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Am I correct in saying that the axis of polarisation must be agreed upon by the two parties prior to the quantum communication line being used? If so this initial communication must be by classical means, and could thus be intercepted. If some 3rd party did intercept this, and knew the setup, could they successfully wire tap the communication without leaving a trace? And practically how does one define a 0º axis on an experiment such as this?

The chances of dying in the first flight you go on it 1/9,000,000. If you go on a second flight, the chance of dying on that flight is also 1/9,000,000, but as you have now been on two flights, and it only takes one to kill you, you say (all are probabilities): dead = ( flight1 crash & flight2 fine ) or (flight1 fine & flight2 crash ) or ( flight1&2 crash ) mathematically you say that as [p(x) is the probability of x]: p(dead) = ( 1/9m * 8,999,999/9m ) + ( 8,999,999/9m * 1/9m ) + ( 1/9m * 1/9m ) = 0.0000002 However then adding a 3rd flight gets a bit more complex, and the expression will get even longer. Luckily there is a much better way of doing this! Instead of all that above, we could simply have said: p(dead) = 1 - p(live) [that is to say, probability of being alive or dead is 1 (it is certain you will either be alive or dead, no other option exists, at least not within the confines of your question), so the probability of being dead is one subtract the probability of being alive. e.g. if it is 1% likely you die, then it is 99% likely you live] = 1 - ( flight1 fine & flight2 fine ) = 1 - ( 8,999,999/9m * 8,999,999/9m ) = 0.0000002 Getting back to your question (I word this the long way to help you understand better) we now have a much better formula. If you went on 3 flights now we can say: p(dead) = 1 - p(live) = 1 - ( flight1&2&3 fine ) and so on for more and more flights. To express this mathematically in a slightly different way, using an exponential, instead of lots of 'and' probabilities (i.e. multiplications): p(dead) = 1 - [ p(one flight fine) ]number of flights so for x flights: p(dead) = 1 - (8,999,999/9,000,000)x So for 100 flights: p(dead) = 1 - (8,999,999/9,000,000)100 = 0.00001 Interestingly, assuming your original statistic to be correct, if a pilot flies every 2nd day of his career (~150days/yr) which lasts 40yrs, the probability of him being in one crash (exponential used it 150*40) is 0.0006. Hope that all helped!

Can you please give more information? I have no idea what this "engineers day on september 15th" is. What branch of engineering? What kind of "articles" do you want? On what subjects and to whom are they targeted? That is to say do you want a research level paper on advanced aeronautical engineering, or a colourful posted teaching F=ma to children?

The entire universe cannot increase in energy, that would violate the law of conservation of energy. The fluctuations, on average, all cancel each other out.

I struggle with revising, it makes me bored and irritable and I sadly have little self-motivation to do it, although I do of course force myself to do something. And I hate memorising formulae and constants, it's pointless and I can't do it! Same for those mathematical proofs where you need to remember special little tricks to do it.

Anything with an integral spin is a boson. Anything with a half-integer spin is a fermion. Fermi-Dirac stats apply to all fermions, and talks about the distribution of fermions. The first thing which comes in to my mind is how electrons (which are fermions) follow the Pauli exclusion principle, which says that every electron in a system (e.g. orbitting a nucleus) must be in a different quantum state. Here's the relevant Wiki page. If you have more questions if you could make them more specific than "explain Fermi–Dirac statistics" it would be easier for us.

I don't. But I know that Basshunter made a song called Dota, look it up on youtube!

... No comment.

Why on earth has one person voted for "Yes, travel over a distance cannot be completed, or begun" when that's clearly a load of rubbish based purely on the fact they moved their mouse to click the option and then 'Vote'.

You should be aware of the fact that the integral of 1/x is defined as ln[x]. If you're used to integrating xn to xn+1/n+1, for n=-1 you'll notice that the bottom of the fraction becomes 0, so the fraction as a whole is undefined, hence this method cannot be used. The integral of 1/P(t) is ln[P(t)]. Also, just in case, the log is in base e. loge is often written as ln. It's just an abbreviation - they're the same thing.

