Problem Solving Strategies

krauser126

04:25

@JohnRennie Good morning. Was wondering if youre busy. I wanted to know why does attenuation happen exponentially (i.e. : Light traveling through a liquid vs. through stacks of acrylic blocks)

And how are physical photons affected differently when traveling through a liquid vs solid layed blocked (i.e.: acrylic). I know both allow light to reduce in intensity exponentially but idk why and how the mechanisms are different.

John Rennie

@krauser126 exponential attenuation happens because a photon has a constant probabibility of interacting with the atoms/molecules it passes through.

user8718165

@JohnRennie hello

krauser126

@JohnRennie Right I understood that, but wouldnt that mean it happens linearly?

John Rennie

So we can define a characteristic length L for which the photon has a 50% chance of being absorbed.

user8718165

@JohnRennie how can a changing magnetic field create an electric field?

John Rennie

04:29

@user8718165 give me a moment to answer @krauser126's question ...

user8718165

@JohnRennie Okay...no problem

John Rennie

@krauser126 so after a length L 50% of te original photons remain unabsorbed. Then in the next length L half of those photons are absorbed so after 2L 25% of the photons remin. Then after 3L 12.5 and so on.

So the fraction remaining is $2^{-x/L}$ where $x$ is the distance travelled.

krauser126

Oh alright. Is this the same regardless of the medium that light travels through? Whats different between light traveling through, say, water vs. a stack of acrylic blocks?

John Rennie

Different materials have different absorption lengths, but they all obey the same exponential equation.

Offhand I don't know how the absorption lengths of water and acrylic compare.

krauser126

When you say photons have constant probability, do you mean no matter how far the photons go, the probability of interacting with a single air molecule stays the same?

John Rennie

04:36

@krauser126 Yes. The past history of the photon doesn't affect its chances of interacting with an air (or whatever) molecule.

krauser126

Ok cool. Thanks :)

John Rennie

@user8718165 hi

user8718165

yes sir

how do changing magnetic fields create electric fields

John Rennie

Magnetic and electric fields are not two different things. There is a single field called the electromagnetic field.

Depending on how we interact with it the EM field can look like an electric field, a magnetic field or a mixture of both.

So it's not really meaningful to ask how a changing magnetic field creates an electric field because that isn't really how it works. When we wave a magnet around we create an electromagnetic field that has both magnetic and electric components.

user8718165

when a charge is oscillating it creates a changing electric field which in turn creates a changing mag field. How does this happen? Is there some intuition?

@JohnRennie oh okay

John Rennie

04:43

@user8718165 That isn't true. When a charge is oscillating it creates a changing electromagnetic field.

user8718165

it creates both changing electric and mag fields?

John Rennie

Understanding how an EM field can appear as both an electric and a magnetic field is a bit complicated because you need to understand special relativity.

user8718165

I understand about length contraction. Is that enough sir?

John Rennie

@user8718165 depending on what you do with the charge you can create a range of different behaviours. For example a charge moving at a constant velocity creates a constant magnetic field. A charge that is oscillating creates an oscillating magnetic field.

Electromagnetic tensor

In electromagnetism, the electromagnetic tensor or electromagnetic field tensor (sometimes called the field strength tensor, Faraday tensor or Maxwell bivector) is a mathematical object that describes the electromagnetic field in spacetime. The field tensor was first used after the four-dimensional tensor formulation of special relativity was introduced by Hermann Minkowski. The tensor allows related physical laws to be written very concisely. == Definition == The electromagnetic tensor, conventionally labelled F, is defined as the exterior derivative of the electromagnetic four-potential, A,...

@user8718165 this is how you describe the EM field using special relativity.

It's ... complicated :-)

user8718165

@JohnRennie tooo much

@JohnRennie can't we talk about it just without using a lot of math? I'm not sure. If its not possible sorry

John Rennie

04:54

@user8718165 there are simplified explanations. For example you can explain how a current, i.e. moving electric charges, creates a magnetic field by using length contraction. Let me see if I can find an example ...

user8718165

@JohnRennie yes sir, I know about it

John Rennie

@user8718165 ak OK.

user8718165

How Special Relativity Makes Magnets Work

►

John Rennie

But bear in mind that this is a simplified description and can be misleading. To really understand what is going on you do need to understand the maths.

user8718165

@JohnRennie Yes sir...I'll learn it later surely because most of the math needed for this is what I don't know

John Rennie

04:57

You won't learn this stuff unless you do a physics degree, and even then you won't learn it until your second or third years. It's pretty advanced stuff.

user8718165

@JohnRennie when a charge accelerates (oscillates), it creates a varying magnetic field and electric field. I basically understand this. The charge continues to accelerates and this process continues which gives rise to the EM wave. My book states something like this. Sir, does my understanding make sense?

