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Simple Quetion

  1. Aug 7, 2004 #1
    I have an Einsteinian meter rod traveling at .5c which is 1 meter per 6.6667-09 seconds.

    How do I discover how far the rod will travel in 1.9245-09 seconds?
     
  2. jcsd
  3. Aug 7, 2004 #2

    Ivan Seeking

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    (meters/second) X seconds = meters

    so 1.9245/6.6667 =
     
  4. Aug 7, 2004 #3
    .288673557 meters in 1.9245-09 seconds.

    Thanks, Ivan. I'm just begining to calculate with sci notation and get dazzled by the new appearance of things (very slow brain for math).
     
  5. Aug 8, 2004 #4

    HallsofIvy

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    Hold on for a moment: is that 6.66667-09 (which would be a peculiar calculation) or 6.66667x10-9 ("scientific notation"). Those are very different numbers!
     
  6. Aug 8, 2004 #5

    Zurtex

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    A couple of years ago Edexcel wrote in the exams book that the gravitational constant = 6.67300 × 11-11
     
  7. Aug 8, 2004 #6
    It would look like this written out:

    .0000000066667

    I've been using 6.66667-09 to indicate this in a thread and no one has said anything about it, although everyone else is writing it the way you do. I just figured they were interchangable.

    If not, then I'm curious to find out what interesting thing I have been indicating without realizing it.
     
  8. Aug 8, 2004 #7

    jcsd

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    6.66667-09 ≈ 0.0000000384432

    6.66667 x 10-09 = 6.66667e-09 = 0.00000000666667
     
  9. Aug 8, 2004 #8

    Gokul43201

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    You have been indicating the reciprocal of the 9th power of 20/3 (or 6.66667)...but that would be a very unusual representation, given the context.

    So, everyone understood what you meant, or what you meant to mean.
     
  10. Aug 8, 2004 #9
    I think you need to use SR in this question, since distance contracts when travelling at high speed.

    [tex] d=vt [/tex]

    [tex] d\sqrt {1- \frac{(0.5c)^2}{c^2}}=(0.5c)(6.66667 x 10^(-9)) [/tex]

    and solve for d from then on.
     
  11. Aug 9, 2004 #10

    uart

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    Yes, using relativist speeds in this problem (to confuse the issue) didn't seem like a particularly good idea for someone struggling with basic scientific notation.

    But if this is indeed meant to be a relativistic problem then it is inadaquately stated. The speed of 0.5c is measured relative to who? This time of 1.9245 x 10^(-09) is measured relative to who? If the speed, time and distance are all measured in a consistant reference frame then the distance travelled (in that same reference frame) will still be sucessfully computed by d = vt.
     
  12. Aug 9, 2004 #11
    I can state the same problem in terms of non-relativistic speeds:

    If I have a dragonfly which is flying straight and steady at a rate of 1 meter per .0493 seconds, how do I determine how far it will fly in 1.9245x10-9 seconds?
     
  13. Aug 9, 2004 #12

    HallsofIvy

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    d= vt=1 1 m/(0.0493 sec)*1.9245x10-9= (1.9245/0.0493)x10-9= 25.4349 x10-9.

    Since the "least accurate" number (0.0493) had 3 significant figures, I would write that as 2.54x10-8 meters.
     
  14. Aug 10, 2004 #13
    Thanks HallsofIvy. I'm interested in the way you expressed this, with the single exponent outside the parenthesis. It makes me wonder how you would put it if both had an exponent, say if .0493 had been 4.93x10-7.
     
  15. Aug 10, 2004 #14

    plover

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    [tex]
    \frac{1.9245 \times 10^{-9}}{4.93 \times 10^{-7}}\ =
    \ \frac{1.9245}{4.93} \times \frac{10^{-9}}{10^{-7}}\ =
    \ (1.9245/4.93) \times 10^{-2}
    [/tex]
     
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