Morning to all :) @user8718165 @Aladdin @JohnRennie @Nobodyrecognizeable

@yuvrajsingh morning.

Morning :-)

@JohnRennie hi.

@Nobodyrecognizeable morning :-)

@JohnRennie 7.

That's just the lensmakers equation.

@JohnRennie ie $f_2/f_1=(n_1-1)/(n_1-n_2)$; n_1 = ri of glass?

The lensmakers equation for a thin lens is:

@yuvrajsingh Good morning Yuvraj :)

$$ \frac{1}{f} = \left( \frac{n_1}{n_0} - 1 \right) \left(\frac{1}{R_1} - \frac{1}{R_2} \right) $$

Good morning sir :-) @JohnRennie

@user8718165 morning :-)

@JohnRennie OK.

@Nobodyrecognizeable In this case $R_1$, $R_2$ and $n_1$ are remaining constant, and you're just changing $n_0$ from air $n_0=1$ to water $n_0=4/3$.

@JohnRennie ok

@JohnRennie 8

If you Google for the lensmakers equation you'll find lots of info about it, but this equation is all you need to remember.

@JohnRennie I'm not that good at googling :p.

@JohnRennie please have a look at 8

What's the problem with Q8? It looks straightforward.

@JohnRennie it gives me a.

You've forgotten about the second plastic-air interface :-)

@JohnRennie please elaborate. At first intensity is going off due to reflection then what?

At the first air-plastic interface the intensity is reduced to 0.8 then the intensity is reduced by another factor of 0.98 so the light arriving at the plastic-air interface has an intensity of 0.8 x 0.98 = 0.784.

But then the intensity is reduced again at the plastic-air interface so the light emerging has a lower intenity.

@JohnRennie sir I need your help

So 0.8*0.784

Yes

@JohnRennie done. Have a nice day professor. Thanks for the help.

:-)

@user8718165 yes?

@JohnRennie can I remove both my name and number from truecaller database?

@JohnRennie hello sir

I have no idea. What's the truecaller database?

@JohnRennie sir I saw that the name which has been associated with your number (in contacts) by majority of the truecaller users is displayed to someone (you may not know) who tries to contact you.....and that name is stored in the database with that corresponding number in the database

Is there an intuitive explanation for why capacitive reactance (perhaps using the hydraulic analogy?) is inversely proportional to the frequency of the voltage source?

@user8718165 true caller does a simple thing if you want I can explain. @user8718165

I remember thinking about this a year ago but I can't recall anymore.

I Googled TrueCaller. I assume you mean the Swedish company that maintains a database of telephone numbers.

@JohnRennie yeah sir...even I never used it....didn't you also use it:)

No, I've never used it.

@JohnRennie that's a very bad app...it violates right to privacy by exposing your name :-(

I have no idea if the company will honour requests to remove your number from its database. I would guess not.

When you make account on a true caller, the app ask to access your contact, so he collect the contact data including name and number. Your name and number might be save in some other phone so it is impossible to remove you contact from true caller, that's reality my friend working in bangalore with true caller he told me this. @JohnRennie @user8718165

@yuvrajsingh thanks yuvraj, even if I don't have the app installed, will others be able to view the name and stuff?

Yes, because if I am your friend, and I have you contact number with phone, and I have an account on true caller, it is impossible to remove your contact number. @user8718165

@psa the charge is proportional to the voltage, and and current is $dQ/dt$

So the current is proportional to $dV/dt$

Higher frequencies have a larger value of $dV/dt$ and therefore a higher current for any given voltage. Hence the reactance is smaller.

thanks!

@JohnRennie hi

i guess you can think of a capacitor like a rubber membrane and an inductor like a water wheel or something too

@yuvrajsingh okay got it :( BTW thanks for giving info on this.

@psa yes, in the hydraulic analogy the capacitor is modelled by an elastic sheet.

I guess you can look at it this way. For a given current the charge transferred (the stretching of the sheet) is proportional to the period of the oscillation so the higher the frequency the smaller the charge transferred.

Since the energy is proportional to $Q^2$ this means the energy needed reduces with increasing frequency.