John Rennie

Yes, that's a good description of what happens.

user8718165

@JohnRennie all these will take me a long time. I just want to have a good understanding of physics and apply that in real world with the help of computers(I don't know how I will do that) and I don't think I'm going for a physics degree. Rather I love programming:-)

John Rennie

I have always felt that the best reason for doing something is that you love it.

I was fascinated by physics from an early age.

But then personal computers didn't exist when I was your age or maybe I'd have become a programmer as well :-)

user8718165

05:12

@JohnRennie All these days I used to think that once the charge had oscillated for some time and then had been stopped, the electric and magnetic fields previously produced by the charge would be sufficient to sustain the EM wave. Today I learnt that the charge has to oscillate constantly because its the charge that produces the fields and each bit of the field is generated by the charge

@JohnRennie you are a good programmer. you wrote some good codes on this chat site before. I'm unable to find those.:-)

John Rennie

@user8718165 If you oscillate a charge for one second you produce an EM wave $c$ metres long ($c$ is the speed of light). When you stop oscillating the charge that wave doesn't disappear.

user8718165

@JohnRennie I don't get it

John Rennie

Let me draw a diagram ...

user8718165

wouldn't the wave disappear because the source has stopped producing EM waves?

@JohnRennie :-)

John Rennie

Before you start oscillating the charge there's no wave.

When you start oscillating the charge it creates a wave that move off to the right at a speed $c$

When you stop oscillating the charge it stops creating a wave, but the wave you have already created keeps on moving away from you.

user8718165

05:22

Okay sir got it...just like a rope?

John Rennie

Yes

You get a wave pulse:

Wave packet

In physics, a wave packet (or wave train) is a short "burst" or "envelope" of localized wave action that travels as a unit. A wave packet can be analyzed into, or can be synthesized from, an infinite set of component sinusoidal waves of different wavenumbers, with phases and amplitudes such that they interfere constructively only over a small region of space, and destructively elsewhere. Each component wave function, and hence the wave packet, are solutions of a wave equation. Depending on the wave equation, the wave packet's profile may remain constant (no dispersion, see figure) or it may change...

user8718165

so it will travel up to infinity...

John Rennie

... or until it hits something :-)

user8718165

why does light instantly snap when we switch off the torch? Is it that the photons hit the wall at 'c' and hence instantly give out their energy in some other form? If the wall weren't there the generated wave pulse would travel and travel?

@JohnRennie okay sir...got it

I mean is the light disappearing thing the same phenomena

John Rennie

If you were in a vacuum then the light from the torch doesn't disappear when you turn it off. The light emitted by the torch carries on off into the universe.

But when you shine a torch you aren't seeing the light it emits. You are seeing the light that gets reflected back towards you.

user8718165

05:29

@JohnRennie tyndall effect??

John Rennie

@user8718165 the Tyndall effect is one form of light scattering, but there are lots of others. When you shine the light on a white wall you're seeing the light reflected off particles in the paint.

krauser126

@JohnRennie Do you know why I cant upload any pictures here anymore?

Used to have the option right next to "send" but now its gone

user8718165

@JohnRennie got it sir. Thanks a lot:-)

John Rennie

@krauser126 on a PC or a phone? What browser?

krauser126

PC

Google Chrome

Same as before

John Rennie

05:44

I don't know I'm afraid. Can you try on a different PC to see if it's something on your PC?

Or maybe try with Internet Explorer and see if that shows the same.

Ajay Mishra

06:00

Hi @JohnRennie , I guess you know of sliding chain problem with friction?

John Rennie

@AjayMishra yes ... ?

Ajay Mishra

In the problem I have to find the work done by friction by the moment is slides off completely, I thought my equatiom should be Change in GPE = $W_f$ + Change in kinetic energy. Am I right?

John Rennie

@AjayMishra yes, that seems reasonable

YUSUF HASAN

@JohnRennie Hi, are you free rn?

