The h Bar - 2017-01-05 (page 1 of 3)

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12:11 AM

@Kaumudi.H My coworker would like to start a flip-flop/sandal museum and has recommended me as the "honorary secretary-president".

Importantly, we need Ghandi's sandals as the primary attraction.

Can you help?

2 hours later…

1:48 AM

1

Q: Using Delta Dirac function as a mathematical tool in Green's functions

So, I was studying green's functions and in general I understood that if I have an operator $\mathscr{O}$ that acts of a function $h_1(\vec{r})$ such that $$\mathscr{O}h_1(\vec{r})=h_2(\vec{r})$$ Then all I need to do is to find the function, $g(\vec{r})$, on which the operator acts to yield the...

electromagnetism electrostatics mathematical-physics greens-functions

^The question has answered duplicate at math.SE, and is about physics only insofar as this particular distribution is of particular interest to physics. Is this on- or off-topic? Duplicating posts from math.SE here seems rather pointless to me.

off-topic i'd say

user228700

2:13 AM

@DanielSank Interesting. How can I help?

user228700

(Also, are u or are u not kidding me?)

@Kaumudi.H, hello =)

2:38 AM

@Kaumudi.H get Ghandi's sandals.

@heather homework questions

@DanielSank, agh!!!

^ 2$^\text{nd}$ worst title in site history

i don't want to know/remember what the worst was

user228700

@heather Hi :-)

user228700

@DanielSank A replica?

user228700

2:42 AM

This is an interesting question but perhaps not worth spending too much time on:

user228700

scienceforums.com/topic/…

good night everyone

user228700

Bye :-)

@Kaumudi.H original.

user228700

2:58 AM

@DanielSank bbc.com/news/uk-england-shropshire-22612506

user228700

user image

user228700

^ A pair of Gandhi's sandals sold for £19,000

user228700

Gandhiji is truly the father of our nation. Even you should've suspected that his sandals wouldn't just be lying around.

user228700

If you want, I will find a replica for u in 5 months, is that okay? (I am still lead to wonder--are u or are u not kidding me?)

3:14 AM

@Kaumudi.H I am kidding, yes.

user228700

3:49 AM

@DanielSank -________-

user228700

4:55 AM

I was studying the limitations of dimensional analysis when I came across the following point in my textbook:

user228700

> "We equate the powers of $M$, $L$ and $T$ hence we get only three equations. So we can have only three variables. So dimensional analysis will work only if the quantity depends only on three parameters and not any more than that."

user228700

I don't understand this at all. So what if the physical quantity in question has dimensions involving...say, $A$? Wouldn't dimensional analysis work just as well? What point are they trying (and failing so spectacularly) to get across?

5:40 AM

@Kaumudi.H are you a whale?

user228700

@DanielSank Last time I checked ("Free Willy"), whales are very agreeable characters. On the other hand, that emoticon denotes a feeling of annoyance.

Pissed off layman

Try >8(

ie it's one of my favourites

user228700

@Pissedofflayman Will use next time (:-P)

Pissed off layman

:D

user228700

5:58 AM

@JohnR: Morning :-)

Morning.

Did you satisfy yourself about constants, dimensions, etc or is that still an outstanding issue?

Pissed off layman

Top of the mornin'

Irish today? :-)

user228700

@JohnRennie No, I think I'm done with that.

@Kaumudi.H Just as well. I'm not sure I understand it myself :-)

How about:

1 hour ago, by Kaumudi. H

I don't understand this at all. So what if the physical quantity in question has dimensions involving...say, $A$? Wouldn't dimensional analysis work just as well? What point are they trying (and failing so spectacularly) to get across?

user228700

6:03 AM

@JohnRennie :-P Work is done for the day?

Good morning

Tuning fork A when sounded with a tuning fork B of frequency 480 Hz gives 5 beats per second.
When the prongs of A are loaded with wax, it gives 3 beats per second. Find the original frequency
of A.

In above question does we need second condition

@Kaumudi.H Are you asking me if my work is done for the day? At 6 a.m.? That would be a short working day :-)

user228700

@JohnRennie :-P Oh, right, of course.

user228700

@JohnRennie Yeah, that's still confusing but I highly suspect that my textbook has just messed up its wording again :-\

@JohnRennie could you help me in that

6:05 AM

@koolman Yes. You're told the difference between $f_A$ and $f_B$ is 5Hz, but not whether $f_A > f_B$ or $f_A < f_B$.

