<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:
786.6 N
Explanation:
mass of car, m = 912 kg
initial velocity of car, u = 31.5 m/s
final velocity of car, v = 24.6 m/ s
time, t = 8 s
Let a be the acceleration of the car
Use first equation of motion
v = u + a t
24.6 = 31.5 + a x 8
a = - 0.8625 m/s^2
Force, F = mass x acceleration
F = 912 x 0.8625
F = 786.6 N
Thus, the force on the car is 786.6 N.
Answer:
force becomes one - ninth
Explanation:
According to Coulomb's law in electrostatics, two charges can exert a force of attraction or repulsion on each other which is directly proportional to the product of two charges and inversely proportional to the square of distance between them.
Here both the charges remains same but the distance is variable.
So, we can say that
.... (1)
Where d be the distance between the tow charges
As the distance between two charges increases by factor of three, let the new force be F'.
.... (2)
Divide equation (2) by equation (1), we get
Thus, the force becomes one - ninth times the initial force.
In order to measure the size of the electrical current flowing in the circuit,
the current must pass through the meter.
Answer:
Spring constant in N / m = 6,000
Explanation:
Given:
Length of spring stretches = 5 cm = 0.05 m
Force = 300 N
Find:
Spring constant in N / m
Computation:
Spring constant in N / m = Force/Distance
Spring constant in N / m = 300 / 0.05
Spring constant in N / m = 6,000
