Problem Solving Strategies

Jasmine

5:52 AM

@JohnRennie Morning :-)

John Rennie

@Jasmine morning :-)

Jasmine

What is the best way to approach problems for finding time of collision

John Rennie

Usually you write $x_1(t)$ and $x_2(t)$ for the two objects, set $x_1 = x_2$ and solve for $t$.

Jasmine

One particle is thrown from a tower and other is dropped from other tower directed towards each other , and thief police problem

@JohnRennie what is the approach using relative velocity

John Rennie

I can't see how using relative velocity will help you solve the problem.

Jasmine

5:59 AM

A police is directed towards thief both having mutually perpendicular velocities initially but police is always directed towards thief. We need to find time when police catches the thief

And the thief runs in a straight line

John Rennie

Jasmine

Yup

What are the various approach for this question

John Rennie

There's a trick for doing these types of questions, but I can't remember it ...

I'll have to go away and think about it. Monday mornings are very busy for me and in a few minutes I have to start work and I'll be working for several hours.

yuvraj singh

Alesha here, can we do this by assuming time t, when both are collide.

user434058

@JohnRennie I know how to do this!!

Jasmine

6:10 AM

@JohnRennie Ok sir no problem :)

John Rennie

@FakeMod do you want to take over then, and I'll start work.

user434058

First lets get an insight for the problem, see this desmos.com/calculator/vjxup1qglc

Jasmine

@FakeMod @yuvrajsingh please provide the gist of your approach

user434058

Play with the graph, don't worry about the equations yet.

user434058

@Jasmine alright, here is the gist

yuvraj singh

6:13 AM

@Jasmine there are many to solve this particular, problem as you started with a projectile problem, so it would be easy for us to go through that. First Let fake mod, show his efforts.

Jasmine

@yuvrajsingh yup

I need to leave in few minutes for some work

user434058

Wait how do I upload an image!

Jasmine

@FakeMod are you on phone

user434058

Yup

yuvraj singh

@Jasmine let us start, suppose u standing on ground.

Now you thrown, a ball at angle theta towards a tower.

Jasmine

6:16 AM

Open desktop version and you will see upload image @FakeMod

yuvraj singh

At same time a ball is thrown from the tower, let say horizontally.

Jasmine

@yuvrajsingh go on I am reading

user434058

i.stack.imgur.com/SckOW.jpg here you go

yuvraj singh

Now you need to find the time when they collide.

user434058

@Jasmine

Jasmine

6:18 AM

@FakeMod familiar approach thanks though

I wanted to know the relative velocity approach

user434058

@Jasmine 👍

Jasmine

@FakeMod preparing for JEE

user434058

Yup

user434058

@Jasmine You?

yuvraj singh

So initially the ball thrown from the tower has the theta initially zero.

Jasmine

6:19 AM

@FakeMod which class

@FakeMod yea

yuvraj singh

Yes?

user434058

12th

Jasmine

@FakeMod cool

@yuvrajsingh yup, go on

yuvraj singh

OK, now we use the relative concept.

Jasmine

@yuvrajsingh need to leave , share your approach will see later

yuvraj singh

6:22 AM

Let the ball thrown from. The ground has two component, $vcosx$, $vsinx$, .

Horizontal component of both projectile will be same throughout journey.

So ball which was thrown from top of a tower with velocity u.

Now let say they collide at time t.

So we get distance =speed. Time.

So. At this time. Their horizontal component are equal so $u=vcosx$.

Now I you can find the time, by putting this value in the tow equation.

I got a question similar to police thief,. Which you was asking

Sorry typo were.

@Jasmine we can discuss it when you want.

@Jasmine ah, sorry, in projectile, we if the height of tower let 10m then vertical distance traveled by 1 which is on ground is let say x, and other h-x.

Or you can use the relative like u-vcosx will be the velocity h be height, t=h/(u-vocsx)

Jasmine

7:21 AM

@yuvrajsingh cool thanks :-)

yuvraj singh

7:33 AM

Sorry in the last part it must be sin component instead of cos component, I hope you understand @Jasmine

yuvraj singh

7:57 AM

@JohnRennie if you get free please. Check this answer.

11

Q: Is it possible for more than two particles to be entangled in a quantum way?

So I know that two particles can be entangled in a quantum way, but is it possible that more than two particles be entangled in a quantum way? Most descriptions provide with two-particles cases, so I wonder. (It's hard to think of three particles entangled in spin, or so.)

yuvraj singh

9:35 AM

@JohnRennie hi.

John Rennie

@yuvrajsingh hi :-)

I read your answer but it feels like it's just been copied and pasted from some other article rather than an answer you wrote yourself.

yuvraj singh

Actually the later part, which I have taken from a website.

Initially para was. written by me.

So what should I do update it? @JohnRennie

Please sir.