Well that's not a very good image upload site then, is it!? Use Photobucket or Imageshack instead.

Is it possible to solve ln[x]=x-2 directly?

The most significant difference between inhaling and exhaling is that most of the oxygen we inhale is absorbed by our lungs, and so we don't exhale it, and that we do exhale CO2, which is a waste product of the energy making processes which occurs inside our body. Although the question doesn't ask, the beginning of the question talks about pH, so I guess I should point out that CO2 is acidic.

Well your explanation is, in a very simplistic way, about right. A photon hits an atom and knocks an electron up to a higher energy level. This "knock up" is exactly equal to the energy of the photon. The photon no longer exists, it's energy has gone into the atom (more specifically the electron). But the atom is no longer stable, because the electron is in a high energy level, so the electron falls back down to where it was before. This realises energy exactly equal to that "fall down", which is equal to the "knock up", which is equal to the photon's energy. 1) I don't have anything to add to what Klaynos answered above 2) no, because different things will absorb/emit photons in different ways. This comes down to the structure of the material. So for example my blue wall reflects most blue light (i.e. light hits the wall, then "bounces" off it), other colours are all absorbed. Whereas light that hits the glass in my window will be absorbed by the glass, but then emitted in the same direction it was travelling. So glass is transparent, whereas a wall isn't.

I know that TSR (The Student Room) has this feature, although note that TSR and SFN run on different forum software. vB, which SFN use, does not necessarily have this feature, although having said that it's quite possible they do. More to the point, as you can see just below my post, it has a "Location:" field. As long as you complete this field in your profile it will display it with your posts. Sure, graphics are always quicker to scan than text, but the text is simple in that it is non obtrusive. [edit] I agree with Klaynos (post below mine)!

Lieutenant Blabbermouth indeed! [/confused]

With smoking: when people decide to give up it's because they truly understand it is for the best of them and everyone around them. It's often taken them a while to understand and accept this, and to actually go through the hard process of giving up. When they come across people who still smoke, maybe they consider themselves to have "seen the light" (i.e. what they consider to be the "correct" path) and advise the smoker, who in their eyes is on the "wrong" path, to stop smoking. And they'll use the "logical" arguments, which originally convinced them to change, to convince the other person. Giving up smoking is a big deal to an addict. I guess that if they give up then they've got a lot of strong will power, something must have stimulated this, and I think it's this something (the sudden discovery of the "correct" path) that people like to share. They share for two reasons: firstly for the pleasure of sharing and helping others (in a manner similar to why we at SFN help people with work, and friends help each other with personal issues) and secondly to show off their achievement (e.g. in the case of successfully managing to give up smoking). I think that the above can be applied to many situations. Hence I use "" around potentially controversial terms (e.g. personal opinions) - because if you say take this argument and apply it to religion you've got to be careful when saying what is "correct", "logical" and "wrong" etc. I hope this is what you were looking for in this thread! It's a slightly unusual thread; although should make for an interesting discussion.

hmm, a hard question. I think whether you quit comes down to you a lot, your determination, and just how you find it (which you can't know until you try it). I should point out I know nothing about the University of Washington. In the UK some unis and some subjects basically do nothing in the first year, whereas others will have an action-packed first year, topped off with continual assessments and final exams which all count towards your actual degree - like mine, grr! At the end of the day at some point during your degree it will get hard and there will be not-nice exams etc., but that's just part of uni life, and everyone does it, so don't fear it, don't be put off by it, but it would be silly to pretend it won't happen. What I don't get is that you say you're acing physics/calc, but you're not in a good calc class... doesn't your physics class include calc? How can you be acing that if you're struggling with calc? More than that I'd wait for a mech eng person to post, because I'm a physics student. From what I know not acing calc would be worst for a physicists than a mech eng student, but that doesn't really help you.

If you go Control Panel > Network Connections > right click on LAN > untick IPv6 and then OK, does it not save the setting? I don't seem to have an IPv6 service (in services.msc), and there's no mention in msconfig (other than a help service), so I'm unsure why it would turn itself back on. What OS are you using? And how do you find the stream rate of YouTube videos?

