@pi-π I got $a_b=0.753846$ but not exactly $0.75$. Is that an issue? Given this value it would be easy to find other accelerations as well as tension in the strings.
I used $g=9.8$ itself to be on the safer side. If used $10$ it gives a slightly higher value for the above acceleration.
@pi-π I hope I did it correctly. I just wrote $F=ma$ equation for each block after assuming an arbitrary tension in each string. Then computed the acceleration of B by relating it with that of A.
In the beginning of the exercise, were you asked to round off to 2 decimal places or something like that?
It is said that, due to end effects, the assumed poles of a bar magnet are slightly inside the ends of the magnet. The distance between the locations of the assumed poles is called the magnetic length of the magnet. The distance between the ends is called the geometrical length.
The magnetic l...
@GuruVishnu I believe that if that post didn't answered your question then you should have asked a new one. Cause new posts tend to get more attention than old ones.
@JohanLiebert Thanks for your message. I thought of that idea. I didn't do so as I felt my new question might be closed as a duplicate of this one. Further, this one has a lot of views. Now, I don't know whether I made the right decision or not.
I incorporated my question into it through my latest edit.
I thought electrons only emmit light when they are going down energy levels. And I though they emmit the highest wavelength the shorter the energy level they go down from. So From that I thought it was D but apparently its wrong :/
Suppose in II the friction is high enough for the wedge not to move. We can always make this the case by making the ball speed slow enough. Then the ball will do this:
The only thing moving is the ball, and clearly $v_x$ changes so $p_x$ must change.
In I when the ball hits the wedge there are three forces: 1. the force on the ball 2. the force on the vertical face of the wedge 3. the frictional force on the base of the wedge