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Everything posted by grifter

  1. I was just trying to "dumb it down" for Mr Jones, by relating photons interacting with charge upon a particle to the photons "bouncing" of of particles, it was probabally still a bit wide of the mark even for a "dumbed down" aproach, never mind, eh.
  2. ha ha ha, yea: "see famous physicist float with three women on a plane used by NASA" the email below reads "cheap Viagra" But on a slightly more serious note, I think its great that the master of gravity itself got as close as you can (without jumping in a tin can filled with enough fuel to blow a hole in the globe...) to experiencing zero G I was in a lecture in Paris the other day where the speaker was the acting doctor on a vomit commit flight in Russia where apparently the safety briefing went as follows: the pilot is speaking to a English interpreter: regarding a small red warning light, a few parachutes and an escape hatch. the interpreter relays the information as follows if the red light goes on and the escape hatch opens; put on the parachute fasten it securely, and jump out the hatch if the red light doesn't go on but the pilot opens the hatch; put the parachute on, don't worry about fastening the cross link, just jump out as fast as you can oh and if the pilot puts the red light on and the hatch doesn't open......[at this point the interpreter begins to argue(shout...) with the pilot] who eventually turns around and says.....oh don't worry about that. that was one hell of a lecture.
  3. I understand most of the high-level concepts of physics on this forum, yet, not why Antony-Jones posts the same questions re-worded, or even a different question, with little relevance. so, even if what I'm about to post makes no sense, I must ask that Mr Jones please accepts it, for the good of humanity. light can be seen as a wave or as a particle, that particle being the photon. "light" (be it waves or particles.) exists even when one cannot see it. "light" can only be seen by the naked eye when real "pieces" of light (photons) hit the retina. to understand why light cannot "bounce off" light you must think of it the light as a wave...... as waves have no mass, hence no surface area, there is nothing for the light to "bounce" off. this is, to quite a large degree, simplified, in the hope that it at least enables you to see the basic concepts of how "light" works. QED(ish)
  4. Cambridge Relativity has a great page about black holes, which really does start at the beginning, so depending on how much you already know it may prove usefull: http://www.damtp.cam.ac.uk/user/gr/public/bh_home.html
  5. well V8 cars are rarely seen as needing turbos, but it is perfectly acceptable to add a turbo to a v8 car, they're mainly bolted on to v4s and v6s to "boost" top speeds and increase mid-range rev' performance. Turbo engines are mainly used for racing (especially drag racing) due to their better performance at top speeds, due to the non-linear power delivery of turbos (turbo-lag); many people choose superchargers for v8's to get more linear power delivery, but to be perfectly honest if your buying a v8 engine (for racing) you've probably got enough money to bi-turbo it therefore significantly increasing power, and top speed. as far as entropy is concerned, Fredrik and Atheists posts covered that perfectly so, no need to repeat what they've said.
  6. I'm not sure if Quark gluon plasma (QGP) can be a superfluid, although it will become quark-gluon liquid which does have very low viscosity, but im not sure if its quite zero.
  7. grifter

