Recent content by nu_paradigm

Relative speeds reaching c?? Just an idea... Suppose i am traveling at the speed of slightly more than c/2 with respect to ground and there is a stationary plane mirror in which i can view myself approaching the mirror. Now won't my relative speed with respect to my image be more than c...

f(x) = \left {\begin{array}{cc}\sin x,&\mbox{if} \ x \epsilon \ [0, \frac{\pi}{2})\\ \cos x, & \mbox{if} \ x \epsilon \ [\frac{\pi}{2}, \pi)\end {array}\right

Just trying... \int_{0}^{\frac{\pi}{2}} \log \cos \Omega \ d \Omega f(x) = \sin x, \mbox{if} \ x \ \epsilon \ (0, \frac{\pi}{2})

\int_{0}^{\frac{\pi}{2}} \log \cos \kappa d \kappa

Correct ! Stefan, your answer is perfectly correct :approve: . Go on... your turn to ask a question.

I think you are getting confused :confused: as to which radius is required to be found out. Take the two charges to be point, dimensionless charges. You have to find the radius of the circular motion which they undergo after collision. Gokul has already outlined all the steps. Only the...

Hey is anybody attempting this??

Suppose we have a sample A of 30000 kg of ice and a sample B of 1g of ice(1ml of water) both at absolute zero, then if we start heating them, so that sample A reaches 0^{0}\ C and sample B reaches 100^{0}\ C, we can calculate the heat they will have, \mbox{For Ice at sample A,}...

Yes, you have to explain how they move after collision and with what radius. And also, please remember, the electric force between the charges is to be neglected.

It's given that they collide inelastically, i.e, they stick together.

Ok... this is not the best question I cud've put up, but here goes... :smile: [FONT=Book Antiqua]A charged particle +q of mass 'm' is placed at a distance 'd' from another charged particle -2q of mass '2m' in a uniform magnetic field B as shown(perpendicularly into the plane). The particles...

I think so... I'm taking the axis at the point of contact, right, so the moment of inertia will have to be about an axis perpendicualr to the ring and tangential to it, so i used the parallel axis theorem to get the M.I about that axis. i.e. M.I = MR^2 + MR^2

Hey Nishant, did u actually try an iron coin? And is the orientation of the coin important ? I'm asking all those whose coins went up! Coz mine didnt. I'd like to know if any other coins apart from a 2p coin works, and is it only a 60 W bulb which did the trick? My guess for the explanation is...

Let me edit my answers: Hence, total torque acting = - r x Mgcos(30) - r x Mgsin(30) (Take -ve to be anticlockwise, as per Siddharth's diagram) Now r = [R^2 + (1/3pi)^2]^1/2 = [0.25 + (0.0112)]^1/2 = 0.511 m and angle between r and mgcos30 : (theta) =...

Hey one mistake i just noticed I've taken the heavier side to be on top. The only difference that would make is that the torques wud add up and both will be in anti clockwise direction. Also the angles will have to be recalculated. Stupid blunder on my part. But still the reasoning is the same.