@JohnRennie hello.

Are you busy @JohnRennie

@yuvrajsingh hi, I need to work for a few minutes but hopefully it shouldn't be too long.

@yuvrajsingh I'm free now.

Hi sir @JohnRennie

I have question related to com

@yuvrajsingh what's the question?

Sir I have square sheet whose, side is 4R,4R, so the Square constitution of four mini square, if I remove a circle of mass m an radius R, from he square sheet from both side, new Centre of mass will be. @JohnRennie it is given mass of sure sheet is 2M

There's a sneak trick for doing this.

Okay can you draw a diagram

So that it is clear, @JohnRennie

Suppose you have two objects then the centre of mass of the two objects is:

$$ x = \frac{m_1x_1 + m_2x_2}{m_1 + m_2} $$

OK so far?

Yup, sir.

When you cut out a circle of mass $m$ it is like adding a negative mass of mass $-m$.

Ok

Suppose we take the centre of the sheet as the origin, so the origin is at the centre of mass.

Then we cut out a circle of mass $m$ centred at a distance $x$ from the centre of the square sheet. The centre of mass is now:

$$ x = \frac{-mx}{2M - m} $$

(where $2M$ is the initial mass of the sheet)

Com of 2M also at $x$ @JohnRennie

It should be at origin for 2M, am I right sir

Yes, the centre of mass of the sheet before the circle is cut out is obviously at the centre of the sheet. So if we take that as our origin then $x$ for the sheet is zero. That's why it doesn't appear in my equation.

Continue sir

That's it. There's nothing more to say.

Can you check my reasoning please, can I @JohnRennie

I assume corner to origin

And let say initial Centre of mass $$x_(cm) $$, to be at (2R,2R)

So initial Centre of mass of circle and remains part of square is at the location.

So there's the initial sheet with the COM at the centre. So where is the circle cut out?

In the left region @JohnRennie

Of radius R

If we put the origin at the COM, so the position of the CM is (0,0) then where is the centre of the circle? At what coordinates?

With regards to origin is at (R, 0)

Like that?

Yes @JohnRennie

So we have a mass of $2M$ (the sheet) at $x=0$ and a mass of $-m$ (the cutout) at $x=R$. Yes?

Ok

$$ x_{com} = \frac{2M.0 - mR}{2M - m} $$

And assuming the sheet is uniform thickness we can find $m$:

$$ m = \frac{\pi R^2}{16R^2} 2M $$

Sir if this mass is placed somewhere can we apply the same approach to solve it. @JohnRennie

Yes. All we are doing is finding the centre of mass of two objects. The only trick is that because one of the objects is a cut out we give it a negative mass.

So add we add and subtract the same for different x right sir @JohnRennie

@yuvrajsingh the $x$ variables are just the positions of the COMs of the masses involved. In this case the two $x$ values are zero for the sheet and $x=R$ for the circle. If the circle is in a different position you just change $x$ to wherever its centre is.

Got it, sir I have small question sir when light reflect from floor does it intensity reduce, @JohnRennie

I guess that depends on what your floor is made of.

My sir told the no floor is perfect, so there is some light which distorted and some light will get absorb @JohnRennie

Nothing is a perfect reflector. Even a polished silver mirror only reflects 99% of the light. Some light is always absorbed.

Can you explain how a solid angle, works I need to find the intensity of sun light on earth. @JohnRennie

It's simpler than you think.

Consider a sphere of unit radius, then the solid angle subtended by the sphere is $4\pi$ steradians. OK so far?

Problem is that how a solid angle calculate, it look funny but I do not how actually that conical angle is define.

That's what I'm about to explain :-)

Sorry sir. Yes

Now draw some shape on the surface of the sphere. It doesn't matter what the shape is - a square, a circle, a wavy shape - all that matters is that the area of the shape is $A$.

Then the fraction of the sphere area that the shape covers is $f = A/(4\pi r^2)$.

