'What did you find most challenging about mastering this skill?'
Ahh- What I found most challenging about the passing skill is hitting the ball the right direction. The last time I tried to pass the ball it hit someone in the face. But when I got a hang of it, It was easy until middle game. I tried to hit the ball and I faceplanted into the bar..-
'What did you find easiest about mastering this skill?'
The easiest? Well for me, nothing was easy. I couldn't even serve the ball without hitting it thw wrong direction!! I can't serve or pass a ball without it going the wrong direction or the ball hitting someone in the face! It gotten easier and easier as the game went particuallarly well. My team did end up winning thanks to one of my teammates. I can't play volleyball well at all. It was easy in some cases but hard at the same time.
Hope this helps you! :)
This question involves the concepts of the time period, orbital radius, and gravitational constant.
The distance of the satellite above the Earth's Surface is "10400 km ".
The theoretical time period of the satellite around the earth can be found using the following formula:
where,
T = Time Period of Satellite = 
R = Orbital Radius = ?
G = Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = Mass of Earth = 5.97 x 10²⁴ kg
Therefore,
![\frac{(21600\ s)^2}{R^3}=\frac{4\pi^2}{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(5.97\ x\ 10^{24}\ kg)}\\\\R = \sqrt[3]{\frac{4.66\ x\ 10^8\ s^2}{9.91\ x\ 10^{-14}\ s^2/m^3}} \\R = 1.675\ x\ 10^7\ m = 1.68\ x\ 10^4\ km](https://tex.z-dn.net/?f=%5Cfrac%7B%2821600%5C%20s%29%5E2%7D%7BR%5E3%7D%3D%5Cfrac%7B4%5Cpi%5E2%7D%7B%286.67%5C%20x%5C%2010%5E%7B-11%7D%5C%20N.m%5E2%2Fkg%5E2%29%285.97%5C%20x%5C%2010%5E%7B24%7D%5C%20kg%29%7D%5C%5C%5C%5CR%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7B4.66%5C%20x%5C%2010%5E8%5C%20s%5E2%7D%7B9.91%5C%20x%5C%2010%5E%7B-14%7D%5C%20s%5E2%2Fm%5E3%7D%7D%20%5C%5CR%20%3D%201.675%5C%20x%5C%2010%5E7%5C%20m%20%3D%201.68%5C%20x%5C%2010%5E4%5C%20km)
Now, this orbital radius is the sum of the radius of the earth (r) and the distance of satellite above earth's (h) surface:
R = r + h
1.68 x 10⁴ km = 0.64 x 10⁴ km + h
h = 1.68 x 10⁴ km - 0.64 x 10⁴ km
<u>h = 1.04 x 10⁴ km = 10400 km</u>
Learn more about the orbital time period here:
brainly.com/question/14494804?referrer=searchResults
The attached picture shows the derivation of the formula for orbital speed.
It is true!!
One au (Astronomical Unit) is the distance from the Earth to the Sun.
Hope I helped!!
Answer:
Speed is 1.962*10^-3m/s
Explanation:
Given that
Mass of proton==1.67*10^-27
Acceleration due to gravity (g)=9.81m/s^2
Angle of inclination A=90°
Charge (q)=1.6*10^-19
Magnetic field (B)=5*10^-5T
From lorentz force
It can be seen that
F=w =mg=qvBsinA
We're Velocity (v)=mg /qBsinA
V=1.67*10-27*9.81/1.6*10^-19*5*10^-5*sin90
V=1.962*10^-3m/s