Ajay Mishra

After solving the differential equation, i am getting answer as a logarithmic expression, which does not matches with the answer.

John Rennie

06:10

@AjayMishra @YUSUFHASAN I need to work for about half an hour but I'll be around most of the morning.

YUSUF HASAN

John Rennie

@AjayMishra I would have to see the question.

Ajay Mishra

Okay.

YUSUF HASAN

Okay, I am uploading the question anyway @JohnRennie , please see it asap

Ajay Mishra

A chain of mass $m$ = 0.80 $kg$ and length $l$ = 1.5 m rests on a rough-surfaced table so that one end hangs over the edge. The chain starts sliding of the table all by itself provided the overhanging part equals $\eta$ = $\frac{l}{3}$. What will be be total work done performed by the friction force on the chain by the moment it slides completely of the table?

John Rennie

06:44

@YUSUFHASAN hi, you there?

YUSUF HASAN

@JohnRennie Yes I am here

John Rennie

YUSUF HASAN

This delta S you have marked is for the universe or system? @JohnRennie

John Rennie

We are told the process from P direct to R is adiabatic, so $dQ_{rev} = 0$, and reversible so we know the entropy change for this process is zero.

@YUSUFHASAN I've labelled the entropy change of the universe. I'm not sure we can say what the entropy change of the system is because we don't know what is happening to the system.

It looks like a PV diagram, though the question doesn't say what the diagram is.

YUSUF HASAN

okay..So entropy change of the universe for process PQ is greater than zero cause it is irreversible?@JohnRennie (Yeah, I also assumed it to be a PV diagram)

John Rennie

06:51

Anyhow the entropy of the system is a state function so it is the same at R regardless of what path was taken from P to R.

YUSUF HASAN

Wait a minute..I am confused..Isn't entropy always a state function, regardless of whether we consider universe or system?

@JohnRennie

John Rennie

I'd guess the entropy of the system increases in step P->Q, so it must decrease in step Q->R. So it's option (C).

@YUSUFHASAN true. I highlighted of the system to make clear I was now talking about the system not the whole universe.

The diagram (whatever it is) only shows us the state of the system and not the state of the rest of the universe.

YUSUF HASAN

@JohnRennie Okay...So this conclusion was drawn using the idea that entropy of system when we reach at R must be same, right?

John Rennie

Yes.

YUSUF HASAN

Okay..But I have a new doubt..According to the second law of thermodynamics..entropy of universe always increases in an irreversible process, while it remains constant(=0) is a reversible process, right? Suddenly, entropy of universe doesn't seem like a state function to me..@JohnRennie

John Rennie

06:59

@YUSUFHASAN what the diagram doesn't show is that the state of the rest of the universe is different at R depending on the path taken.

kyle campbell

Hi, I have a question I cannot figure out: You wind a string around the outer rim of a solid disc with a radius of 10 cm and a mass of 2.8 kg. The disc sits at on a frictionless horizontal surface. You pull on the string with a force of 10 N for 3 s. What is the final kinetic energy of the disc? This includes both translational and rotational kinetic
energy.

I have tried to simply do K = 1/2mv^2 + 1/2I(\omega)^2 but numerically I'm getting half of the correct answer. The moment of inertia for a solid disc, as far as I know, is 1/2mr^2 about the z-axis. Is this not correct?

John Rennie

@YUSUFHASAN The diagram is only showing us P and V (or whatever it shows) for the system. It doesn't show us what the state of the rest of the universe is.

YUSUF HASAN

Okay...So that means that we don't really know whether entropy of the universe is a state function or not unless it is specified?@JohnRennie

John Rennie

The entropy of the universe is a state function.
However we don't know what the state of the rest of the universe is.
Just because the state of the system is the same at R regardless of which path we follow, that doesn't mean the state of the rest of the universe is the same when our system reaches the state R.

YUSUF HASAN

Okay..Just one last thing... how did you judge that whether the entropy of the system is conserved at R? @JohnRennie

kyle campbell

07:04

Also, the final speed was just found with impulse but that's trivial.

John Rennie

@kylecampbell you know the linear impulse delivered because it's just force times time. The angular impulse is the linear impulse times the radius of the disk. The linear impulse has to be equal to the change in linear momentum, and the angular impulse has to be equal to the change in angular momentum.

