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YumYum247

  • one year ago

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  1. YumYum247
    • one year ago
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    @aaronq can i please take just a few mintues of your time?!?!?!? Pleaase check my work.

  2. YumYum247
    • one year ago
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    This is how i did it....... Given: 30Hz, t = 0.20sec, d = 12m To find the velocity, i need to find the wavelength first. Since V = fXLambda V = Displacement in distance/ Displacement in time V = 12m/0.20sec V = 60sec Now that we have our Velocity of the wave, we can now substitute both the given and reseultant values into the Formula V= F X Lambda Lambda = V/F Lambda = 60m/sec/30Hz = Lambda = 2 Therefore the wavelength is 2meters long. To find the Number of Loops, we need to take the inverse of Frequency and plug-in what's already given to us.....so F = N/Change in time 30 Hz (given)= N/0.20sec 0.20sec X 30 = N 6 = N Therefore the number of loops are 6.

  3. Astrophysics
    • one year ago
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    What is takes sec, is that suppose to be .20 sec?

  4. YumYum247
    • one year ago
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    already given to us...0.20sec

  5. YumYum247
    • one year ago
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    i've lablled all the given values at the top.

  6. Astrophysics
    • one year ago
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    Ok I see, well in your question it didn't mention that so just making sure.

  7. YumYum247
    • one year ago
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    ^_^

  8. Astrophysics
    • one year ago
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    So you need to do \[v = f \lambda = (30 Hz)(12m) = 360 m/s\], good I see you have \[v = \frac{ \Delta d }{ \Delta t } \implies \Delta d = v \Delta t\] which is, \[\Delta d = (360m/s)(0.20s) = 72m\] so far you agree?

  9. YumYum247
    • one year ago
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    how did you come up with "12"

  10. YumYum247
    • one year ago
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    "is used to make waves in a rope 12m long", i though the length of the rope is 12m long -__-

  11. Astrophysics
    • one year ago
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    Actually you maybe right

  12. Astrophysics
    • one year ago
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    I guess I was reading it wrong and thought the wavelength is 12m but I guess the rope is!

  13. YumYum247
    • one year ago
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    ok so once i have the velocity and the displacement, do i foolow the same procedure to find the wavelength.......λ = F/V

  14. Astrophysics
    • one year ago
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    So, we should use \[v = \frac{ \Delta d }{ \Delta t } = \frac{ 12m }{ 0.20s } = 60m/s\] this seems better, I think the question is just worded weirdly so it caused the confusion.

  15. Astrophysics
    • one year ago
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    \[\lambda = \frac{ v }{ f } = \frac{ 60m/s }{ 30 Hz } = 2m\] that makes more sense

  16. YumYum247
    • one year ago
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    that's what i was wondering.........360m/sec.....that's like sound breaking speed!!! :")

  17. Astrophysics
    • one year ago
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    Yeah it was a bit confusing when I was writing it out

  18. YumYum247
    • one year ago
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    yep that's what i got.....λ = 2m

  19. Astrophysics
    • one year ago
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    Now you can just divide that by the length of the string, so we have \[\frac{ 12m }{ 2m } = 6\] that is our wavelength in the string

  20. Astrophysics
    • one year ago
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    amount of wavelengths*

  21. YumYum247
    • one year ago
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    yep....that's right...i got the same.....but hey do you know the trick of solving these kinds of peoblems?!?!?!?!

  22. Astrophysics
    • one year ago
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    There's no real trick, just this problem is worded oddly as you noticed haha

  23. Astrophysics
    • one year ago
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    Lots of practice I suppose

  24. YumYum247
    • one year ago
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    no i'm talking in general....tho, cuz i see some of the next problems i have in the book seem kinds hard.....

  25. YumYum247
    • one year ago
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    is there any difference between solving for fixed end and free end?!?!??!?

  26. Astrophysics
    • one year ago
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    There's no trick to it, just takes practice, read the question carefully, put all your knowns/ unkowns down and see what you can use :)

  27. Astrophysics
    • one year ago
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    Well I would have to see the question to see what exactly you're asking

  28. YumYum247
    • one year ago
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    ok....Thank you sooooooooooooooomuch, Luv you to death....!!! Please stay cute!!! :")

  29. Astrophysics
    • one year ago
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    Haha np :) have fun!

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