Ok

And the solid angle subtended by the shape is just this fraction times the $4\pi$ steradians subtended by the whole sphere i.e. the solid angle is:

$$ \theta = 4\pi \frac{A}{4\pi r^2} $$

If I understand correct small A will makes $ (4\pi r^2)$.@JohnRennie

I'm not sure what you are asking. $4\pi r^2$ is the area of the sphere that we have drawn the shape on.

I have one more question @JohnRennie

@yuvrajsingh yes?

I have an accelerated electron pass through slit, will it show a diffraction.

@JohnRennie

Yes

The electron will have a de Broglie wavelength $\lambda = h/p$ and it will show the same diffraction pattern as light of the same wavelength.

Will this pattern be same as electron with very sow velocity. @JohnRennie

The de Broglie wavelength is inversely proportional to the velocity, and different de Broglie wavelengths will give different fringe spaces, though in all cases the diffraction pattern for a single slit will be a sinc function.

What kind of difference we will see in fringe pattern

@JohnRennie

I'm just Googling for the exact equation for single slit diffraction.

OK. Suppose we have a slit of width $d$ and a screen a distance $L$ from the slit. Then we measure distance $x$ from the central maximum in the diffraction pattern. The equation for the amplitude of the diffraction pattern is:

$$ A(x) = \frac{\sin\left(\frac{\pi d}{\lambda} \frac{x}{L}\right)}{\frac{\pi d}{\lambda} \frac{x}{L}} $$

This has minima where the argument to the $\sin$ is a multiple of $\pi$ so we get minima when:

$$ \frac{\pi d}{\lambda} \frac{x}{L} = \pi n $$

$$ x = n \frac{\lambda L}{d} $$

Ok

So the fringe spacing is proportional to the de Broglie wavelength.

Why we included the phase change of wave, @JohnRennie

Phase change of the wave?

No sir, I thought it was double slit, so phase difference it was typo. I got your explanation.

I'm not sure what that means. I haven't included any phase changes anywhere in the working.

One more question

One more question @JohnRennie

@yuvrajsingh yes?

@JohnRennie hi

@Aladdin hi :-)

@JohnRennie hi

@Aladdin hi

@JohnRennie hi

@yuvrajsingh hi :-)

Please. Come. In the chat room that we have discussed about mass @JohnRennie

All are saying that, your statement is incorrect @JohnRennie

@yuvrajsingh inertia is a poorly defined term. When people get into arguments about it that usually just means they have duifferent definitions for it.

@JohnRennie how do u sinplify this

imgur.com/a/cbLA3Gu

Help

@Aladdin I'm eating lunch now, but $AC + ¬A¬C$ is ¬XOR.

@JohnRennie ping after lunch

If u are still free

@JohnRennie are u free

@Aladdin hi

@JohnRennie hi

@yuvrajsingh hi

Can I ask @JohnRennie

Well, Aladdin was first ...

@JohnRennie imgur.com/a/bh2lsaq

Can u help with 2

I'll have to leave this until tomorrow. I was out late last night and I'm worn out now.

Ahh OK. Sorry for disturbing u. We can discuss tomorrow

Yes for sure, you can delete the chat,.

Good morning all. I am trying to figure out how to approach a problem. My goal is to see if I can build a chiller for liquid. In this case water and beer. I'd like to chill it as it flows through a coil of copper tube (I'll look at adding fins to increase the surface area later). I'm just not sure what all I need to consider. I'm not even sure if this is the right place to start.

Deleted

I'm guessing I need to focus first on flow rate, surface area, and the difference in temp that I want to achieve.

Are there some basic formulas for this?

@BrandenBoucher hi. This chat room is for helping with exam questions. We don't know anything about cooling beer :-)

@yuvrajsingh Are you OK now?

@JohnRennie, I'm not sure I totally believe that last part :). But gotcha. I had a feeling this wasn't the right place to be asking.

So JR SIR Thank you so much yes I am fume.

Well, I will take my leave and wish you all a good day!

@yuvrajsingh OK. I'll leave you to talk to your family then. I'll be around tomorrow as usual if you want to chat.

Sir, yes I we will discuss it tommorow.

OK. Sleep well and feel better in the morning.

Good night. @JohnRennie bye.

Bye :-)