YUSUF HASAN

Because by the diagram you knew it was a state function.. or because you took it to be?

John Rennie

@YUSUFHASAN entropy is always a state function.

YUSUF HASAN

Okay...I got it finally!! Thanks @JohnRennie sir!

kyle campbell

@JohnRennie yes, I agree, and that's how I found the angular speed (which is easy to relate to the linear speed). however, I'm still getting half of the correct answer.

John Rennie

07:05

The diagram is showing us the state of the system, and that state is the same at R regardless of the path.

@YUSUFHASAN I have to admit I find it really easy to get confused about whether entropy is the entropy f the system or the total entropy. You just have to be really careful.

@kylecampbell it sounds as if you're doing the problem the same way I would, and the moment of inertia of the disk is indeed $\tfrac{1}{2}mR^2$.

I can have a go at the calculation later, but right now I need to work for a bit.

kyle campbell

@JohnRennie yes, I'm actually a tutor (in 4th year) but I'm a bit confused about that one

Ok! thanks

YUSUF HASAN

@JohnRennie My real doubt which persistently bugs me in these questions is that whether I am applying the second law's postulates( irreversible has entropy change positive, reversible has zero) to the system,or universe..But I guess for problem-solving, I can apply that without actually considering whether we are referring to system or universe, @JohnRennie?(You have seen plenty of problems, so just wanted to know your opinion)

John Rennie

The second law always, always, applies to the whole universe.

kyle campbell

In the end, you should have K = 1/2mv^2 + 1/2(1/2mr^2)(v^2/r^2) = 3/4mv^2, no?

John Rennie

@YUSUFHASAN In an irreversible change the entropy of the system can increase or decrease depending on what that change is. But in an irreversible change the entropy of the universe always increases.

YUSUF HASAN

07:12

Okay...Thanks for the clarification! @JohnRennie

kyle campbell

07:49

@JohnRennie Is the disc perhaps slipping because it's lying flat? If that's the case, it would make sense why v =/= r(\omega)

It seems from the calculation of both linear and angualr impulse that \omega=2v

John Rennie

@kylecampbell I get the linear KE as $F^2 t^2 /(2m)$ and the angular KE as $F^2 t^2/m$

kyle campbell

@JohnRennie Right, I just clued into that. Physically, then, where in the problem does it mention that there is slipping? Or is that something that you would have to find out yourself from the calculation?

John Rennie

Slipping? The question tells us that the contact between the disk and the table is frictionless so effectively it's an isolated disk

kyle campbell

From what I know, v=r(\omega) won't apply if there is slipping. In this case, v =/= r(\omega) so that's where my confusion is.

That is, if this is purely rolling motion, then v =/= r(\omega). But I guess that's not the case because it's lying flat?

John Rennie

The equation $v = r\omega$ gives the tangential speed relative to the centre. It's true that for a rolling disk the tangential velocity relative to the centre is only equal to the linear velocity of the centre when there is no slipping, but I don't see the relevance here. This isn't a rolling disk.

kyle campbell

07:59

Right, I don't know what I was thinking.

Thanks for your help!

user8718165

@JohnRennie sir how do you write things in bold face?

John Rennie

@user8718165 **this is bold** gives this is bold

user8718165

@JohnRennie and why did that the first 'this is bold' not turn into bold despite using double asterisks. I can't do that

John Rennie

@user8718165 use the back quote character ` to surround the text

**this is bold**

Hmm, I can't get the backquote character to display

`**this  is  bold**`

Aha. See above.

user8718165

08:15

thanks sir and you always write equations in La Tex over here. But they don't show up as formatted equations. Does it show up for you?

John Rennie

Have a look at chat.stackexchange.com/faq in the section "How do I format my messages?"

Chat doesn't display MathJax by default. You have to mess around a bit to get it to display. Are you using a phone or a PC? As I recall it's straightforward to get MathJax to work on a PC/laptop but really hard to get it to work on a phone.

user8718165

@JohnRennie I'm using a desktop computer sir

Abcd

9

A: How can I enable MathJax in chat?

Update 2017-05-01 The MathJax CDN retired and the javascript-URL idea is not so easy any more, because of browser security. (Chrome stips away any leading javascript: when pasting into the URL line. SE modified the javascript: link so that it does not work.) So here is my take. I modified the ...

