A few sample physics questions for Reiku

Discussion in 'Free Thoughts' started by James R, Jan 23, 2012.

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  1. AlphaNumeric

    AlphaNumeric Fully ionized Registered Senior Member

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    You mean expression or notation, K isn't an equation.

    Besides, part of being a good mathematician/physicist is being able to spot structures you've seen in similar forms and to make connections. Doing physic is more than just "This formula is the one on page 7 of my lectures notes but with k=2, therefore the answer will be as in my lecture notes but with k=2". It's about being able to properly formulate a problem or phrase it in such a way as to highlight relevant properties/points of interest.

    Firstly, not necessarily given not everyone learns everything from a university course (arguably no one learns everything). Secondly you should know it, given you're claiming you could handle this level of material easily. You're failing to spot simple things like harmonic motion, despite throwing around the quantum harmonic oscillator or Fourier transforms in other threads. It's a sign you don't understand the mathematical structures you're posting, you're just parroting them. Even if you could parrot things perfectly the fact physic courses (and physics in general) aren't going to hand you exactly the same expressions or notation each time you'd still struggle because its to test your understanding rather than equation recognition abilities,

    You're doing it again. You're ignoring a long list of very basic and fundamental mistakes you've made and picked something, anything, you can tenuously say "Look, I was doing good stuff there", as if that 5% magically negates the 95% you botched.

    How terribly mature. What an excellent role model for the children you baby sit you must be.
     
    AlphaNumeric, Jan 25, 2012
    #41

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  3. James R

    James R Just this guy, you know? Staff Member

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    You're right.

    I actually missed the minus sign in that eigenvector that shouldn't be there.

    That is, \(\begin{bmatrix} 0.6\\-0.8\end{bmatrix}\) is not an eigenvector, but \(\begin{bmatrix} 0.6\\0.8\end{bmatrix}\) is.

    So, subtract a mark or two from Question 1 as well.

    Anything is possible.
     
    James R, Jan 25, 2012
    #42

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  5. CptBork

    CptBork Valued Senior Member

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    There are plenty of Java and Flash apps out there to do the trick, very easy to find on Google. Matlab and other similar packages can also do it, although I have my doubts he'd be sufficiently competent to obtain or use them.
     
    CptBork, Jan 25, 2012
    #43

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  7. funkstar

    funkstar ratsknuf Valued Senior Member

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    I strongly doubt that he came up with the solution himself, anyway. \(V_1 = \begin{bmatrix}-0.707\\0.707\end{bmatrix}\) is an extremely peculiar choice.
     
    funkstar, Jan 25, 2012
    #44
  8. Trippy

    Trippy ALEA IACTA EST Staff Member

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    0.707 = \(\frac{1}{sqrt{2}}\) so his answer then is:
    \(V_1 = \begin{bmatrix}-\frac{1}{sqrt{2}}\\\frac{1}{sqrt{2}}\end{bmatrix}\)
    Which means very little to me - my Matrix mechanics/math is buried very deep in the dim dark alcohol addled corners of my brain (along with a bunch of Biology, sorry Herc).
     
    Trippy, Jan 25, 2012
    #45
  9. Guest254

    Guest254 Valued Senior Member

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    James R, there's no need to be quite so defensive. Finding it unbelievable that a certain member still posts here does not the vendetta make...

    Trippy, funkstar -- the eigenvector solver has just normalised the vector for him.
     
    Guest254, Jan 25, 2012
    #46
  10. James R

    James R Just this guy, you know? Staff Member

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    funkstar:

    That's a unit vector. Why do you say it is peculiar?
     
    James R, Jan 25, 2012
    #47
  11. funkstar

    funkstar ratsknuf Valued Senior Member

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    Peculiar, in the sense that he obviously did not arrive at it by hand. It's approximately a unit vector, but not precisely so, which red flags a numerical solution. Without showing the work, this is useless, imo.
     
    funkstar, Jan 25, 2012
    #48
  12. Reiku

    Reiku Banned Banned

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    11,238
    For your information, he was not picking on me. He offered to see if a personal theory was right, which I decided I would take up the challenge.


    But on this note, you can shove this were the sun don't shine!

    I give up with this thread. In regards to Guests response, yet also with others, and obviously cptbork making fun of me taking time inbetween, obviously he never even saw any mistakes!

    Sianara to this James. We can do this in private if you wish, but this is feeding too many of the hypocrites and ****s.
     
    Reiku, Jan 25, 2012
    #49
  13. Reiku

    Reiku Banned Banned

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    11,238
    Alphanumeric,

    K was an equation in his context. I just never wrote it all out.
     
    Reiku, Jan 25, 2012
    #50
  14. Reiku

    Reiku Banned Banned

    Messages:
    11,238
    a solver... More likely an error on my behalf. I'll write it out properly later when I get back. But that is all I will contribute here.

