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Formula for Circular Motion and Tangential Speed

  1. Jun 25, 2010 #1
    1. The problem statement, all variables and given/known data
    The Earth rotates on its own axis once per day (24.0 hr). What is the tangential speed of the summit of a mountain (elevation 6409 m above sea level), which is located approximately on the equator, due to the rotation of the Earth? Take the equatorial radius of Earth to be 6379 km

    I'm looking for the formula to help solve this problem
     
  2. jcsd
  3. Jun 25, 2010 #2

    Uku

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    This is a very straightforward problem. You could use dimensional analysis to solve it in your head.

    [tex]v=\omega R[/tex]

    You do the rest!
     
  4. Jun 25, 2010 #3
    I used that formula it didnt accept my answer.
    v=WR
    W=2pi/T
    W=2pi/86400s

    V=.000074*63979
    .463894
    I wont accept that onwebassign.net
     
  5. Jun 25, 2010 #4

    Uku

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    Where did you get that weird radius from?

    63979
     
  6. Jun 25, 2010 #5
    i meant to say 6379
     
  7. Jun 25, 2010 #6

    Uku

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    Look at your data again, the radius is still weird.
     
  8. Jun 25, 2010 #7
    Its correct i just dont know what to do
     
  9. Jun 25, 2010 #8

    Uku

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    Why are you given two radiuses then?

    Good night!
     
  10. Jun 25, 2010 #9
    nvrminf thanks so much.. i forgot to change the radius to m
     
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