@JohnRennie so how do we get to know about that

@koolman Well, what would adding wax to the tuning fork do to $f_A$?

@JohnRennie its weight will increase so time period should decrease

And f will increase

Am I correct @JohnRennie

The angular frequency of a simple harmonic oscillator is given by:

$$ \omega = \sqrt{k/m}$$

where $k$ is the force constant and $m$ is the mass. Yes?

Yes

6:15 AM

So what happens to $\omega$ if the mass increases?

Decreases

So what happens to $\omega$ when you add wax to the tuning fork?

Decreases

Anmd $\omega = 2 \pi f$ so what happens to $f_A$ when you add the wax?

Hence frequency should also decrease as $\omega =2\pi f$

6:19 AM

OK. So you know that adding wax will decrease $f_A$, and you're told that when you add the wax the difference between $f_A$ and $f_B$ decreases from 5Hz to 3Hz.

So $f_A - f_B =480$

No. The difference between $f_A$ and $f_B$ is the beat frequency.

Oh sorry $f_A - f_B =5$

i.e. without the wax $|f_A - f_B| = 5$, where I've use the modulus symbol because we don't know which frequency is greater.

f$_A$ = 485

@JohnRennie yeah

Thanks @JohnRennie

user228700

6:24 AM

@JohnRennie Did u read this:

user228700

scienceforums.com/topic/…

@Kaumudi.H I've had a quick read through it. The distinction between a mathematical constant and a physical constant seems a false one to me.

user228700

A false one? What dyou mean?

Well is $\pi$ a mathematical constant or a physical one?

user228700

Both..?

6:27 AM

If you insist on the distinction you'd have to say both.

I'd say there's no difference and that avoids the question.

user228700

Meh, never mind that, then. What about this:

user228700

2 hours ago, by Kaumudi. H

I don't understand this at all. So what if the physical quantity in question has dimensions involving...say, $A$? Wouldn't dimensional analysis work just as well? What point are they trying (and failing so spectacularly) to get across?

OK, yes I saw that and it's more straightforward than you think.

user228700

What dyou mean?

Suppose I have a system where the frequency is a function of four variables:

$$ f \propto A^\alpha B^\beta C^\gamma D^\delta $$

user228700

6:29 AM

Right...

On the left side the dimensions are $T^{-1}$

user228700

Right.

What we do in dimensional analysis is require that the dimensions on the right side are also $T^{-1}$.

user228700

Yes...

And we do that by substituting in the dimensions of $A$, $B$, $C$ and $D$ and raising them to the powers $\alpha$, $\beta$, etc.

user228700

6:31 AM

Right.

This gives us a set of simultaneous equations that we solve to get the values of $\alpha$ etc

But we have only three dimensions, $M$, $L$ and $T$.

user228700

Ohhhhhh.

user228700

Dang it, I am duuumb!

And, in this case, we have four unknowns $\alpha$, $\beta$, $\gamma$ and $\delta$

user228700

Yeah. I understand now. Thank you :-)

6:34 AM

That's why there is a limiotation of having three variables on the right hand side e.g.

$$ f \propto A^\alpha B^\beta C^\gamma $$

Because then it's three equations in three unknowns.

user228700

Of course, unless we do have a fourth unit such as $mol$ or $Cd$ or $A$.

$Cd$? Is that a dimension?

If you mean the drag coefficient that's dimensionless.

user228700

Candela--luminous intensity.

Candela is energy per unit area or somethign like that.

user228700

Yeah, isn't that the same as luminous intensity?

6:36 AM

So it would just be, err, $ML^2T^{-2}L^{-2}$

To be honest I lose track of all the ways to measure light intensity as there seems to be dozens of them all slightly different in obscure ways.

user228700

Sep 19 '16 at 6:15, by John Rennie

@KaumudiHarikumar I lose track of all the various ways you can write the power emitted. Energy per unit time per unit wavelength per unit area per unit solid angle per unit elephant.

:-)

user228700

:-) Okay, thanks!

user228700

Why the heck is the chap. Units and dimensions itself taking so long for me to complete?! >.< Stupid units and dimensions.

