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Everything posted by CanadaAotS

  1. That unfortanetly doesnt help me -_-' Some math showing what I did wrong in get ~3m/s would be good lol
  2. A billion years? They MUST have thought of something to get to the next galaxy. A billion years is a lot of time... and a billion years for technology to develop is an extremely long time.
  3. So, any estimates on how long till we having quantum computers? I'm thinking 15 yrs probably. We'll probably hit the processor speed "wall" soon with our current computer chips, and when that happens funding will be poured into possible alternatives...
  4. Isnt that egotistical It may be possible that we are the most advanced civilization in our galaxy (or at least, any other civilzations are a million or less years ahead). How about inter-galactic travel? statistically, there HAS to be intelligent life in the universe. If not in our galaxy, then perhaps in another. If they had a 1 billion yr head start then I'm sure they could make it intergalactically
  5. exactly... thats it, I'm drawing a diagram! lol EDIT: damn photobucket... it made my diagram alot smaller I dont know if you are able to read it. Anyways the small circle with the black dot is the birds-eye view of the tunnel. It forms a 'ring' and inside the ring charges cancel each other out (2 dimensional version of a sphere). This means that the object falling down the tunnel isnt effected by the sides at all. The bigger circle shows the front view where the only force acting on the object is the vertical pull of gravity. By the way, the object would (apparently) be going at 3 m/s when it reached the center. I think this is wrong though... check out my thread where I try to figure it out here
  6. The strong force does not hold electrons to their orbits, the magnetic charge (pos neg) between the proton and electron does. And if the green squiggles are magnetic, then why do they connnect protons to neutrons? the strong force does that...
  7. Yay! thanks so... [math]v = \sqrt{\frac{GM_e}{{r_e}^2}}[/math] I wasnt THAT far off. [math]v = \sqrt{\frac{(6.673 \mbox{ x } 10^{-11})\cdot (5.9742 \mbox{ x } 10^{24})}{6378100^2}}[/math] [math]v = 3.13 ms^{-1}[/math] wtf?!? Is this right? I thought you'd be going ALOT faster...
  8. You cant have 2 of the strong force poles "working" and the others just sitting back in the nuclei...
  9. I'm pretty sure you wouldn't be able to walk on the cylinder walls for the same reason that gravity wouldn't let you walk on the walls of a sphere from the inside. Try looking at how the forces work from birds eye view (only looking laterally). What you get is the object inside a ring. This is the 2D analogy of charges canceling out in a sphere. This means that laterally, no forces are effecting the object. When you look at it from a front view, the earth is still pulling down to the center so all you have is a vertical force.
  10. yah I realize that now that I look. result is the same though, just screwed up the units
  11. Well I've been told that my whole change in gravity is wrong lol. Apparently gravity decreases linearly from the surface to the center so... [math]\frac{dF_g}{dt} = \frac{F_g}{r_e} = \frac{686N}{6378100m}[/math] So force of gravity is dropping at -1.0756 x 10^-4 N/s Acceleration = Force / mass [math]a = \frac{F_g}{m_o}[/math] [math]\frac{da}{dt} = \frac{\frac{dm_o}{dt}*{F_g} - \frac{dF_g}{dt}*m_o}{{m_o}^2}[/math] I think I'm still right with change in mass being 0. plus I can substitute 1.0756 x 10^-4 N/s for [math]\frac{dF_g}{dt}[/math] [math]\frac{da}{dt} = \frac{1.0756\cdot 10^{-4} Ns^{-1}}{m_o}[/math] Again things simplify nicely the [math]m_o[/math]'s cancel out leaving me with a nice equation. When I put 70 kg in for mass of object I get da/dt equaling [math]1.5365*10^{-6} ms^{-3}[/math] And now I'm stuck again lol... I know I have to take an integral somewhere here because I have to get from acceleration equation to velocity.
