Answer:
The value is 
Explanation:
From the question we are told that
The period of the asteroid is 
Generally the average distance of the asteroid from the sun is mathematically represented as
![R = \sqrt[3]{ \frac{G M * T^2 }{4 \pi} }](https://tex.z-dn.net/?f=R%20%3D%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7BG%20M%20%2A%20T%5E2%20%7D%7B4%20%5Cpi%7D%20%7D)
Here M is the mass of the sun with a value

G is the gravitational constant with value 
![R = \sqrt[3]{ \frac{6.67 *10^{-11} * 1.99*10^{30} * [5.55 *10^{9}]^2 }{4 * 3.142 } }](https://tex.z-dn.net/?f=R%20%3D%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7B6.67%20%2A10%5E%7B-11%7D%20%20%2A%201.99%2A10%5E%7B30%7D%20%2A%20%5B5.55%20%2A10%5E%7B9%7D%5D%5E2%20%7D%7B4%20%2A%203.142%20%7D%20%7D)
=> 
Generally

So

=> 
=> 
In telecommunication systems, Carrier frequency is a technical term used to indicate: ... The frequency of the unmodulated electromagnetic wave at the output of a conventional amplitude-modulated (AM-unsupressed carrier), or frequency-modulated (FM), or phase-modulated (PM) radio transmitter.
the area bounded by the line and the axes of a velocity-time graph is equal to the displacement of an object during that particular time period
Thank you
Answer:
Solution given:
height [H]=25m
initial velocity [u]=8.25m/s
g=9.8m/s
now;
a. How long is the ball in flight before striking the ground?
Time of flight =?
Now
Time of flight=
substituting value
- =

- =2.26seconds
<h3>
<u>the ball is in flight before striking the ground for 2.26seconds</u>.</h3>
b. How far from the building does the ball strike the ground?
<u>H</u><u>o</u><u>r</u><u>i</u><u>z</u><u>o</u><u>n</u><u>t</u><u>a</u><u>l</u><u> </u>range=?
we have
Horizontal range=u*
<h3>
<u>The ball strikes 18.63m far from building</u>. </h3>