I wholeheartedly agree :-(

user228700

6:46 AM

Thanks for ur sympathy :-(

user228700

@JohnR: Are u up for some more thermo. by any chance? (::smiles sheepishly:) Or perhaps some surface chemistry...

Thermodynamics I can do. It's when you get to electrodynamics I might run and hide :-)

Oooh colloid science!

user228700

:-) Yes. When I finish this wretched chapter, I will get to either thermo. or surface chemistry...must decide which one to finish first.

user228700

7:13 AM

... almost there...

user228700

Someone please gimme a high-five 0/

user228700

(My face is very long--that's why the "0" instead of the "o" :-P)

user228700

(Oh, the high-five is because I just finished this stupid chap.)

user228700

:-( Nobody? Okay...

long face means sad in UK slang ...

user228700

7:23 AM

Oh, no, that's not what I meant. My face is literally quite long.

I didn't think you'd be sad to have finished the chapter on units :-)

user228700

Wow, there's a crossword puzzle (the question, as always, is effed up-they switched "Down" and "Across" for some. So lovely -__-)

user228700

Anyhoo, onto better things!

user228700

7:49 AM

@JohnR: (::sheepishly::) Are u familiar with Le Chatelier's principle?

Didn't we discuss Le Chatelier's principle (at some length) a while ago?

user228700

We did...but we didn't really get anywhere 'cause I was scared of free energy back then :-P

user228700

U kept explaining it to me in terms of free energy and I kept saying "Oh God oh God oh God" and eventually, we dropped it :-P

I can't remember where we got to wihth it. Anyhow, what was the question?

user228700

One second...

user228700

7:55 AM

Alright, before that, dyou have any idea about the real gas equation? With the Van der Waals constants and all..?

user228700

Gah, never mind.

user228700

@JohnR: For how much longer will u be around..?

Another four hours at least ...

user228700

Phew. Okay. I'll go have lunch, then.

@JohnRennie Thermodynamics is good stuff

8:07 AM

@BalarkaSen Hmmm

is that an agreement or a disagreement or neither? :)

That is a recommendation for you to seek urgent psychiatric help :-)

there exists worse things than thermodynamics that I think are good stuff...

8:33 AM

John Rennie, you there?

I'm here!

You answered my ideeal-gas question so can i ask some further help.

OK

So basically, if we try to think about these things intuitively:
P~n only for constant T and V.
P~T only for constant n and V.
P~1/V only for constant n and T.
V~T only for constant n and P. -

Is it okay to say something like this:

P~n, because if we scale the number of particles by x the momentum in the system
gets scaled by x.

P~T, because if we scale the kinetic energy by x then the particles move
√(x) faster, but because they move that much faster also the time it takes
to the other side of the container is scaled down by √(x). So for force:
Δ(m*√(x)*v)/(Δt/√(x)) which is ofcourse x*Δm*v/Δt.

P~1/V, because if the Volume is scaled up by x^3 any distance between two ends
of the container would be scaled up by x so the time between two ends would be

Or am i thinking it as too simple.

That's basically correct.

8:37 AM

So the thought process is ok, and i didn't doo anything dumb?

Just bear in mind that the microscopic model is meant as a way to help you get an intuitive feel for what is going on.

When you start doing microscopic calculations properly it all gets very complicated!

But isn't it supposed to be the model to explain what is happening?

That's the subject of statistical mechanics.

Yeah, ik it might get complicated, when you start to do it properly.

So it's ok to make a statement like that?

Yes, I think your statement is fine.

8:40 AM

If i try to think the V~T proportionality like that, it doesn't seem to follow though.

But is that because of the expansion, or maybe it's more complicated some other way?

Also do you mean to say that this statement would be fair to say in the field of statistical mechanics.

But maybe in some other field it would be more complicated.

I'm still a noob though, so idk what i'm talking about.

user228700

@JohnR: Please do ping me when u're relatively free again :-) ...it seems like u've a lot on ur plate right now.

I find it hard to believe that a particle can't cross speed of light

Well you have to, right?

Isn't it one of the assumptions in general relativity.

You have to assume it right? =)

a mathematical formula can't decide the practical exceptions

But it's an assumption, so you have to believe it.

=)

8:46 AM

@LuBu if you increase the temperature the particles are going to move faster and the pressure is going to go up. You need to work out the volume increase to bring the pressure back to it's initial value. If you do that you'll find V/T is a constant.