I'm assuming you're familiar with integration, but just not how to do this specific problem. Firstly if you can picture the graphs it will help: You need to work out the points where the two graphs intercept. That will give you the upper and lower bounds. Set: y=ln[x]=x-2 and solve for x. But err, I can't see how to solve that myself. ln[x] = x-2 eln[x] = x = ex-2 = exe-2 xe-x = e-2 but then what? if I use my calculator to do it for me then I get: x=0.1586 ; x=3.146. Then you want to integrate for one function, and then subtract the integral of the other. So you start off with the entire area under the higher graph (the ln[x] one) and then subtract the area under the lower graph (the x-2 one) to leave just the area between them. [edit] oh, and to integrate ln[x] what you do is the integral, by parts, of 1*ln[x]. So you integrate the 1 and differentiate the ln[x], as per the "by parts" method of integration. [edit2] also you're gonna have to be careful about where the function crosses the y-axis, else you might get 0 area (where "positive" area cancels with "negative" area). Do you know what I mean? Like the integral of (x-2) from 0 to 2 is -2, and the integral from 2 to 3 is 1/2. The total area is 2+½, not ½-2, because area has to be positive - hopefully you know what I mean! So maybe if you do the problem in 2 parts, firstly for the positive y values, and then for negative. [edit3] ok, I did the question and got a correct looking answer. If you want the solution I'll post it, but if you want to do it yourself (it's good practice, seriously!) then use the two intercept points I gave you, because I still can't see how to get them without a calculator, and do what I said in [edit2], treating +y and -y as two seperate areas, because under the y-axis will give you a "negative" area.

You can post your solution and we can check it, if you would like.

Agar and growing bacteria fall under the 'Modern and Theoretical Physics' forum section!? Maybe if a mod moves this to a bio section it's more likely to be found by someone who can help!

So a man, who is initially at rest (i.e. u=0m/s where u is the initial speed), falls down 4.6m under gravity (a=9.8ms-2), and you want to know the final speed. Well there are two ways of doing it: 1) you could use kinematic equations. 2) you could use conservation of energy. The gravitational potential energy (mgh) is converted in to kinetic energy (½mv2). You can equate the two, i.e. say: mgh=½mv2, and then rearrange for v=?. There is another thread where someone is asking slightly different questions, but they also use kinematic equations to get the answer, so you might find some useful information there.

This thread is a duplicate with this one.

Klay: it's suvat (sue-vat (rhymes with bat))! Not vuats! fungulo: ok, so if you're totally stuck, here's a good way to start: 1) what do you want 2) what do you have 3) what can help you 4) check the units 5) do it! So for question 1: 1) distance moved whilst under a constant deceleration (i.e. 's' in the above equations) 2) u=105km/h ; v=0km/h ; t=3.8s 3) v²=u²+2as -- nah, doesn't help, don't know 'a' s=ut+½at² -- nah, doesn't help, don't know 'a' ah, so this means I missed out an equation in my last post that I probably should have told you. That is: [math]a=\frac{v-u}{t}[/math], it's only for constant acceleration, but you can see where it comes from, it's saying the acceleration is the difference in speeds divided by the time it takes to do that. So now you can use this to work out the acceleration, and then you can update (2) to include acceleration. Then you can use either of the two equations to calculate s. The acceleration will be negative, because it's deceleration aka negative acceleration. 4) note that the speed is in km/h. You could do the whole question in hours, although be careful because the time is in seconds. It would be easier, and a better, more common and the "official" way of doing it would be to convert that speed into m/s. 105km/h -(*1000 to account for the kilo)> 105,000m/h -(divide by 60 to go from hour to min> 1750m/min -(divide by 60 to go from min to sec)> [math]29\tfrac{1}{6}[/math]m/s. 5) now just stick all the numbers into the equations and type it in on a calculator (or do it in your head) and you'll get your answer. Follow a similar process for the other questions. If you get stuck write down every formula you know, check this thread to see if you missed any, then do the "what to I want" and "what do I have" to see if you can use any of the equations to help you. Sometimes like in this question you'll need to use two formulae.