Follow the instructions given to enable mathjax in chat

user8718165

@JohnRennie saw it now

John Rennie

@user8718165 also see:

121

A: Should chat have TeX support?

There are four bookmarks which are located on this installation page: start ChatJax installs MathJax and starts a loop that renders $\LaTeX$ as needed. This is intended for use in chat, where the contents of the page are not static. Reloading the page will stop the loop, so the bookmark needs t...

That's what I use. As I recall the bookmarks suggested by Abcd also include support for rendering chemical equations.

Abcd

08:26

@JohnRennie Dont you remember you changed yours??

@JohnRennie Are you able to see this in mathjax: $\ce{H2 + O2 -> H2O}$

John Rennie

$\pu{test}$

Ah \pu works and I think that's only in the links you use, so I must have installed them.

Abcd

@JohnRennie Are you able to read the equation above??

John Rennie

@Abcd yes, that renders. So I obviously have your links.

kyle campbell

$G_{ab}=8{\pi}T_{ab}$ just testing myself

John Rennie

@AjayMishra are you around? Do you want to look at that chain problem?

Ajay Mishra

08:48

@JohnRennie , sure.

John Rennie

@AjayMishra it should be straightforward because the friction is just $\int Fdx$ and we don't need to know the time dependence of the motion.

Ajay Mishra

There are not any friction coefficient given in the problem.

How am I supposed to figure out the force?

John Rennie

The question tells you that the chain starts sliding when a third of the length is off the table. That means it slides when the force is $mg/3$

And the frictional force will be $\mu 2/3 mg$

So $\mu = 0.5$

Ajay Mishra

okay.

But then why the problem is not solved through conservation of energy consideration?

I regret! It was quite easy to figure out the value of $\mu$

John Rennie

Conservation of energy must apply, but it seems complicated to me.

What I would do is just calculate the work directly.

Suppose the distance slid is $x$, then the length of the chain left on the table is $1-x$ so the normal force is $F(x) = \mu(1-x)\tfrac{2}{3}mg$

And the work is $\int_0^1 F(x)dx$

Ajay Mishra

09:01

I was solving like this, $$\Delta U = W_f + \Delta K $$
as $K_i$ = 0, $ \Delta K = K_f = \frac{1}{2}m \Bigr(\frac{dx}{dt}\Bigl)^2 $
,$U_o$ = $\int_0^\frac{l}{3} \frac{M}{L}xgdx $ , $U_f$ = $\int_0^x \frac{M}{L}xgdx $ $$ \Rightarrow \Delta U = U_f - U_0 $$

Where x is distance from tip to the table

kyle campbell

Conservation of energy seems a bit complicated because we don't know the final velocity, and since the acceleration isn't constant it's not entirely trivial to find that quantity. Also, correct me if I'm wrong @JohnRennie but since the force of gravity is a conservative force there's no need to integrate to find the work done by gravity if you consider it acting through the centre of mass of the chain.

I think John's proposal is the more straightforward way.

Ajay Mishra

Why centre of mass? Here, clearly the GPE is function of x

kyle campbell

But why invoke an integral for something that you could easily find with the relation $-\Delta U_g = W_g$?

Ajay Mishra

@kylecampbell How you gonna find work at any instant?

kyle campbell

I wouldn't, that's what you would need to find $W_f$ for in the first place. That's why I don't see the use in using conservation of energy for something that you could calculate explicitly.

But, continue if you think you can figure it out that way (not saying you can't).

Ajay Mishra

09:11

how you would differentiate between these two situation, [assuming you're not going to use integrals]

imgur.com/a/3NeVerS

I use conservation of energy because it is very powerful way to figure out stuffs.

kyle campbell

I never said integration isn't necessary for this problem. I'm only saying that since the force of gravity is conservative, you don't need an integral to calculate the work done by it. I agree that it's a powerful principle, but sometimes there's an easier way to do it.

Looking at your terms, if you found your $\Delta U$, how do you plan to find $\Delta K$?

Ajay Mishra

$ v = \frac{dx}{dt} $

kyle campbell

Ok, then proceed.

