<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>
There would be infinite acceleration, and the body would have no mass <span />
Explanation:
The given data is as follows.
q = 6.0 nC = 
inner radius (r) = 1.0 cm = 0.01 m (as 1 cm = 100 m)
So, there will be same charge on the inner surface as the charge enclosed with an opposite sign.
Formula to calculate the charge density is as follows.
.......... (1)
Since, area of the sphere is as follows.
A = 
........... (2)
Hence, substituting equation (2) in equation (1) as follows.
= 
= 
or, = 4.77 
Thus, we can conclude that the resulting charge density on the inner surface of the conducting sphere is 4.77 .
<h3>
Answer: 130 newtons</h3>
===============================================================
Explanation:
We'll need the acceleration first.
- The initial speed (let's call that Vi) is 8.0 m/s
- The final speed (Vf) is 0 m/s since Sam comes to a complete stop at the end.
- This happens over a duration of t = 4.0 seconds
The acceleration is equal to the change in speed over change in time
a = acceleration
a = (change in speed)/(change in time)
a = (Vf - Vi)/(4 seconds)
a = (0 - 8.0)/4
a = -8/4
a = -2
The acceleration is -2 m/s^2, meaning that Sam slows down by 2 m/s every second. Negative accelerations are often associated with slowing down. The term "deceleration" can be used here.
Here's a further break down of Sam's speeds at the four points of interest
- At 0 seconds, he's going 8 m/s
- At the 1 second mark, he's slowing down to 8-2 = 6 m/s
- At the 2 second mark, he's now at 6-2 = 4 m/s
- At the 3 second mark, he's at 4-2 = 2 m/s
- Finally, at the 4 second mark, he's at 2-2 = 0 m/s
Next, we'll apply Newton's Second Law of motion
F = m*a
where,
- F = force applied
- m = mass
- a = acceleration
We just found the acceleration, and the mass is fairly easy as all we need to do is add Sam's mass with the sled's mass to get 60+5.0 = 65 kg
So the force applied must be:
F = m*a
F = 65*(-2)
F = -130 newtons
This force is negative to indicate it's pushing against the sled's momentum to slow Sam down.
The magnitude of this force is |F| = |-130| = 130 newtons
Answer:
6.23x10^6Pa
Explanation:
Data obtained from the question include:
F (force) = 490N
r (radius) = 0.005m
A (area of the circlular heel) =?
P (pressure) =.?
First, we'll begin by calculating the area of the circlular heel. This is illustrated below:
Area of circle = πr^2
Area = 22/7 x (0.00)^2
Area = 7.86x10^-5m^2
Pressure is simply force per unit area. It represented mathematically as
Pressure = Force /Area
Pressure = 490/7.86x10^-5
Pressure = 6.23x10^6N/m2
Recall: 1N/m2 = 1Pa
Therefore, 6.23x10^6N/m2 = 6.23x10^6Pa
Therefore, the woman exert a pressure of 6.23x10^6Pa on the floor