    Killing a tree

    hmmm, electrocution...well that would depend on the moisture content of the tree, what tree is it ?
  8. to be perfectly honest space can be easily defined using Mr. H. Simpson’s theory of a doughnut shaped universe.......... no seriously: the easiest way to think of the universe (in its expanding state) is as a sheet of rubber, draw three dots (think of them as galaxies) on the rubber (in its relatively un-stretched state) then begin to stretch the rubber, un-surprisingly the "galaxies" will begin to move apart, this same little miracle occurs if you draw dots, sorry "Galaxies" on a balloon, then blow the balloon up.... Anyway there are my two pence.
  9. ha ha yeast and sugar in the cement eh' i think ill give that one a go ohhhh and sodium iodide, and hydrogen peroxide. now thats good!
  10. Wright's is a nice simple calculator to use; I haven't come across that before, thanks. P.S. if anyone likes python (the programming language.... as oppose to the snake; ) There is a python CosmoCalculator which can be found here http://www.astro.ucla.edu/~wright/CC.python (written by James Schombert )
  11. thats great, I will use that But does anyone have any, even rough, values for the following (as being a few thousand pounds of thrust out isn't a real issue, don't feel that the values need to be correct, so long as they are accurate so when they are put in the equation I will get an answer to the correct power of 10!!) Mdot = Exhaust gas mass flow Ve = Jet velocity at nozzle exit plane Ae = Flow area at nozzle exit plane Pe = Static pressure at nozzle exit plane Pamb = Atmospheric pressure (101.325 kPa) Thanks
  12. Well I have a feeling this is gunna be a headache to answer !!! any help would be sooooooooooooo apreciated: I'm in the process of working out the thrust produced by a rocket, I'm using the following equation: (i apologise for not using latex) Fg=Mdot x Ve + Ae(Pe-Pamb) where: Mdot = Exhaust gas mass flow Ve = Jet velocity at nozzle exit plane Ae = Flow area at nozzle exit plane Pe = Static pressure at nozzle exit plane Pamb = Atmospheric pressure (101.325 kPa) I can get Mdot with: (A star * pt/sqrt[Tt]) * sqrt(gam/R) * [(gam + 1)/2]^-[(gam + 1)/(gam -1)/2] where: Astar = the area of the throat. pt = the total pressure in the combustion chamber, Tt = the total temperature in the combustion chamber, gam = the ratio of specific heats of the exhaust, R = the gas constant. (8.314472(15) J • K-1 • mol-1 I can calculate the exit pressure (pe) and exit temperature (Te) from the isentropic relations at the nozzle exit: pe / pt = [1 + Me^2 * (gam-1)/2]^-[gam/(gam-1)] Te / Tt = [1 + Me^2 * (gam-1)/2]^-1 once i've got Te i can calculate the exhaust exit velocity (Ve) with Ve = Me * sqrt (gam * R * Te) and then its back to: F = mdot * Ve + (pe - 101.325 kPa) * Ae any aprox Values for the above would help sooo much !!! and any actuall examples or work-through would also be a mahoosive help!!! Thanks guys
  13. im in pretty much the same position as you, my physics coursework is a total nightmare !!! (i live near Bristol too)
  14. thats a good point, a wmd on earth isn't necessarily a wmd on an asteroid. an ionized particle beam sounds like a good idea but it would take a mean piece of kit to produce a beam... not to mention one with a big enough radius to do any serious damage to a 400m asteroid, then there's the problem of aiming the damn thing to hit a crazy spinning dirty snowball, good luck Nasa...... what about a chemical oxygen iodine laser, (as modeled by the Boeing YAL-1) that's capable of knocking out a missile, so perhaps theres hope for the earth yet...
  15. thats great, thanks Hotcommodity, you really helped clear that up! it was really bugging me
  16. no, I doubt it, it would take something a bit more monumental, that was just an example of a split forming, as soon as it is proved false (as i believe it to be) suddenly everyone who claimed it existed will suddenly blend in to he background!
  17. Yea my last post was just me being a luddite lol I totally agree, I mean, look what happened to the new Mac after it was released, there was a huge prize fund set up by various individuals on offer to the first person to "crack the Mack" if that mentality is true for all (most) new technological advances, it seems to suddenly defy the point of spending £billions on R&D just to have someone crack & "rip off" your final product, the simple way of avoidance is constant evolution i.e. new software updates released on a daily/weekly basis, in doing so cracking the product becomes a pointless exercise, as the firm will already have a solution in place, I agree with the single point of failure argument posted above, there will always be an exception, this is why as consumers we should buy responsibly, i.e. products with extremely good counter-measures, hopefully a deterrent to most considering stealing our "identity"
  18. No matter how hard company's try. Bored spotty teenagers will always crack it, no matter how much blasted quantum entanglement you use, it's the inevitable, we should go back to the old days, you write the information on on a piece of paper, you are the only one with the piece of paper, you do not copy the paper, the paper remains with you, no one knows you have the paper, you use the information on the paper, you burn the paper, problem solved...
  19. ha ha reminds me of the film armageddon and in answer to your question a shockwave is a nonlinear pressure wave, carrying energy which is able to propagate through a medium ( S L & G ). Across the shockwave - within the medium it is traveling there is allways an increase in pressure, temperature and density, (which is what you would feel as a shockwave travels "through" you) hope this helps
  20. thanks for the answer hotcommodity im sorry the question i posted was screwed up, i have edited it and the correct one is now at the top
  21. I think Physics will remain in its current state until a theory turns up that is revolutionary enough to create another field entirely, just as string theory is currently causing so much hype, and "splitting" us in to two distinct groups, those who believe it and those who don't, all it takes is a theory........
  22. Brønsted-Lowry denotes that Acids are proton doners and bases are Proton acceptors....fine: So why is it that a weak acid like HCLO4 can be made by its conjugate pairs H+ and CLO4- but a strong acid like H2SO4 or HNO3 will not i.e. a strong acid can be split in the following way HCL >>> (H+)(Cl-) but (H+)(Cl-) will not join to give HCL
  23. I know what you mean, this question is tricky, the way I did it was similar to the one posted above, I just outline the reaction N2(g) + O2(g) --> 2NO(g) (don't forget lightning/ large amounts of energy above the arrow) i then went on to discuss how the newly formed NO reacts with O2 to form NO2 and then how that reacts with H2O to form HNO3(aq) + NO(g) you might also want to include how nitric acid is formed, with the OH-group attaching to the N-atom through a single bond, H-O-N(=O)=O it all balances quite neatly.......honest anyway, good luck guys
  24. Okay I'm guessing your doing the open book (I am to) In answer to your question; I'm pretty sure ozone is represented as O=O-O hope this helps!
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