I believe the answer is 153.8 m.
Answer:
1) 883 kgm2
2) 532 kgm2
3) 2.99 rad/s
4) 944 J
5) 6.87 m/s2
6) 1.8 rad/s
Explanation:
1)Suppose the spinning platform disk is solid with a uniform distributed mass. Then its moments of inertia is:

If we treat the person as a point mass, then the total moment of inertia of the system about the center of the disk when the person stands on the rim of the disk:

2) Similarly, he total moment of inertia of the system about the center of the disk when the person stands at the final location 2/3 of the way toward the center of the disk (1/3 of the radius from the center):

3) Since there's no external force, we can apply the law of momentum conservation to calculate the angular velocity at R/3 from the center:



4)Kinetic energy before:

Kinetic energy after:

So the change in kinetic energy is: 2374 - 1430 = 944 J
5) 
6) If the person now walks back to the rim of the disk, then his final angular speed would be back to the original, which is 1.8 rad/s due to conservation of angular momentum.
Answer:
there are two way to get mate and i gave them sepaert explation
Explanation:
55
N
Explanation:
Using Newton's second law of motion:
F
=
m
a
Force=mass
×
acceleration
F
=
25
×
2.2
F
=
55
N
So 55 Newtons are needed.
Answer link
Nam D.
Apr 6, 2018
55
N
Explanation:
We use Newton's second law of motion here, which states that,
F
=
m
a
m
is the mass of the object in kilograms
a
is the acceleration of the object in meters per second
F
=
25
kg
⋅
2.2
m/s
2
=
55
N
Answer:
It would because the shape of the rocket is designed to be able to slice through the air as smooth as possible and now you may be thinking that air is already smooth but when you try to push something as large and heavy like a rocket then the shape of the rocket will be very important. The bottom of the rocket is flatter then the top so it is not designed to fly smoothly through the air. So the rocket would fall vertically downward(If it was still in one piece)because of it's shape. It is easier for the top of the rocket to go smoothly through the air then the bottom.
Explanation:
I am 90% sure this is correct but if I'm not please tell me