Is that so.

I think i thought about it a lil and didn't seem to compe to the appropriate result.

I'll see what i can do.

@AlanWatch see:

34

A: What is so special about speed of light in vacuum?

Special Relativity is based on the invariance of a quantity called the proper time, $\tau$, which is the time measured by a freely moving (i.e. not accelerated) observer. The proper time is defined by: $$ c^2d\tau^2 = c^2dt^2 - dx^2 - dy^2 - dz^2 $$ This is similar to Pythagoras' theorem as lea...

It comes down to a fundamental symmetry of spacetime called Lorentz invariance.

@Kaumudi.H We can start now. I'll flex my thermodynamic muscles in preparation :-)

user228700

Aright, then! Gimme a minute...

user228700

Wokay, this shouldn't take very long. Will u please tell me if what I've written here is correct:

user228700

> "If volume is increased, pressure decreases and hence, the reaction will try to shift in which pressure increases, i.e. In the direction in which the number of moles increases"

8:56 AM

Yes, that's a fair statement of Le Chatelier's principle.

user228700

Alright. Previously, I used to insist that since the volume increased, the reaction should shift in the direction in which the number of moles decreases because we want the volume to go back down again.

reaction doesn't shift..it is the equilibrium that shifts forward and backward

user228700

Terminology meh. In this case, it doesn't make that much difference to say "reaction" instead of "equilibrium" does it? What does reaction shifting even mean outside of this context?

Ok, I can see why you might be confused. However volume isn't something that directly affects the reaction while pressure is. Because pressure is related to the densities and energies of the reacting molecules.

user228700

> "Because pressure is related to the densities and energies of the reacting molecules.

user228700

9:00 AM

Why are these two factors so important?

I suppose I mean pressure and temperature are intrinsic properties.

user228700

Pressure is an intrinsic property? That's not so intuitive...

if you increase pressure reaction will move forward

@Kaumudi.H yup they both are an intrinsic property

@Kaumudi.H isn't it? If you have gas at 1 atm it doesn't matter if you look ata cubic micron or cubic metre (or cubic mile) the pressure is still 1 atm.

user228700

:-| Huh. That does seem to make sense...

9:04 AM

Likewise temperature

what was the question?

user228700

It's easier to understand why temperature is an intrinsic property.

Temperature will vary only if anyone of them becomes constant

user228700

Alright, so @JohnR: What was it you were saying..? "OK I can see why you might be confused". I understand that pressure is an intrinsic property but how is this fact related to my original statement..?

Well reactions are basically molecules colliding. If you change the pressure and temperature you change the density and energy so you change the numbers of molecules colliding per socond, and you change the collision energy. Both factors will affect the reaction kinetics.

user228700

9:08 AM

OK..?

But when you talk about changing the volume what does that mean for the reaction? It only affects the reaction indirectly by changing the pressure and/or temperature.

user228700

Riight...

user228700

Gimme a minute, please...

It's the intrinsic properties that directly affect the reaction because they are, by definition, some quantity per mole i.e. per molecule.

user228700

Right, okay. Do u agree with this statement:

user228700

9:16 AM

> "When the whole (balanced) chemical reaction is multiplied with a constant $c$ throughout, the new equilibrium constant becomes $K^c$. Your rate hasn't necessarily increased; the kinetics of the reactions are just the same. The equilibrium constant serves as a way to analyse how many moles of your product you are gaining, compared to how may you are losing. If you increase the number of moles of your reactant, your product amount has to increase correspondingly"

user228700

^ This is supposed to be a justification as to why the constant changes to $K^c$ on multiplying everything with $c$.

Hmm.

Suppose you have $A + B \to C + D$ so the equilibrium constant is $$K = \frac{[C][D]}{[A][B]} $$

user228700

OK...

Do you guys see that text formatted.

If you change the reaction to $2A + 2B \to 2C + 2D$ then you'd have a new constant: $$ K' = \frac{[C]^2[D]^2}{[A]^2[B]^2} $$

@LuBu See:

18

A: Any chance of MathJax in chat?

As a workaround while this request is pending, there exist several client-side workarounds that can be used to enable LaTeX rendering in chat, including: ChatJax, a set of bookmarklets by robjohn to enable dynamic MathJax support in chat. Commonly used in the Mathematics chat room. An altern...