Ajay Mishra

Finally, problem reduces to $$ t = \int\frac{1}{\sqrt{x^2-a^2}}dx $$

From, there it is possible in principle that, a function can be made such that $ x = f(t) $

John Rennie

@AjayMishra that's kind of complicated though. Personally if there's an easy way to solve a problem I'll always choose the easy way :-)

kyle campbell

09:29

@JohnRennie Off topic, but, do you think it's necessary to do a class in differential geometry if you're trying to learn GR? Is there any other math classes specific to GR that are probably useful as well?

John Rennie

@kylecampbell GR is 99.99999% diff geo so you certainly need to know it. The only problem is that if you take a diff geo course in a maths department most of it won't be of use to you as a physicist. You really want a diff geo for physicists course.

kyle campbell

@JohnRennie How does this sound: "The course is on differential geometry of curves and surfaces in $\mathbb{R}^3$, with an emphasis on the surface theory. We will introduce some basic concepts, such as the curvature of a curve, the tangent plane,
the differential of a map, differential forms, the first and the second fundamental forms, mean curvature and Gaussian curvature, Gauss map and vector fields, parallel transport, and geodesics. We will study Gauss’ Theorema Egregium and the Gauss-Bonnet Theorem."? That's straight from a math department unfortunately.

Probably worth waiting for more physics-specific stuff?

John Rennie

@kylecampbell I'd probably ask in the h bar as there will be people there who have done that sort of course (I've never done a diff geo course). My understanding is that mathematicians take a very different approach to diff geo and I'm not sure how much help it will be.

If there's no charge I guess you could audit the course for a few days and see how it goes ...

I'd probably just look for a suitable book

kyle campbell

Fair enough, that's kind of what I have also heard from others as well.

user8718165

10:32

$hello$ @JohnRennie

John Rennie

@user8718165 hi

user8718165

just added mathjax support:-)

John Rennie

$\large{Cool}$

user8718165

Thanks @JohnRennie

Dante

11:58

@AdvilSell

Advil Sell

@Dante Yup ?

Dante

Are you able to download papers?

resonance.ac.in/answer-key-solutions/JEE-Main/2019/…

yesterday's

Advil Sell

@Dante Yup , they are opening now

Dante

Damn, not opening here

Let me try clearing cookies

Btw

shift-1 physics last question.

How did they find out the extension?

Advil Sell

@Dante Opening

wait

@Dante They have given in question that initial amplitude is A0

Dante

12:08

Wth bad request again

Advil Sell

@Dante try incognito

Dante

Send SS of question

Advil Sell

Dante

Not this one

I don't remember which paper exactly

But question 28/29

Elastic potential energy one

Advil Sell

@Dante Nope , No such question in 1st shift , 29th is reynolds number question

Dante

12:14

Check second

Ajay Mishra

Catalpult one ? @Dante ?

There was 2 question regarding extension in 8th april first shift.

Dante

Yeah

Catapult

Ajay Mishra

I used energy conservation.

To find out k, I thought that that was wrong, but on checking answer keys it seems right.

Dante

@AjayMishra What's the extension in the band?

Ajay Mishra

$ x = \frac{mv^2}{k} $

But in the problem, We were given to find out that young modulus.

Dante

12:19

Oh wait

Could you please post the question here?

In case you've downlaoded the PDF

Ajay Mishra

When was your exam btw?

Dante

It's on 12th

Was yours yesterday?

Ajay Mishra

yup. First shift, that why I know this question

Dante

Oh nice! You must've done well!

Ajay Mishra

Didnt expected damped oscillation.

Couple of silly mistakes. But sure of 270 +

There was, I guess, question with wrong options too.

Dante

12:22

Oh, cool!

@AdvilSell Post the question man -_-

Advil Sell

Dante

Ok, how do you find extension in the band here?

I mean, do you calculate the approximate extension? Pretty sure calculating exact value is not feasible.

@AjayMishra @AdvilSell?

Advil Sell

@Dante Yeah they approximated delta L to be 0.2 meter

Dante

How did they do that?

Advil Sell

@Dante Just took it by reading the question the extension is given as 20 meter , But I think the question/diagram must have been diffrent

Dante

12:37

Hmm

Ajay Mishra

13:04

@AdvilSell there were not any diagrams.

Advil Sell

@AjayMishra Oh is it , was the language of question same as given

?

Ajay Mishra

Nope.

It was personified, that a boy pulls that.

Advil Sell

@AjayMishra Oh...Okay

Transcript for

Problem Solving Strategies