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Everything posted by hobz

  1. Suppose I have an antenna transmitting a signal. Around the antenna I have several receiving antennas. Now I vary the energy of the transmitting antenna so that it corresponds to only one photon. If a receiving antenna absorbs the photon, what happens to the electromagnetic wave propagating throughout space in the, say, the oppsite direction?
  2. I have constructed augmented matrix [math] \left[ \begin{array}{ccccccc|c} \frac{4}{5} & -1 & 0 & 0 & 0 & 0 & 0 & 1\\ 0 & \frac{4}{5} & -1 & 0 & 0 & 0 & 0 & 1\\ 0 & 0 & \frac{4}{5} & -1 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & \frac{4}{5} & -1 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & \frac{4}{5} & -1 & 0 & 1\\ 0 & 0 & 0 & 0 & 0 & \frac{1}{5} & -1 & 0\\ \end{array}\right] [/math] and I would like to find all solutions. How would I go about doing that?
  3. Pretty good explanations. Thanks! I suppose that punch cards work in a similar way, by converting the holes to ones or zeros in some process.
  4. I was wondering how and when an instruction is turned into Volts. suppose I have: add ax,bx Now that is compiled into some bit sequence like 01010111 (e.g.). How do these bit get turned into 0 V and 5 V? What component is responsible for this? Thanks.
  5. It was not meant as an answer. I am not sure what formal entails. My guess would be; the formal way of saying 'rigid'
  6. Perhaps [math]\partial x = \mathrm{d}x[/math] and the soft 'd' is only to emphasize that we are talking about a partial derivative?
  7. A good explanation! So, [math]\frac{\mathrm{d} y}{\mathrm{d}x}[/math] might produce something that relates [math]y[/math] and [math]x[/math], while [math]\frac{\mathrm{d} y}{\partial x} = y[/math]?
  8. The definition of the partial derivative of [math]f[/math] with respect to [math]x[/math] is [math]\frac{\partial f}{\partial x} = \frac{f(x+\mathrm{d}x,y,z)-f(x,y,z)}{\mathrm{d}x}[/math] right? While [math]\mathrm{d}x[/math] has a meaning, an infinitesimal change in [math]x[/math], what does [math]\partial x[/math] mean?
  9. I see. I will give both a look. Thanks!
  10. Could you give a practical example of this?
  11. I am currently reading about the divergence and I am curious what is gained by reformulating stuff like Gauss law from integral form into differential form? The expressions become shorter and nicer, but surely there must be another reason?
  12. Most likely in phase? I was under the impression that if emission occured due to an incident photon, the emitted photon would ALWAYS be in phase. Supressed by what?
  13. Interesting. What's the math like? And what about the Heisenberg uncertainty stuff?
  14. I am currently examing this phenomenon but I have not found a single text describing the cause of the emission. What governs the need for the excited electron to emit a coherent photon when stimulated by an incident photon? Why must it have same phase, direction and wavelength? (I guess these may vary due to the uncertainty, so that the wavelength might be slightly different and the direction as well?)
  15. I guess energy is conserved since the frequency of the photon doesn't change. So the incident momentum gets mapped, so to speak, to a new momentum (direction) depending on the materials properties? If the absorption/re-emission is true, then isn't that contradicting that only photon of specific wavelengths can be absorbed in a given material? I read somewhere that the absorption is virtual; the photon is not absorbed per say. Instead it interacts with the atom of the material that tries to absorb it, and depending on how different the wavelength of the photon and the energy required for absorption is, there is a time delay explaining the slowing of the photon as it goes through the material. Is this true?
  16. If that is the case, what causes the re-emitted photon to have a certain direction based on its frequency?
  17. Without defining the term "bug" more ridigly, I would suggest that the calculator that comes bundled with Windows XP is useful and bugless.
  18. That seems to agree with my speculation on the piston/flywheel intertia.
  19. I see. So why does the torque decrease?
  20. Have a look at this page: http://www.howstuffworks.com/horsepower1.htm Just before the "Torque" infobox, it states: "[...] convert torque to horsepower you simply multiply torque by rpm/5,252." After the infobox, there is a curve rpm/(hp & torque) which clearly doesn't demonstrate a linear relationship. I am guessing that the torque goes down (as the rpm goes up) due to the fuel mixture not exploding fast enough as the piston already is moving down due to flywheel inertia. Is this correct? If this is so, then the increase in hp as rpm goes up is less efficient that when the torque is high, right?
  21. I don't see your problem? Usually the blade are angled so that the turbine faces the wind. Thus the wind direction will face the upper side of the blades.
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