9:22 AM

Oh.

Also the result of my primitive brain:

V~T, because if Temperature is scaled by x then the force is scaled up by x.
For the pressure to stay constant the area would need to be scaled by x.
If the area is scaled up by x then the volume is scaled up by √(x)^3.

@Kaumudi.H and obviously $K' = K^2 $

Should i think about the work done in the expansion.

user228700

@JohnRennie Well, yeah.

But in effect all you've done is change your units. For $K$ the units are moles per litre, but for $K'$ the units are half-moles per litre

user228700

Wait, what? :-o

9:25 AM

Or did I mean twin-moles per litre ?

user228700

I'm not getting any of this.

Hmm, I'm not entirely sure I understand what I'm saying :-)

user228700

What? :-P

user228700

Brb

It hadn't occurred to me before that you could multiply your equation through by a constant and thereby change the equilbrium constant ...

It seems a strange thing to do.

@LuBu don't forget the time taken for the molecules to cross the box they ae in.

9:29 AM

But don't i account that in the scaling of the force.

user228700

@JohnRennie Oh :-| Okay, I'll do some semiconductor physics for awhile, then.

user228700

(BTW, the source for my original statement is the 2nd answer to this question:

user228700

7

Q: Why does the equilibrium constant change on scaling the stoichiometric coefficients of a reaction?

I read in my textbook that if we multiply a chemical reaction by some factor(let's say $b$) its new equilibrium constant becomes $K^b$.But I don't understand why this happens.. What is the difference between the reaction $\ce{A} + \ce{B} = \ce{C}$ and $b \ce{A} + b \ce{B} = b \ce{C}$ ? Aren't th...

physical-chemistry equilibrium

@Kaumudi.H some people have too much free time :-)

user228700

:-|

user228700

9:42 AM

What dyou mean?

Why did it occur to anyone to multiply the equation by a constant and see what happens?

Obviously nothing happens because you aren't changing the reaction.

user228700

It's a common question on tests...

So it comes down to working out how come the maths gives you a new value for $K$. Fun to do if if you have nothing better to do e.g. if you've already cut your toenails and watched paint dry :-)

@Kaumudi.H Really?

user228700

@JohnRennie Yes, which leads curious students to wonder exactly what is the physical significance of this constant, if it changes every time u tweak the stoichiometric coefficients.

user228700

Oh, crap, I just added an incorrect spelling of the word "significance" to my dictionary on this computer (by mistake, of course). HOW TO REVERSE THIS PLEASE HELP!

9:48 AM

howtogeek.com/246686/…

user228700

Yep, I found that, thanks! :-D

user228700

Phew. Okay.

I didn't know about the Chrome spell checker. Gosh, the things you learn in the physics chat.

user228700

:-P I have added many words to the Google dictionary under my Gmail account. It's very handy.

user228700

Apparently, I added "CREEEEEPY". God only knows why. There. No red line :-P

9:51 AM

spellcheckers can be tedious sometimes

@Kaumudi.H I'm not sure I want to know why you added that word to your dictionary :-)

user228700

I don't remember anyway :-P

user228700

Wow. Many HP spells too :-P

user228700

CRUCIO.

gibbsfreeenergiatus!

user228700

9:53 AM

FeSO4.7H2O. WHY, like WHY did I add that? Sigh, past me.

@Kaumudi.H I don't remember that spell from HP ...

user228700

Lol :-P

ferrosulphatio!

user228700

Oh God x'D

Turns people green and slightly acidic.

user228700

9:55 AM

Hexaflexagon. Now there's a handy word:

user228700

user image

I must have decided I know how to spell :-)

user228700

:-P How were u not aware of this?

I came up with this:

V~T, because if Temperature is scaled by x then the velocity is scaled by √(x).
Force: Δ(m*√(x)*v)/(d/√(x)v). If i assume Volume is scaled by x then:
Δ(m*√(x)*v)/(cuberoot(x)/√(x)v)

for Pressure: Δ(m*v)/(cuberoot(x)*x) wouldn't stay constant?

