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Mach 3

  1. Dec 8, 2009 #1
    Here is an answer from a book of mine: From the force diagram we have [tex]N-mg=(\frac{mv^2}{R})e_r[/tex]. The acceleration that the pilot feels is [tex]\frac{N}{m}=g+(\frac{mv^2}{R})e_r[/tex]. I'm confused though; what happened to the g being divided by the last term as well?
  2. jcsd
  3. Dec 8, 2009 #2


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    Hi delve! :smile:
    I'm not following you …

    the equation is equivalent to N/m = g + (v2/R)er

    what do you mean by g being divided by something? :confused:
  4. Dec 8, 2009 #3
    Actually, you just helped me realize what I was confused about. Thank you!
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