Explanation:
It is given that,
Density of asteroid, 
Mass of asteroid, 
We need to find the diameter of the asteroid. The formula of density is given by:

V is the volume of spherical shaped asteroid, 



r = 2441311.12 m
Diameter = 2 × radius
d = 4882622.24 m
or

Hence, this is the required solution.
Answer:The Doppler effect is a change in the frequency of sound waves that occurs when the source of the sound waves is moving relative to a stationary listener.
As the source of sound waves approaches a listener, the sound waves get closer together, increasing their frequency and the pitch of the sound. The opposite happens when the source of sound waves moves away from the listener.
Explanation:
If gravity had no effect on a ball after you threw it ... and there also
were no air to slow it down ... then the ball would continue traveling
in a straight line, in whatever direction you threw it.
That's the heart and soul of Newton's laws of motion ... any object
keeps moving at the same speed, and in a straight line in the same
direction, until a force acts on it to change its speed or direction.\
If you threw the ball horizontally, then it would keep moving in the
same direction you threw it. But don't forget: The Earth is not flat.
The Earth is a sphere. So, as the ball kept going farther and farther
in the same straight line, the Earth would curve away from it, and it
would look like the ball is getting farther and farther from the ground.
Good morning.
We calculate the acceleration with the <em>Torricelli equation</em>:

We see that:

Now:

Now we can calculate the
resultant force that makes that acceleration of
0.3 m/s² with the 2nd Law of Newton:

We have a force of
26 N → and a friction force
F ←. Adding those vectors, he have a force
6 N →. Therefore:
26 - F = 6
F = 20 NWe have a friction force of 20 N. We calculate the kinect coefficient with the formula:

Since we are in a horizontal plane, we hava that
N = P = mg = 200 NTherefore: