Recent content by terbum

The acceleration you solved for was the acceleration for the system. So, BOTH M1 and M2 have an acceleration of 0.83 m/s^2. Yeah, I had a tough time with this topic when I learned it, too. It does get a little confusing. I guess if you want to learn it, keep doing problems. Practice never...

The contact force is merely the mass of M1 times its acceleration. Remember, the contact force is the force giving M1 its acceleration. Do you see why?

No, let me explain this differently. Now, draw the free body diagram of M2. There's the 4.55 N force acting on it, but there is another horizontal force acting in the opposite direction, the contact force. Newton's Third Law states there are equal and opposite forces, so this applies to the...

If the force on M2 was also applied to M1 like you said, M1's acceleration would not be the 0.83 m/s^2 you calculated. So, a different force must be causing this acceleration...

Try drawing a free-body diagram of M1, to make this easier. What forces are acting on it? You also know its acceleration, so what forces are causing this acceleration?

You can solve for the enclosed charge, since the elementary charge of an electron (-1.60*10^-19 C) and the number of electrons is known. Just use Q/epsilon to solve for flux.

Weights aren't measured in kilograms. Are those values you gave supposed to be the weights, measured in Newtons, or, are the masses, measured in kilograms?

Yup. Now use that velocity to solve for the range. You have the kinematics equations, but the easiest way is to just use the range equation.

Think about the vertical velocity of the ball at 1.33 m, the top of its upward path. It's moving from going upwards to downwards, right? So, at that instant, what is the velocity (or, what is vf, the final velocity)?

You have a few options to solve for the speed of the ball coming out of the cannon from the vertical case: 1.) Conservation of energy. All of the kinetic energy of the ball turns to potential energy at 1.33 m above ground. You can set up something that way. 2.) The kinematic equation...

Pressure is F/A. Manipulating it gets you F=PA