g = 12.5 m/s²
on earth:
ge = 9.8 m/s²
Since,
ge < g
Therefore,
<u>The gravity on this planet is stronger than that of earth.</u>
Answer:A piece of driftwood moves up and down as water waves pass beneath it. However, it does not move toward the shore with the waves. What does this demonstrate about the propagation of waves through a medium?
A) Waves transmit energy but not matter as they progress through a medium.
B) Waves transmit matter but not energy as they progress through a medium.
C) Waves do not transmit matter or energy as they progress through a medium.
D) Waves transmit energy as well as matter as they progress through a medium.
Explanation:
A piece of driftwood moves up and down as water waves pass beneath it. However, it does not move toward the shore with the waves. What does this demonstrate about the propagation of waves through a medium?
A) Waves transmit energy but not matter as they progress through a medium.
B) Waves transmit matter but not energy as they progress through a medium.
C) Waves do not transmit matter or energy as they progress through a medium.
D) Waves transmit energy as well as matter as they progress through a medium.
The direction of work.........
Answer:
The acceleration of the satellite is 
Explanation:
The acceleration in a circular motion is defined as:
(1)
Where a is the centripetal acceleration, v the velocity and r is the radius.
The equation of the orbital velocity is defined as
(2)
Where r is the radius and T is the period
For this particular case, the radius will be the sum of the high of the satellite (
) and the Earth radius (
) :
Then, equation 2 can be used:
⇒ 
Finally equation 1 can be used:
Hence, the acceleration of the satellite is
