<span>373.2 km
The formula for velocity at any point within an orbit is
v = sqrt(mu(2/r - 1/a))
where
v = velocity
mu = standard gravitational parameter (GM)
r = radius satellite currently at
a = semi-major axis
Since the orbit is assumed to be circular, the equation is simplified to
v = sqrt(mu/r)
The value of mu for earth is
3.986004419 Ă— 10^14 m^3/s^2
Now we need to figure out how many seconds one orbit of the space station takes. So
86400 / 15.65 = 5520.767 seconds
And the distance the space station travels is 2 pi r, and since velocity is distance divided by time, we get the following as the station's velocity
2 pi r / 5520.767
Finally, combining all that gets us the following equality
v = 2 pi r / 5520.767
v = sqrt(mu/r)
mu = 3.986004419 Ă— 10^14 m^3/s^2
2 pi r / 5520.767 s = sqrt(3.986004419 * 10^14 m^3/s^2 / r)
Square both sides
1.29527 * 10^-6 r^2 s^2 = 3.986004419 * 10^14 m^3/s^2 / r
Multiply both sides by r
1.29527 * 10^-6 r^3 s^2 = 3.986004419 * 10^14 m^3/s^2
Divide both sides by 1.29527 * 10^-6 s^2
r^3 = 3.0773498781296 * 10^20 m^3
Take the cube root of both sides
r = 6751375.945 m
Since we actually want how far from the surface of the earth the space station is, we now subtract the radius of the earth from the radius of the orbit. For this problem, I'll be using the equatorial radius. So
6751375.945 m - 6378137.0 m = 373238.945 m
Converting to kilometers and rounding to 4 significant figures gives
373.2 km</span>
Answer:T=1316.21 N
Explanation:
The tension has two components: Vertical and Horizontal. The
horizontal component is ma, the vertical component is mg. Using
Pythagoras theorem, we can find the tension as:
T=((ma)^2 (mg)^2)^(1/2)
So
T=((129*2.84)^2 (129*9.8)^2)^(1/2)
T=1316.21 N
Answer:
d = 1700 meters
Explanation:
During a rainy day, as a result of colliding clouds an observer saw lighting and a heard thunder sound. The time between seeing the lighting and hearing the sound was 5 second, t = 5 seconds
Speed of sound, v = 340 m/s (say)
Let d is the distance of the colliding cloud from the observer. The distance covered by the object. It is given by :
d = 1700 meters
So, the distance of the colliding cloud from the observer is 1700 meters. Hence, this is the required solution.
Answer:
<em>b. Observe the radio waves coming from all dark matter; from the strength of the radio waves from each cluster, estimate the amount of dark matter needed to produce them.</em>
<em></em>
Explanation:
The universe is thought to be made up of 85% dark matters. <em>Dark matter is called dark because it does not appear to interact with the electromagnetic field, which means it doesn't absorb, reflect or emit electromagnetic radiation, and is therefore difficult to detect. This means that option b is wrong since radio wave is an electromagnetic wave</em>. Dark matter is a form of matter that makes up about a quarter of the total mass–energy density of the universe. Dark matter was theorized due a variety of astrophysical observations and gravitational effects that cannot be explained by accepted theories of gravity unless there were more matter in the universe than can be seen.
Complete Question
Part of the question is shown on the first uploaded image
The rest of the question
What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.
Answer:
The net force exerted on the third charge is 
Explanation:
From the question we are told that
The third charge is 
The position of the third charge is 
The first charge is 
The position of the first charge is 
The second charge is 
The position of the second charge is 
The distance between the first and the third charge is
The force exerted on the third charge by the first is
Where k is the coulomb's constant with a value 
substituting values
The distance between the second and the third charge is
The force exerted on the third charge by the first is mathematically evaluated as
substituting values
The net force is
substituting values
