Answer:
pretty sure its B if it isnt im so so sorry
Explanation:
No...........................................................................Work for them urself
Answer:

Explanation:
From the question we are told that:
Initial Speed 
Time 
Angle
Generally the Newton's equation for motion is mathematically given by



Answer:
<h3>What is the angular speed of the earth around the sun? </h3>
It takes the Earth approximately 23 hours, 56 minutes and 4.09 seconds to make one complete revolution (360 degrees). This length of time is known as a sidereal day. The Earth rotates at a moderate angular velocity of

<h3>
What is the tangential speed of the earth? </h3>
The earth rotates once every 23 hours, 56 minutes and 4.09053 seconds, called the sidereal period, and its circumference is roughly 40,075 kilometers. Thus, the surface of the earth at the equator moves at a speed of 460 meters per second--or roughly 1,000 miles per hour.
To solve this problem we will begin by finding the pressure through density and average depth. Later we will find the Force, by means of the relation of the pressure and the area.

Here,
h = Depth average
= Density
Moreover,

Replacing,


Finally the force


