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4:44 AM
@JohnRennie Hi :)
 
@KevinN hi :-)
 
When we say potential difference between A and B, what does it actually mean?
I'm kinda confused
I mean the electric potential between A and B keep changing
 
Potential difference is just a work done per unit charge.
The graph shows a field strength as a function of distance, so we can write this as $E(x)$. OK so far?
 
So when we say between A and B, what does it really indicate?
The workdone in moving charge B to A?
or vice versa?
@JohnRennie ok
 
And the force on a charge $q$ in an electric field $E$ is $F = qE$.
 
@JohnRennie Yes
 
So if we move a small distance $dx$ the work done is $dW = qE(x)dx$. Yes?
@sheltonBenjamin I'll have a look at that in a bit ...
 
@JohnRennie yes :)
 
And the total work done if we move from the point $x=A$ to $x=B$ is calculated by integrating this:
$$ W_{ab} = \int_a^b q E(x) dx $$
And if we set the charge $q$ to unity we get the work per unit charge, which is just the potential.
$$ V_{ab} = \int_a^b E(x)dx $$
 
@JohnRennie What I'm a bit puzzled about is, let's say, the electric potential at a certain point is the work done needed to bring that test charge to that point from infinity. When we say the electric potential between A and B, which point is it actually referring to? Is it the work done needed to move a test charge from A to B?
 
4:54 AM
Potential doesn't have an absolute value. The only thing we can ever measure is a potential difference.
 
So potential difference between A and B is basically the work done needed per unit charge to move it from point A to B?
 
It is often convenient to choose the potential to be zero at infinity. The potential at some point A is the difference between the potential at A and the potential at infinity, but choosing the potential to be zero at infinity makes this difference just equal to $V_a$.
@KevinN yes. It is that simple.
 
@JohnRennie Ah alright thanks :)
@JohnRennie When we say the potential is zero at infinity, is it really zero or approaching zero?
 
"Infinity" is a risky thing to mess around with. What we would actually say is that the limiting value of the potential as the distance increases is zero.
Typically we'd get $V = -kq/r$ which is never actually zero but tends to zero as $r \to \infty$.
 
Ah I see :)
 
5:00 AM
Sir i want to learn infinite potential well for excercise in ie irodov chapter atomic model do i have to learn quantum mechanics also to learn the concept?
 
@sheltonBenjamin the maths behind the infinite potential well is easy so you can certainly learn how to do it without needing to fully understand quantum mechanics.
I would say go for it.
 
5:59 AM
@sheltonBenjamin this is my FBD:
 
 
4 hours later…
9:37 AM
@satan29: Hi! It seems I was bit too late to see this message of yours. I was online but I didn't hear your ping as I had muted the speakers and was on a different tab. Now coming to your question, I don't have anything valuable than this message by John Rennie sir.
On looking at the main site, the following question showed up:
4
Q: Why does an increase in pressure raise melting point?

user58953I've heard that if you increase pressure in a system that has ice in it, then its melting point will increase. However, I would've thought that it would decrease the melting point as the ice would heat up more easily due to the air molecules colliding more frequently with the surface of the ice, ...

Might be useful. I thought I would find some other reasons there. But they also describe this based on the anomalous expansion of water on freezing.
By the way, there seems to be a similar outcome for an increase in mechanical pressure. See: Regelation
You might find this video by Veritasium to be interesting too. This is with reference to the second kind of pressure.
 
 
1 hour later…
10:48 AM
Chemistry told me Anamolous expansion is because water gets into Hydrogen bonding ands gets arrange in a very spacious lattice (Each water molecule is tetrahedrally attached to other)

Not sure if Im right in this part, most probably wrong, but In Clausius Clapeyron Relation delta S/ delta V must be +ve though delta V is negative since final entropy is lesser. Ice's possible microstates for the given state variables is much lesser than Water because of the spacius lattice arrangement by H bonds
 
11:17 AM
@JohnRennie another day, amother problem!
the pulley is massless and so are the strings
i will be better off with a answer, folloed by some discussion
 
 
2 hours later…
12:58 PM
@JohnRennie sir but what does this imply
 
 
2 hours later…
3:23 PM
@GuruVishnu thanks a lot!!
 
4:18 PM
although theres still something i dont seem to get
 
5:08 PM
@satan29 can i have a moment?
 
yes
 
please help me with the above problem
 
right
 
Just a tip, take the wedge as frame for mass m, becomes easier
otherwise we would have to introduce a new variable
$\psi$
or just theta
 
B represents block and w represents wedge
second derivative of xb is second derivative of l1 + cos(37)*second deriv. of l2. yes?
= -a + acos(37)
 
5:19 PM
that is wedge and block have same acc?
 
no, i am just writing acceleration of block.
relative to my origin, its just second derivative of xb.
 
similarly, differentiate the y cordinate twice, (and ofcourse h doesnt change), to get -asin(37)
 
ok
final answer is?
 
you know the y and x acceleration of the block in terms of a now. So write forces in x and y direction.
and you already have an equation for a
(the one you wrote in the image)
so three variables (N,T,a) , three equations. It can be solved easly
 
5:35 PM
is the acceleration of wedge 3g/7?
acceleration of block=3g/35*sqrt(10)
@CaptainBohemian hi!!
 
@satan29 what's up? I just come in to see what people are chatting here.
 
its usually physics (high school) questions
@AbramIvanov if you could confirm the answer...
i tend to make a lot of compuation mistakes
 
@satan29 that's the problem
i don't have a answer
but i get it as $frac{60}{11}$
 
5:50 PM
post your solution, lets see
 
@satan29 Can you wait for a day, offical answer key out this morning
 
 
1 hour later…
6:58 PM
@AbramIvanov what test was this?
 

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