You may use that method, however, you may run into trouble with actually implementing it. I suggest using conservation of angular momentum because it is much simpler (visually). If you use this method:

1. find the final moment of inertia from the different moment of inertia given to you for the kid and the merry go around.

2. Use I_i*w_i = I_new*w_f to find w_f (angular speed), where I_new if the final moment of inertia.

There are actually two forces involved as he is running around the roof top. One of the forces is due to gravity. The other is due to the affects of centripetal force. Since moving in a circle requires a force towards the center of the circle Harry is moving about, Harry (and his feet) will "feel" something pushing him outwards so that he doesn't actually fall straight toward the center. Although the gravitational force will be constant in a uniform circular motion, the force pushing him outwards will not be uniform because it depends on how fast he is moving (F=mv^2/r).

Center of gravity is the point in an object where the force of gravity appears act. In a uniform gravitational field, this will be at the same place as the center of mass. They are not the same thing, but they will often be at the same point in an object. Thus in problems that that place near the surface of earth, the terms are sometimes used interchangeably. Hope this helps!

Well, I believe you should use the idea of what pressure actually is. It is simply applying a force over an area. Therefore, I suggest finding it initially without the person standing on top of the ball and then comparing it with the person standing on top of it. You'll be able to find the mass of the person this way.

The gravitational force and the normal force, I would say, definitely cancel out at a specific point but that is not why there is no work. The reason they say this is because they do not want you to account for friction. As you may know Work is defined to be -F*Δx, where F is the force in a direction and Δx is the change of the object's magnitude of distance in the same direction as the force. However, since you are ignoring friction you do not need to account for frictional force (thus no work is necessary). Keep in mind that there is work going down the ramp from gravity but there is also equal and opposite work going up the ramp as well, therefore you do not need to worry about that either.

If you know that the magnetic field will make the electron go in a circle then imagine it going in a circle just large enough that it will just scrape the wall. It would be easier to imagine it if you drew it along with labeling every information you wrote here (the distances from the wall, velocity, etc.). Then, write newton's second law, force of a magnetic field, and centripetal force equation. You'll see a connection with these 3 equations along with the drawing.

There is not enough information here to solve. This is a 2-D system, so you will need to know at least 2 parameters to solve it. You know one: the horizontal distance. If you know how fast the bullet travels as it leaves the barrel, then the system can be solved.

If all the car's energy is in gravitational potential, then the car would not be moving at the top of the track(since there's no kinetic energy). Can a car sit stationary at the top of the track? Of course not; it would actually fall before it gets that high.

You are on the right track though(no pun intended). Equating the energies is the way to solve this, but you need to take into consideration the kinetic energy at the top of the track.

Hint: the loop is a circle.

"Dark matter" refers to matter that we can conclude must be present but can't see. When you solve this problem, you will find the mass of the galaxy and see that it is not equal to (it's much greater than!) 2* 10^40 kg.

It does not give you the mass because you do not need the actual mass, just the relative masses of the two boxes. Technically, you could assume it is 50kg and the same answer, but I think you will find it more beneficial to call the mass "m" and solve without a value for it.

There will be multiple forces on both blocks, so it will not be as easy as saying "The force on it is__." You will need to draw a force diagram for both blocks.

I hope this helps you get started!