I think it's 3. within an outer arm
Answer:
995 N
Explanation:
Weight of surface, w= 4000N
Gravitational constant, g, is taken as 9.81 hence mass, m of surface is W/g where W is weight of surface
m= 4000/9.81= 407.7472
Using radius of orbit of 6371km
The force of gravity of satellite in its orbit, 
Where
and 

F= 995.01142 then rounded off
F=995N
From the case we know that:
- The moment of inertia Icm of the uniform flat disk witout the point mass is Icm = MR².
- The moment of inerta with respect to point P on the disk without the point mass is Ip = 3MR².
- The total moment of inertia (of the disk with the point mass with respect to point P) is I total = 5MR².
Please refer to the image below.
We know from the case, that:
m = 2M
r = R
m2 = 1/2M
distance between the center of mass to point P = p = R
Distance of the point mass to point P = d = 2R
We know that the moment of inertia for an uniform flat disk is 1/2mr². Then the moment of inertia for the uniform flat disk is:
Icm = 1/2mr²
Icm = 1/2(2M)(R²)
Icm = MR² ... (i)
Next, we will find the moment of inertia of the disk with respect to point P. We know that point P is positioned at the arc of the disk. Hence:
Ip = Icm + mp²
Ip = MR² + (2M)R²
Ip = 3MR² ... (ii)
Then, the total moment of inertia of the disk with the point mass is:
I total = Ip + I mass
I total = 3MR² + (1/2M)(2R)²
I total = 3MR² + 2MR²
I total = 5MR² ... (iii)
Learn more about Uniform Flat Disk here: brainly.com/question/14595971
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Answer:
V=14.9 m/s
Explanation:
In order to solve this problem, we are going to use the formulas of parabolic motion.
The velocity X-component of the ball is given by:

The motion on the X axis is a constant velocity motion so:

The whole trajectory of the ball takes 1.48 seconds
We know that:

Knowing the X and Y components of the velocity, we can calculate its magnitude by:
