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Cap'n Refsmmat

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Everything posted by Cap'n Refsmmat

  1. Right. You can only determine uniquely a side length if you already know some other property of the triangle, like the length of another side.
  2. What do you mean by "quantum energy"? Is this something with independent existence?
  3. The only way you can do this is if you know the length of one side of the triangle. You often know the length of one side and one angle, and want to find the length of the other sides. Trigonometric functions give the ratios you can use to find them. I don't understand what you're asking. As far as I can tell, it's asking you to find find the values of csc and cot and such. The values of those trig functions evaluated at a specific angle are ratios between lengths. They are not side lengths.
  4. What do you mean, one exact answer? Do you mean one exact value for sine, cosine, etc.? The trigonometric functions are ratios of sides, so it doesn't matter how long the sides are, so long as the ratio between them is the same. So [imath]\sin \frac \pi 3[/imath] has the same value, regardless of whether the hypotenuse is 1 unit or 23 units long, since the ratio between the two sides will be the same.
  5. I suppose so; I must've been taught by fashionable physicists. There's still the issue of units. If the equation is [math]E=m(c+(s-c)^2)[/math], then there are two parts: [math]E=mc + m(s-c)^2[/math] The left term does not have units of energy, while the right term does, so the equation doesn't work out. One could interpret the missing parenthesis differently, as [math]E=m(c+(s-c))^2[/math], but that's just algebraically equivalent to [math]E=m(s)^2[/math]. morgsboi, which version of the equation did you mean?
  6. I'm not sure I understand your question. The unit circle is merely a helpful way to remember the values of trig functions at different angles; the side lengths can indeed be any sizes that form the same ratio. Could you clarify your question?
  7. The full form of the equation for particles in motion is [math]E^2 = m^2 c^4 + p^2 c^2[/math], where p is momentum. Your equation will need to reduce to this when [math]v<c[/math].
  8. I think My Content may limit itself to threads started in the last year, even if you've posted in them more recently. I don't know that I can change this easily, unfortunately.
  9. What does she find lacking in the current sites?
  10. Hm. Would you need a timer at each end, or would an effect be visible if you bounced the signal back and detected it at the same place it was emitted from?
  11. What sort of nanosecond resolution do you need for a signal that travels up and down by a mile, for example?
  12. It reflects no interest because no theory of evolution says humans evolved directly from chimps. You might try perusing TalkOrigins: http://talkorigins.org/faqs/faq-intro-to-biology.html http://talkorigins.org/faqs/comdesc/
  13. A lady performer in yellow Entertained a salacious young fellow He thrust a reward Into her G-string, and Lord! The string fell right off her cello.
  14. Use the full editor and go down to "Options" and click where it says "click to configure post options." You can select an icon there.
  15. That explains why you were a bit... overenthusiastic, after coming over from Phi's place.
  16. Pftt. You know I only said it was unintentional to avoid embarrassing you in front of the others.
  17. I have to admit you were right about the cheese, though. I'm still cleaning it out of my shoes.
  18. I seem to recall you at least left John an interesting voicemail message...
  19. The de Broglie wavelengths of most particles are very, very small, so they should be correspondingly high-energy as aether waves. But you didn't answer my question. Would I detect radiated EM waves from objects at their de Broglie wavelength?
  20. When you post, you can select an image to be displayed at the top (like the smiley here). Is that what you saw?
  21. Well, the short version would be that blike and fafalone founded the site originally. They knew each other (attended the same school) and set the site up. Subsequent moderators and administrators were simply trusted members of the SFN community, rather than their own personal acquaintances. For example, I became an admin in 2006 when blike needed help setting a few things up and knew I could help. We use this commercial software: http://www.invisionpower.com/products/board/ Haven't you had a chat with John on the phone?
  22. But radiated aether waves are EM waves, aren't they? So wouldn't I detect radiated EM waves from objects at their de Broglie wavelength?
  23. I've met dave and a few SFN members before, but not the rest of the staff. I'd certainly like to, though. Shh. They're secret.
  24. Heh. You have to admit those were some pretty awesome pictures, though.
  25. But you said: So the particles radiate aether waves (which is EM radiation) at the speed of light, at the wavelength h/p given by the de Broglie equation?
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