John Rennie

I think the problem is that until you do QM at university it's hard to understand what entanglement actually means. So you are trying to describe something you haven't learned about.

yuvraj singh

10:01 AM

Agree, but I explain in the first para which I know a bit, but still on you say, I have deleted the answer.

@JohnRennie

Advil Sell

@JohnRennie Hi !

is there a way to do this analytically ?

John Rennie

10:33 AM

@AdvilSell hi, sorry, I was on the phone ...

yuvraj singh

@JohnRennie, sir I have a question?

Are you free now?

John Rennie

10:49 AM

@yuvrajsingh yes

Advil Sell

@JohnRennie it's cool

can you answer that ?

yuvraj singh

@JohnRennie let advil finish.

John Rennie

@AdvilSell The transistor saturates when $R_cI_c = V_{cc} - V_{be}$. Yes?

Advil Sell

yup

John Rennie

So $I_c = (5 - 1)/1000 = 4mA$

Advil Sell

10:56 AM

yes

i got that

John Rennie

And $\beta I_b = I_c$ so $I_b = 0.004/200 = 20 \mu A$

@AdvilSell Agreed so far?

Advil Sell

the transistor is in saturation right ? so Ic<(Beta)Ib ?

John Rennie

We're after the point where the transistor is just saturated.

At this point we still have $\beta I_b = I_c$.

Advil Sell

hm , okay

John Rennie

If we increase $I_b$ any further, i.e. move into the saturation region, then the equality breaks down. But up to the saturation point we are OK.

Advil Sell

10:59 AM

yes

John Rennie

So are we looking for the value of $V_{bb}$ that corresponds to this base current?

Advil Sell

yes

John Rennie

Then that's just $I_b R_b = V_{bb} - V_{be}$

$V_{bb} = 20 \mu A * 100000 + 1$

Hmm, that gives $V_{bb} = 3$ which isn't one of the options ...

Advil Sell

yes

John Rennie

Are you sure it's asking for $V_{bb}$? The question says input voltage. Doesn't that mean $V_i$?

Advil Sell

11:04 AM

@JohnRennie How would that effect transistor ? , Isn't it AC ?

John Rennie

I suspect what it means is you adjust $V_{bb}$ to make $I_c = 2mA$ i.e. half the saturation value. Then you calculate the value of $V_i$ that just saturates.

Advil Sell

@JohnRennie I don't get you

John Rennie

@AdvilSell If you look at the output voltage $V_c$ then it can vary from 1V to 5V. It can't go below 1V because that's the BE voltage drop, and it can't go above 5V because that's the max supply voltage. OK so far?

Advil Sell

yeah

John Rennie

So if you're trying to amplify an AC input you want the $V_i=0$ level to be 3V. That allows the amplified AC to swing by 2V either side.

Advil Sell

11:10 AM

okay

John Rennie

i.e. you adjust $I_b$ to give a voltage of 3V when $V_i = 0$.

So you want $I_c = 2mA$ and that gives $I_b = 10\mu A$. OK so far?

Advil Sell

yes

John Rennie

So $V_{bb} = I_b R_b + 1 = 2V$

Advil Sell

yes

but it still isn't matching , right ?

John Rennie

Hmm, it still isn't matching ...

OK, I'm officially baffled.

Advil Sell

11:15 AM

so am I

John Rennie

(c) seems the closest answer ...

Advil Sell

interestingly the answer is B , and probably they refer to Ib and Vbb

because it's the only option satisfying

$V_{bb}-I_{b}(100,000)-1=0$

user8718165

@JohnRennie hello sir :)

@AdvilSell Hii :)

John Rennie

@AdvilSell well, OK, it does satisfy that equation while the other answers don't. But 3.5V would oversaturate the transistor so I don't understand why they say that.

Advil Sell

@user8718165 hey

@JohnRennie Yeah , :( lets just conclude it's a bad question

John Rennie

11:23 AM

I hate to give up on a question, but I don't see where to go with this one.

Advil Sell

@JohnRennie this is what they have done

:|

John Rennie

I can see where they get the difference, because they write $I_cR_c = V_{cc}$ while I used $I_cR_c = V_{cc} - V_{be}$

What I don't see why they are correct and I'm wrong.

Advil Sell

@JohnRennie because probably they are wrong

John Rennie

It's always tempting to think the textbook is wrong. And sometimes it is, but not very often.

4

Advil Sell

how can Vcb be zero ?

@JohnRennie I do agree

John Rennie

11:38 AM

The total voltage drop from the collector to the emitter is $V_{cb} + V_{be}$.

Advil Sell

yes

Ajay Mishra

@AdvilSell JEE main stuff?

John Rennie

$V_{be}$ is a regular forward biased PN junction, so it has the usual voltage drop across it. The question says 1V though I would have thought 0.6V was more typical for a silicon transistor.

Advil Sell

@AjayMishra Not mine

@JohnRennie Yeah , that's what I would have thought

John Rennie

But anyhow, $V_{cb}$ is a reverse biased junction that only allows current to flow due to the carriers created by the base current, and I think $V_{cb}$ can fall to effectively zero when the transistor is saturated.