  12. you must be missing an equation. I'm sure with 3 variables you'd need 3 equations... EDIT: nvr mind if you had 3 equations it wouldnt be inequality, you'd just solve it -_-' lol
  13. Apt title... hehe Anyways, I was trying to do some calculations based on the "Tunnel Through Earth" thread. I wanted to actually find out how fast an object, that was dropped from earths surface to the center, would be going. This involves Physics, but its more math intensive, so I put it in Math. If that rubs anyone the wrong way, then move it Earths Radius: 6378100 m Earth Mass: 5.9742 x 10^24 kg Object (Sphere with app. human mass): 70 kg Now, I'm thinking that throughout the fall the force of gravity is constantly changing. [math]F_g = \frac{G*m_e*m_o}{{r_e}^2}[/math] So I need to find out what [math]\frac{dF_g}{dt}[/math] is. (Force of gravity changing over time) [math]\frac{dF_g}{dt} = \frac{({r_e}^2)*(G*m_e*m_o)' - (G*m_e*m_o)*({r_e}^2)'}{{r_e}^4}[/math] [math]\frac{dF_g}{dt} = \frac{({r_e}^2)*(G*(\frac{dm_e}{dt}*m_o + \frac{dm_o}{dt}*m_e)) - (G*m_e*m_o)*2r_e*\frac{dr_e}{dt}}{{r_e}^4}[/math] Now I think the change in masses are all 0 (mass never changes). I can probably simplify alot with that because it cancels out [math]\frac{dm_e}{dt}*m_o + \frac{dm_o}{dt}*m_e[/math] completely. I'll try fixing it up a bit... [math]\frac{dF_g}{dt} = \frac{-2Gm_om_e\frac{dr_e}{dt}}{{r_e}^3}[/math] Well, that worked out better then I thought it would Not sure how to proceed from here... I think I need acceleration or something so that I can take the integral to get velocity. But I'm in the dark... never took Physics Calculus, just regular Calculus lol Pointing me in the right direction would be great
  14. Thats good. I like it when I'm not completely wrong
  15. I think you are confused with "electron shells" and electron energy levels. An electron shell is a shell of an atom, and only the outtermost shell is effected at any one time, the inner shells are full. However, when an electron loses energy or gains energy it jumps (or lowers) its energy level. I think this is where you got mixed up. (Unless I'm wrong as well lol)
  16. er... CO2 and methane are 2. Water Vapour as well.
  17. pffft fine, reproduce then evolve maybe not the moon bacteria, but theres a very good chance that bacteria left on the rovers can survive and reproduce on mars. The enviroment that NASA created forced them to sustain themselves on the rovers themselves. I'll go look it up in Discover magazine archives and quote it if you like. EDIT: Found it here Requires a membership though, so I'll just quote it.
  18. nah, the current poll is better. Thats what the "I Don't Know" section is for. They could've made "I dont know which to believe" or "I dont know what evolution theory is" as two seperate sections to make it a bit more specific I guess
  19. I love that "Magic Fire" dust that makes rainbow coloured fires
  20. I actually think hes right on this. Or at least barking up the right tree, I do know that in making computer chips they make electron holes or something like that for semi-conductors... What hes saying sounds familar The diagrams dont help at all though And I thought I taught you about the magical world of quote tags alpha?!? lol
  21. I wouldnt be surprised if somehow, the few million bacteria or so that was left on equipment left on the moon and/or mars has somehow evolved itself and is reproducing... Life (especially those damn hardy bacteria lol) tend to adapt their way out of extremely intense enviroments. The rovers that were sent to mars were subjected to extreme sanitization... similar to being out in space actually. They put them in a vacuum, put them through intense radiation etc. to kill off any lifeforms since they didnt want to 'infect' mars with our bacteria Turns out they made an environment perfect for bacteria to live through space... Radiation resistant, vacuum resistant These super bacteria are probably colonizing mars as we speak
  22. you can make your own stress ball with a balloon and corn starch (or flour or sand). It's probably the lowest cost solution. Another one could be silly putty placed in a condom lmao. And (I hope) the condom will be alot more durable then a balloon. haha
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