I use floccinaucinihilipilification on a regular basis

user228700

9:58 AM

Whoa.

pneumonoultramicroscopicsilicovolcanoconiosis

That's a (pseudo?)scientific term though, contrary to floccinaucinihilipilification.

relevant

@BalarkaSen that's a worthless word :-)

obv

I simplified incorrectly for the last line which should be: for Pressure: Δ(mxv)/(cuberoot(x)) wouldn't stay constant?

But what am i doing wrong here if i should arrive to the result that both are scaled by x.

user228700

10:12 AM

This is what my textbook says:

Suppose we increase the volume of our box by a factor of $n^3$. The area goes up by a factor of $n^2$, so the number of collisions per unit area goes down by a factor of $n^2$.

Plus the width of the box increases by a factor of $n$ and that decreases the collision frequency by another factor of $n$. In all the number of collisions per unit area per second goes down by a factor of $n^3$.

Assuming the temperature doesn't change that means the vleocity/momentum of the gas particles doesn't change so the pressure goes down by a factor of $n^3$. So PV remains constant. So far so good.

user228700

> "As we know, a crystal is a regular periodic arrangement of atoms separated from each other by a small distance called the "lattice constant". The value of the lattice constant is different for different crystalline solids, however, it is of the order of linear dimensions of atoms (~$A^o$)"

OK

user228700

What is the "linear dimensions of atoms"?

It just means a length scale of around the diameter of an atom

user228700

10:16 AM

Ohh. What an idiot (me). Thanks!

user228700

Dammit. YouTube videos that cover the Band theory of solids are either over-simplified or too complicated to understand >.<

user228700

(Needless to say that my textbook has done a miraculous job, as always)

user228700

Sheesh. God only knows how many times I've dissed my textbook on this site :-P

stating facts is not a sin

user228700

10:43 AM

@JohnR: Are u busy?

Give me 5 minutes ...

user228700

:-) Okay.

Yeah, that makes sence.

Also in thermodynamics is the Temperature viewed as the average kinetic energy of a particle or proportional to it?

user228700

@LuBu It's not exactly the same but yes, temperature is a measure of the average kinetic energy of particles in the material in question. Yes, proportional.

user228700

$E= 3/2 K_B T$

10:48 AM

In systems which are ideal gas, in any case.

user228700

Where $E$ is the average kinetic energy of the particles in the material, $K_B$ is Boltzmann's constant and $T$ of course, is temperature.

user228700

@BalarkaSen Oh, yes, this is an important point to note, @Lu

And that is the equation for KE which is the same as internal energy for simple system where we only consider the KE.

Right?

Ty.

user228700

@LuBu Give this a read:

user228700

boundless.com/physics/textbooks/boundless-physics-textbook/…

user228700

10:52 AM

TL; DR : The number in front of $K_BT$ simply denotes the number of degrees of freedom available to the particles in the material.

What exactly is a degree of freedom =).

user228700

:-P Google!

Is it a simple concept.

?

user228700

Never mind. Yes, it is simple enough.

Degrees of freedom (physics and chemistry)

In physics, a degree of freedom is an independent physical parameter in the formal description of the state of a physical system. The set of all dimensions of a system is known as a phase space, and degrees of freedom are sometimes referred to as its dimensions. == Definition == A degree of freedom of a physical system is an independent parameter that is necessary to characterize the state of a physical system. In general, a degree of freedom may be any useful property that is not dependent on other variables. The location of a particle in three-dimensional space requires three position coordinates...

@LuBu have a read through that and come back to us with any further questions.

10:55 AM

@LuBu The number of independent directions the atoms of your gas can move. Eg if it's monoatomic it can move in 3 directions (three dimensions of space). If it's diatomic it can move in 3 * 2 (three degrees of motion for each atom) + 1 (flip the molecule around along the axis perpendicular to the bond) = 7 directions.

Im just starting, i guess you could say im taking introductory physics, but the school systems are very different nevertheless.

@LuBu if you're only just starting out in physics then you're doing pretty well :-)

user228700

^ :-)

@Kaumudi.H what did you want to ask before I disappeared for five minutes?

user228700

This:

10:57 AM

Well i'd like to complain about the school system/or the schools in my city/town. Because i'd feel like a school of similiar lvl in another country would be able to provide much more.

=)

But idk.

user228700

user image

user228700

Particularly the last paragraph.

OK, what in particular?

user228700

Just the last paragraph--I don't understand anything.

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