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.
Explanation:
It is given that,
Mass of the box, m = 100 kg
Left rope makes an angle of 20 degrees with the vertical, and the right rope makes an angle of 40 degrees.
From the attached figure, the x and y component of forces is given by :
Let 
and 
is the resultant in x and y direction.
As the system is balanced the net force acting on it is 0. So,
.............(1)
..................(2)
On solving equation (1) and (2) we get:
(tension on the left rope)
(tension on the right rope)
So, the tension on the right rope is 1063.36 N. Hence, this is the required solution.
Answer:
The cocoa was mixed with hot water, but some cocoa settled as it cooled because the solubility went down.
The average speed <em>appears to be</em> (distance) / (time) =
(length of the cable) / (time from when a pulse goes in until it comes out the other end) .
That's 1,200,000 meters/ 0.006 second = 2 x 10^8 = <em>2 hundred million m/sec</em>
That figure is about 66.7% of the speed of light in vacuum.
The reason I went through all of this detail was to point out that this is
NOT necessarily the speed of light in this glass, for two reasons.
1). The path of light through an optical fiber is not straight down the middle. In the original fibers of 20 or 30 years ago, the light bounced back and forth off the inside walls of the fiber, and zig-zagged its way along the length. In current modern fibers, it still zig-zags, but it's a more gentle, up-and-down curved path. In either case, the distance covered by the light inside the fiber is more than the straight length of the cable, and the time it takes it to come out the other end is more than its actual speed inside the glass would have meant if it could have traveled straight through the pipe.
2). This problem talks about an optical fiber that's 1,200km long. There is loss in optical fiber, and you're NOT going to get light all the way through a single piece of it that's something like 745 miles long. It takes electronic repeaters, "boosters", and regenerators every few miles to keep it going, and these devices add "latency" or time delay in the process of going through them. That delay in the electronics shows up as apparent delay through the fiber-optic cable, and it makes the speed through the glass appear to be slower than it actually is.
