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Everything posted by darkenlighten

  1. Awesome question! You should go to http://www.ecocarchallenge.org , it is a competition where 16 universities and colleges in North America design and build hybrid vehicles use the same base platform of a GM donated vehicle (it is now in the 3rd and final year). Advanced Vehicle Competitions have been going on for a while and they are a really good opportunity for students to learn how to build these vehicles, while also informing the public the new technology. I am on the Ohio State Team ( http://ecocar.osu.edu/ ). We are actually building the type of vehicle you mentioned (where we are actually using E85 fuel instead of petroleum). And the thing is, as mentioned, it is being done. The Volt is a perfect example....and the ability to link the combustion engine to the wheels is actually advantageous as certain speeds, such as highway, where the efficiency of the combustion engine usually is better then the recharging process to use just the electric motor as a drive source. The thing from here is letting people understand the technology and what can be done with it...I think GM is on the way to leading the industry towards the next steps to use less petroleum and hopefully none at all.
  2. I realized I was only considering the vertical velocity components and not the velocity vector...so with that from energy conservation its easy to see [math] E_o = E_f \Rightarrow -mgh_o + \frac{1}{2}mv{_o}{^2} = -mgh_f + \frac{1}{2}mv{_f}{^2} \Rightarrow v_f = [v{_o}{^2} + 2gh_f]{^ \frac{1}{2}}[/math] So since [math] v_o [/math] and [math] h_f [/math] are the same for all so yea that's it. Same speed! And to add to "...", mass of the objects will not affect anything. They could all have different masses and nothing will change.
  3. Their speeds will still be different when hitting the ground, due to only the vertical distances and starting velocities (think of potential and kinetic energies if you like)
  4. Unless I'm thinking about this wrong, they will not have the same final velocities, they will be different. I think the thing here is the fact that they will indeed fall as the same rate ( acceleration = g) , but the direction of their velocity is going to effect their final velocities. For example, Labeling (1) Upward at an angle, (2) downward and (3) horizontally...you want to consider only their vertical velocities, which they are not all the same, (1) and (2) only have the same velocities when heading down if (1) is thrown straight upward and (3) always has a 0 downward velocity. So according to this they will not all have the sames velocities when hitting the ground since [math] v_f = v_0 + at [/math]. Unless I'm crazy.
  5. I would like some clarification from Widdekind....What electrostatic force are you referring to? Because the way you are phrasing the question imo doesn't make much sense. What forces acting on what? Re-reading your above post. Are you just talking about the force between two charged particles (Coulomb force) in your original question....if so what does the photon have to do with anything?
  6. Not to mention the energy required to get something like that up to speed. Since the amount of torque needed is related to the angular momentum, which is related to its moment of inertia, which involves mass/radius.
  7. Yes the magnetic field of the earth is always changing, we are currently in a shift to flip flop. And since we know what is what, any devices that would rely on it would just have to be accounted for.
  8. Agreed with swansont, its the torque you are getting from the motors that is causing the shift in movement. You either want an independent gyroscope to counteract the torque or move the motors themselves appropriately, either way this is not an easy feat.
  9. If you are not considering impulse, then the ball will be the same weight, since it is under the same acceleration. If you are considering the fact it is de-accelerating really fast (once it hits the scale), you would need the impulse, but you would need to know how fast it would stop, in order to calculate this.
  10. Transformers use inductors and magnetic fields (essentially Maxwell's Equations), so it depends what you are trying to know? I mean, it is going hold, you just need to apply it properly.
  11. So when you select the axis of the inner contour, what results are you getting, can you post the analysis on that? And what are your current restraints/fixed surfaces?
  12. For the "Define by" option, what can you select. There should be a way to select about the face, the coordinate system you are in should matter.
  13. I was referring to your screenshot. So the moment advanced options. I would like to help, but I dont have specific experience with ANsys, (I have experience with NX Unigraphics, Solidworks and Autodesk Inventor), so I will do what I can. Also this is a great website for engineering: General: http://eng-tips.com/ For ANsys: http://eng-tips.com/threadminder.cfm?pid=569
  14. Usually can you select along the surface and not normal to the surface. Can you show the advanced options? I would say see if there is an option to select about the face. This should be an easy task, you might be missing something or this program is lacking.
  15. Okay I'll give it a whirl. First they are not described as imaginary lines, the lines represent vectors, which shows the direction and magnitude of the magnetic field. To understand the magnetic field you need to understand the electric field. So lets consider electrostatics, no moving charges. First lets start with one charge, call it q. There is an electric field around this stationary charge, now place another charge, -q, there is an attraction between the two, related by [math] \mathbf{F} = q\mathbf{E} [/math]. Where q is the charge of one and E is the electric field of the other. So now we have a relationship for attraction between stationary charges. Now lets speed things up hahah. Let's consider a moving charge, you now have a charge q moving at speed v, this moving charge now has a magnetic field (and still a electric field, but not as simple as when it is stationary). Now place a stationary charge near by this moving charge. The force on the stationary particle is [math] \mathbf{F} = q\mathbf{v}\times\mathbf{B}[/math]. Now take this to larger scales, from single charges, to multiple charges (electrons) say in a wire and you get a relationship of current, force and the magnetic field. Now move this to materials, like a metal, that has electrons around a nucleus, causing tiny loops of currents and you get a total magnetic field in that material. Its more complicated to calculate the magnetic field of a material, but its the same concept. See if that helps.
  16. And this does depend on whether or not they are asking for the electric potential or the potential difference, since those are two different things. It seems like it wants the potential difference, considering it is between two points, even though the questions states electric potential. And also due to the nature of this problem, if it does have an answer, which I believe it should, the two different arcs will have the same potential difference. I would like to wait for the poster's response before I go further.
  17. Yea it's a shame. This forum is a lot better and moderated.

