I meant how can the door close from rod's frame of reference if garage was too small? rod can't get outside garage from behind. The door wont shut no matter how long rod based observer waits
What if we close the gate for the garage when the rod was inside the garage? Also, assume that once the door was closed the rod can't get out so either it bends or breaks.
Actually, I just generalised the length of rod and garage in terms of L, the rest is given as an explanation in Relativity special, general and cosmological by Rindler (2006 edition - 3rd chapter), including the explanation about mechanical pressure travelling through the rod.
I also read...
Homework Statement
Special relativity: A rod travelling with the relativistic velocity of 0.866c moving towards a garage. The length of the rod is L, that of the garage is L/2. From the reference frame of the garage, we find the value of Lorentz factor = 2 we say that rod would fit inside...
Hey everyone, I am thinking something completely different and it is quite possible that I am wrong, so forgive me if I turn out to be wrong. The way I am thinking this problem to be is that, since it is given that balls attain same potential as the plate (as they had a metal film), so from...
Thanks a lot guys. Apparently I should add a word of caution while treating system of charges as dipoles, I got the answer by both electric field and using dipoles method. Winding up both 11th and 12th class isn't the easiest thing to do, I have been using binomial theorem to get approx...
Did you by any chance use, E= KQx/(R^2 + x^2)^3/2 for both rings and substract (or add depending upon you, if you think in terms of magnitude or vectors) and use binomial expansion?
My point is - that is how we deduce electric field due dipole - so using superposition, or result of dipole...
Ah! now I get the result of post no. 9. Now I recall that I did a similar question on calculating electric field inside a cavity in an insulator having uniform charge density. Quite similar. But a question, how does this figure look anything like in post 4 or 6.
Edit: I mean - here we are using...
That was a very misleading question - the link provided by Nidium tells that magnetic field is vertical "at the centre". While a comparison with Solenoid will make it appear as if it will have a constant vertical field
I may be wrong but here is what I think - To take cosine, sine etc. you need to have an angle. Clearly it is changing in the equation -as shown- with time. say at time equals to one second - (notice that second in denominator and that of time are going to vanish with each other) - the angle is...
I tried this question yesterday. But something is pretty off with the description. I tried to google the image of saddle coil as description was way too poor, but the image that I found was way different than what I can make out from question..Dewgale you don't happen to have any image of this...
Homework Statement
Two identical co-axial rings ,(radius R each) are kept separated by a small distance d, one of them carrying a charge +Q and the other a charge -Q. The charges are uniformly distributed over the respective rings. A point charge q is kept on the common axis of the rings, at a...
I didn't put much thought to what was implied by differentiation (sorry I'll try to figure it out.) My book says:
"A function f(x) is said to be continuous at x = a; where a is an element of domain of f(x).
If left hand limit at a = right hand limit at a = f(a)
i.e. LHL=RHL=Value of function at...
I think h is not continous at 0. limit h(0) does exist but limit has to be equal to h(0) which is undefined.
h(0) = infinity/infinty. You had h(0) = ln2 by taking the limit (using a standard formula.)
Finally! Thanks a ton! Yipee! got the answer.
By the way, I was curious from your first post (on this thread) - what does - "mfb" stand for? Your initials?
Praise be unto He. Helium!!
When I was posting the last thread, it was 5am here, and so I was feeling very sleepy, so I thought I am not getting it right now, I should try later. But even now that I have tried when I am wide awake I can't find the two products you mentioned.
As in my last thread
The two factors that I...
I am sorry but I could not pick up the clue. From your description I got.
$$1/4 \lim_{y/2\to 0}[(\frac {e^{y/2}-y/2-1}{y/2})(\frac {e^{y/2}-y/2+1}{y/2})-1]$$
Am I going right?
Homework Statement
Evaluate ##\lim_{x\to 0} \frac{2^x-1-x\log_e2}{x^2}## without using L'Hospital's rule or expansion of the series.
Answer is given to be = ##\frac{(\log_e (2))^2}{2}##
Homework Equations
Squeeze play theorem/ Sandwich theorem, some algebraic manipulations and standard...
Finally, deciphered. Yet another mistake in calculation and hopefully this will be my last thread to this post.
Mistake -1. ##f'(cos\phi)=\frac{2}{\sin\phi}\cdot(\arctan\cot\phi-\arctan\tan\phi)=\frac{\pi}{\sin\phi}##. Since R.H.S. is correct so you made a typo on computer here (started...