@JohnRennie there are a few sums based on that....but those are more or less easy...I'm confused with the 3rd point...though previously there was 1 mole of $O_2$ and after reaction its 1 mole of $CO_2$ but won't that 1 $CO_2$ mole just take up more space and hence volume will increase?
No, because the number of moles of gas isn't changing and (at a given temperature) the volume is proportional to the number of moles of the gas. Assuming ideal behaviour, but both oxygen and carbon dioxide are pretty close to ideal.
@JohnRennie sir in kinetic molecular theory they say that the molecules aren't perfectly hard spheres and their size isn't negligible...Please tell me sir if that's right
We have to write $E = I_0 Z(0) + IZ(\omega)$ where $I_0$ and $Z(0)$ are the DC current and impedance and $I$ and $Z(\omega)$ are the AC current and impedance.
There are different MathJax scripts. I use a script that turns MathJax on so I only have to load it once. But if I leave and reenter the room I have to run it again. But once started it automatically converts new posts to MathJax.
But there are are some scripts that just convert what's on the screen, and if new posts are made you have to manually run it again to convert the new posts.
@Nobodyrecognizeable OK. The point of the question is that when we write $E = I_0 Z(0) + IZ(\omega)$ what we are going to find is that $I_0 Z(0) \gg IZ(\omega)$ so the voltage across the resistor-capacitor pair is mostly DC.
The DC impedance is given by the AC impedance in the limit $\omega \to 0$, but we don't need to do that since the impedance of a capacitor is infinite for DC i.e. DC won't flow through a capacitor. So for DC the impedance is just $R$.
Is that clear, or do you want to do the calculation to show it's just $R$?
The equation for the DC impedance looks like $1/Z = 1/R + 1/\infty$ so we get $Z=R$.
We are told the current is $1 + 0.5\cos\omega t$ (in milliamps) where $\omega=2000$, so the DC voltage is just $E_0 = 2000mV$. I'll write it in mV since we're given the current in mA.
The resistor is $2000$ ohms and the capacitor is $1/j\omega C$. We are told $\omega = 2000$ and $C = 10^{-4}$ so $Z_c = 1/0.2j$ or $|Z| = 5$. OK so far?
Compare that to $E_0 = 2000mV$ and we can immediately see that the AC component of the voltage is a thousand times smaller than the DC component, which is what your book is saying.
@Nobodyrecognizeable the only outstanding question I remember is that one about the two coils, but I don't understand what the question is asking so I don't know how to answer it.
@Nobodyrecognizeable I think it's just a transformer i.e. two coils wound on the same former. So the secondary voltage is just the primary times the ratio of the number of turns.
I would guess the mean velocity is defined as $\int \mathbf v(t)dt/\int dt$
So in this case the mean velocity is a vector pointing from the point $i$ to the point $f$ and with a magnitude equal to the distance $if$ divided by the total time taken.
@JohnRennie Just wanted to ask a personal question: How much time did it take you to go from school to PhD, and did you like studying upto PhD? Was it worth it?
@YUSUFHASAN I took a year off after school to work and save some money, so I started my degree a year after leaving school and the degree lasted three years. Then I went straight into doing a PhD and that also lasted three years.
I loved doing the PhD. It was the best time of my life without doubt.
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All the way through school and a degree you're basically being lectured at and you have to learn what you're told then try and regurgitate it in an exam.
But once you start the PhD you are deciding what you need to do and how to do it. For the first time you're doing what you decide to do not what someone is telling you to do.
@JohnRennie So the time taken for a PhD in the UK is around 3 years,after which you can take up a job as a research scientist in a company,right? And,did it(the PhD)get frustrating at some point? What did u do about it?
I was lucky in that my PhD went very well and I never found myself at a roadblock.
All PhD students have a supervisor who is supposed to help you when things get sticky. The supervisor will be an experienced scientist. Some supervisors are better than others in this respect so it depends who it is.
But in a way hitting a roadblock is not a bad thing because that does happen in real life and figuring out how to deal with it is an important part of learning to be a scientist.
It depends on the country. In the UK in my time (the 1980s) PhDs were three years. You could take longer, and some people did, but the funding was only for three years so you had no money after that.
In the US it's rather different as I get the impression PhDs can go on a lot longer.
The best reason for doing a PhD is that you really, really want to.
It's quite a hard life. I spent the three years of my PhD with barely enough money to live. I didn't buy any new clothes for three years. That was OK because I enjoyed doing the research so much, but if you weren't enjoying the research then you'd probably get fed up pretty quickly.
It was absolutely amazing. I met loads of amazing people and got to play with cool technology. I used a Rutherford backscattering accelerator and also worked at a synchrotron, both of which were amazing.
See how much you like doing the degree. You almost certainly have some project work to do in the final year and that will give you some idea of what doing a PhD is like.
Most jobs are basically managerial. That is your employer needs something to be done and it's your job to do it. Whether it's a science job, or engineering or working for a bank, the job will require you to analyse the problem, work out a strategy for addressing it then get it done.
The skills involved will depend on the job, but skills can be learned when you need them.
So doing a science degree doesn't necessarily mean you can only get a science job. Employers will view a deree as evidence that you're a smart guy who can work hard.
And that makes you attractive as an employee.
It's impossible to say now what you'll find you enjoy doing in three or four years. I ended up working in an area (colloid science) that I didn't even know existed when I started my degree.
I've just looked on the McKinsey web site and it says:
> Our undergraduates join us from many backgrounds—there is no single “right” major or course of study. Our people do share some common qualities including excellent academic performance, leadership abilities, and experience working on or off campus. We look for strong problem solvers with potential—we will teach you the rest.
I would divide the ball into cylinders centred on the axis of the ball. Then the charge will be the charge density multiplied by the area of the end of the cylinder, and the dipole will be the charge multiplied by the length of the cylinder.
Something like this. This is a cross section so you're seeing the top and bottom slices through a cylindrical shell of inner radius $x$ and thickness $dx$