That's effectively what I used when I said $I_cR_c = V_{cc} - V_{be}$.

Advil Sell

11:41 AM

@JohnRennie Vce for a generally is 0.05V to 0.2V isn't it ?

John Rennie

Can't remember.

Advil Sell

@JohnRennie sparkfun.com/datasheets/Components/2N3904.pdf

$V_{ce(SAT)}$

John Rennie

Can $V_{ce}$ be less than $V_{be}$? I would have guessed not, but I don't speak from a position of any great expertise.

Advil Sell

isn't The base collector also forward biased in saturation ?

John Rennie

Aha, I think $V_{ce}$ goes to zero at saturation. I assumed it went to $V_{be}$ but my Googling suggests this is wrong.

In that case we could get $I_cR_c = 5V$ so $I_c = 5mA$ and $I_b = 25\mu A$.

OK, that makes sense.

Advil Sell

11:51 AM

@JohnRennie Can you share some reading material ?

yuvraj singh

@JohnRennie hi.

John Rennie

... on phone.

Advil Sell

okay just drop the material here , I would read it , Thanks btw.

yuvraj singh

@AdvilSell are you preparing for jee.

Advil Sell

No

The doubt is of one of my friend

yuvraj singh

11:54 AM

Mean branch change?

Advil Sell

nah

John Rennie

12:05 PM

@AdvilSell hi

I haven't got any reading material, I just Googled for Can vce be less than Vbe

yuvraj singh

@JohnRennie hi.

John Rennie

@yuvrajsingh hi

yuvraj singh

Actually, it is regarding the brightness of the stars.

John Rennie

Yes ... ?

yuvraj singh

As far as I know earth doesn't have asteroid belt.

John Rennie

12:17 PM

Correct. The Moon is the only natural satellite of the Earth.

yuvraj singh

So my question is all the stars have unifrom brightness

If no, then why?

John Rennie

No. Stars come in all different sizes and bigger stars are brighter than smaller stars.

We typically plot stars on a Hertzsprung-Russell diagram:

Hertzsprung–Russell diagram

The Hertzsprung–Russell diagram, abbreviated as H–R diagram, HR diagram or HRD, is a scatter plot of stars showing the relationship between the stars' absolute magnitudes or luminosities versus their stellar classifications or effective temperatures. More simply, it plots a star's luminosity (brightness) against its temperature (color). The diagram was created circa 1910 by Ejnar Hertzsprung and Henry Norris Russell and represents a major step towards an understanding of stellar evolution. The related color–magnitude diagram (CMD) plots the apparent magnitudes of stars against their color, usually...

The brightness is plotted on the y axis and it varies by many orders of magnitude.

yuvraj singh

Yes, but my next question is regarding milky way.

John Rennie

OK ... ?

yuvraj singh

Is each of tier in the milky way has fixed brightness.

Sorry unifrom brightness.

John Rennie

12:24 PM

Is each of tier in the milky way a typo? Did you mean each type of star?

yuvraj singh

there should be some cutoff magnitude. I don't know exactly what's the best cutoff, but I would guess somewhere around 4.0 to 5.0 apparent magnitude. If the cutoff is 4.0, then that gives us 3 tiers to look at (1 to 2, 2 to 3, 3 to 4). Are each of those tiers uniformly distributed?

@JohnRennie

John Rennie

The apparent magnitude is how bright the star looks to us here on Earth, so it's a function of two factors:
1. how bright the star is i.e. the absolute magnitude
2. how far away the star is

And just to be confusing higher magnitudes, i.e. more positive numbers, are dimmer.

The brightest star as seen from Earth is Sirius with a magnitude of -1.46.

The dimmest star visible with the naked eye has a magnitude of +6.5, but far dimmer stars can be seen with telescopes.

The dimmest objects visible are very distant galaxies with an apparent magnitude of about +30.

yuvraj singh

OK.

@JohnRennie have you heard about artificial sun.

John Rennie

No ...

yuvraj singh

xinhuanet.com/english/2019-11/26/c_138584676.htm

@JohnRennie

John Rennie

12:42 PM

@yuvrajsingh ah, OK, that's an experimental hydrogen fusion reactor. Yes I did know about that.

yuvraj singh

They are saying, that they will make sun which produce 10times more hest than original one.

John Rennie

Lots of people are trying to build fusion reactors, but no-one has yet succeeded. The European attempt is called Iter.

iter.org

yuvraj singh

@JohnRennie I have updated my profile but still it is not reflecting in chat room.

John Rennie

It can take up to an hour for a change made to your profile to show here. How long ago did you make the change?

Admiral kay

An hour ago.

John Rennie

12:48 PM

Ah, it just changed. Your name now appears as Admiral.

Transcript for

Problem Solving Strategies