Jump to content

5614

Senior Members
  • Posts

    6408
  • Joined

  • Last visited

About 5614

  • Birthday May 5

Profile Information

  • Location
    London, UK
  • Interests
    Science, electronics (esp. high voltage!), computers, badminton, go-karting, gadgets (lasers!) etc.
  • College Major/Degree
    Imperial College London; Physics Masters degree
  • Favorite Area of Science
    Physics / Maths / Computers
  • Occupation
    Student

Retained

  • Genius

Recent Profile Visitors

The recent visitors block is disabled and is not being shown to other users.

5614's Achievements

Genius

Genius (11/13)

28

Reputation

  1. 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?
  2. 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!
  3. 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?
  4. 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.
  5. 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.
  6. 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.
  7. I don't. But I know that Basshunter made a song called Dota, look it up on youtube!
  8. 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'.
  9. 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.
  10. Well that's not a very good image upload site then, is it!? Use Photobucket or Imageshack instead.
  11. Is it possible to solve ln[x]=x-2 directly?
  12. 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.
  13. 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.
  14. 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)!
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.