    You just substitute the values of lambda into the matrix equation \((A-\lambda I)C = 0\) where C will be our column vector. And \(I\) is our identity.

    \(C = c_1\)

    thus substituting

    \(\lambda = 0\)

    into the original matrix...

    In fact I need to go. I'll be back and do this properly.
     
    Reiku, Jan 25, 2012
    #51
  15. CptBork

    CptBork Valued Senior Member

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    I didn't even waste my time checking, I already know you're full of hot air.
     
    CptBork, Jan 25, 2012
    #52
  16. Trippy

    Trippy ALEA IACTA EST Staff Member

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    No it wasn't, it was a constant:
    \(K=\frac{4\pi G \rho}{3}\)

    4 is constant, \(\pi\) is constant, \(\rho\) is constant, G is constant, 3 is constant...

    Or is there something that you know that you're not telling us that means when you mutliply and divide a bunch of constant numbers by each other you get a variable result.
     
    Trippy, Jan 25, 2012
    #53
  17. Guest254

    Guest254 Valued Senior Member

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    1,056
    In an ideal world I would, of course, take the same position. Unfortunately the world isn't ideal and Reiku is evidence of this. He has done the same thing again and again. In fact he's it doing it right now: "I know what you're talking about..." in reference to this post. Personally I think Reiku has been given enough chances and it doesn't look like he want to change his spots!
    Go-go gadget Google!
     
    Guest254, Jan 25, 2012
    #54
  18. Reiku

    Reiku Banned Banned

    Messages:
    11,238
    Whether K is a constant or not, K has been made equal to other constants which makes it an entire equation.
     
    Reiku, Jan 25, 2012
    #55
  19. Trippy

    Trippy ALEA IACTA EST Staff Member

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    It is a constant.

    \(K=\frac{4\pi G \rho}{3}=\frac{4*3.142*6.674E^{-11}*5515}{3}=1.542E^{-6}\)

    K is simply a convenient way of expressing the most accurate form of the numbers.

    It is a constant, not an equation.
     
    Trippy, Jan 25, 2012
    #56
  20. Bells

    Bells Staff Member

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    24,270
    Reiku, you deemed yourself knowledgeable enough to apparently write a book. You carried on as if actual physicists knew less than you.

    These are first year exam questions, questions you boasted 'shouldn't be too hard', and claimed that with what you knew, 'first year would be a breeze'. In fact, you were offended when you were advised by those who do know more than you that you are incapable of the work.

    So why are you upset Reiku? You have no problem with being corrected, remember? So why are you angry that James told you question 2 was wrong and broke it down for you to explain how and where you went wrong? Or are you upset because the other questions from the first year exam, which is based on assumed knowledge from high school physics, are above your level of knowledge? Or are you upset that it is becoming blatantly obvious you actually are incapable of even the most basic maths, as much as you complained about being told that previously? Or are you upset that you can't 'come across as confident' in this thread and it is becoming very clear that you don't even understand the very basics, instead of your boast that you understand a lot about the subject? Or does it bother you that your inability to answer first year questions shows you to be dishonest about your confidence on this forum?

    So what is your issue now Reiku?

    You asked for this. This is supposed to be easy for someone who works on this stuff for 4 hours a day for a hobby, isn't it?

    Just to remind you, Reiku...

    Guess what?

    You just failed a first year 'college' question that James posted after you asked him to throw you some questions. He picked the easiest ones he could find because he did not want to overwhelm you. He gave you as long as you wanted to answer them, instead of a few hours like first year students would get. And you failed one question and it seems can't answer the other ones and so you get angry and abusive?
     
    Bells, Jan 26, 2012
    #57
  21. Reiku

    Reiku Banned Banned

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    11,238
    What the fook are you talking about? :bugeye:
     
    Reiku, Jan 26, 2012
    #58
  22. Reiku

    Reiku Banned Banned

    Messages:
    11,238
    Yes well done.

    Now, tell what is this F=Mg? :bugeye:
     
    Reiku, Jan 26, 2012
    #59
  23. Trippy

    Trippy ALEA IACTA EST Staff Member

    Messages:
    10,890
    F=mg is an equation.
    F is a variable
    And K is still a constant (for the same reason that F is a variable).

    Get it yet?

    You claim that you would which through first year, but you can't even diffirentiate between a constant, a variable, and a function?

    K is a constant.
    \(K=\frac{4\pi G \rho}{3}\) is the precise formulaic statement that describes the exact value of that constant (by relating it to other constants), which can be approximated as 1.542x10[sup]-6[/sup]
    f(K) would be a function, if Density were allowed to vary* (as it does in the real world), but in that case K still would not be a constant, it would be a variable (specifically, a dependent variable).

    *Adddendum:In the event that density is constant, and f(K)= 1.542x10[sup]-6[/sup], f(K) is still a function, it's just a function with a constant value.
     
    Trippy, Jan 26, 2012
    #60
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