  18. It looks like it very well could be. Since the main information you know is the fact that the loop has an EMF of 5V, which means the potential difference is 5V and the potential difference is [math] V(b) - V(a) = - \int_a^b \mathbf{E} \cdot d\mathbf{l} [/math] Edit: Or you could just say [math] \varepsilon = \oint \mathbf{E} \cdot d\mathbf{l} [/math], where [math] \varepsilon [/math] is the EMF. So the electric potential is described fully as [math] V® = - \int_o^r \mathbf{E} \cdot d\mathbf{l} [/math], where o is the origin point of choosing.
  19. First, it does matter because I'm afraid you are not understanding the difference between the wave function and its probability, they describe two different things. And I know the wave function can be the physical description of the wave. Just to be clear then, what is your meaning of a probability wave, to make sure we aren't arguing semantics? And yea I did say that, because just because something can be represented by all possible states, doesn't mean it is in all possible states.
  20. So you are not reading what I am writing. The wave function is the physical description of the wave, but the probability wave is the probability of it being at a certain place. Have you taken a Quantum Mechanics class yet?
  21. hey! its sad frostcloud took the toll it did, but oh well. it happens I guess.

  22. Okay yea so you do not know what a probability wave is. Of course light acts like a wave and a particle, that is known. The probability wave is not the description of the physical wave-like nature. I think people get confused when one talks about the wave function and a probability wave. The wave function describes a physical system, which doesn't have to be necessarily the position of a particle. A probability wave is the mathematical description of the probability of that system doing a certain thing, such as an electron's position. So no light is not a probability wave, nothing is a probability wave. Unless I'm skewing what you and the poster mean by a probability wave. Because as far as I know it, it is the wave function squared or [math] |\Psi|^2 [/math]
  23. I'm very well aware of the double split experiment. Saying they take every path possible seems over the top to me, but nonetheless, to the original question, light is not a probability wave.
  24. [math] \mathbf{a} \cdot \mathbf{b} \cdot \mathbf{c} [/math] is not equal to [math] a(\mathbf{b} \cdot \mathbf{c})[/math] Because you cannot dot a vector and a scalar, you can multiply them, but not dot them. This is kinda trivial, but the reason I say it is if there was maybe some type of proof or equation that had a similar form, you would not be able to an operation like this. Or whatever you are trying to say but either way [math] \mathbf{a} \cdot \mathbf{b} \cdot \mathbf{c} [/math] this cannot work due to the reason above.
  25. Probably not that much then, do you know how many volts per degrees the TEC can produce? It doesn't seem like you would be able to produce enough from using body heat, due to heat transfer from you to the metal. You most likely would not be relying on just what the body can generate directly, so the way the device works will greatly effect what you can get. Just because you convert your temperature into energy via KT or of the like, it is somewhat arbitrary until it is applied to something.
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