Recent content by ItsAnshumaan

Thank you for the help :D

When p is 2, the y co-ordinate will be double of what it should had been in normal cosine graph. Hence I'm assuming that the amplitude of the wave will increase.

This is not a homework question but a general doubt. Suppose we have a function y = pcosx, where 'p' is an arbitrary constant. So my question is how will the graph of this function change with different values of 'p'? This doubt can also be extended for other functions like y = pex, y = p...

Homework Statement The electric field and the electric potential at a point are E and V respectively. (a) If E=0, V must be 0 (b) If V=0, E must be 0 (c) If E≠0, V cannot be 0 (d) If V≠0, E cannot be 0 Homework Equations [/B] E = V/d The Attempt at a Solution [/B] I basically substituted...

Thanks Chestermiller, TSny and Biker for your help, now I am able understand the pulley problems to the core!

Thanks a lot! Turns out that I had copied the equation wrong. Yes it worked, and the answer I got is 7mg/6, perfectly the 3rd option! :)

Oh yes, so that force must be the tension. In that case T = f, or T = mg/2 + m/2(3g/2 + m a1) So the earlier equation i had for the block was, T - mg = m a1.. So a1 = (T - mg)/m Putting the value of a1 in T = mg/2 + m/2(3g/2 + m a1), I am very unfortunate to get the value T = {mg(4-m) /...

Sorry that would be f - mg/2 = (m/2)(3g/2 - a1). But I am unable to draw the relation between f and T..

Oh right finally understood, 5m/s2 upwards. So then, 3g/2 = a1 + a2, that means acceleration of the man is; a2 = (3g/2 - a1). Then for the man, the relation would be: f - mg = (m/2)(3g/2 - a1). Is this correct now? (f is the frictional force)

Yes, the value of the acceleration of the man with respect to the rope would be basically 1 m/s2 upwards. Okay so with that knowledge, I can say that the man is moving up with an acceleration (3g/2 + a1) m/s2, and the overall sum of all the forces acting on the man would be equal to (m/2)(3g/2...

The mass is actually moving up, I can figure out that the only equation we can get out of that mass is: T - mg = m a1 (Here I am considering a1 as acceleration of the rope and a2 as the acceleration of the man relative to the rope). So the man's acceleration with respect to ground must be; a2 =...

OK here is what I tried to do So I am not able to figure out what to do with the force which is helping the man to climb up.

I am a student of India, currently studying in my 12th grade. I have a huge passion for Physics, Chemistry, Maths, and Computer Science. Even though I have not mastered any of these subjects, but I am really interested in learning them to core! :) I am also preparing for JEE mains, an entrance...

Homework Statement In order to raise a mass 'm' a man ties it to a rope and passes the rope over a friction less pulley. He climbs the rope with an acceleration 3g/2 relative to the rope. If the mass of the man is m/2 and the mass of the rope is negligible, the tension in the rope is a) 3mg/2...