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Homework Help: It's that time again: Help by tonight

  1. Oct 6, 2003 #1
    Thanks in advance for the help. I'm stuck on problem #12(please see the attached file). I did this:
    r= cos(26)*2.8(is this right?)
    r= 2.52
    a= mg/(cos(ang))
    r*a=v(squared)

    a= 5.9(9.8)/cos(26)
    a= 64.3

    v= sqrt(2.52*64.3)
    v= 12.7

    What's wrong with this?
     

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  3. Oct 6, 2003 #2

    HallsofIvy

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    "r= cos(26)*2.8(is this right?)"

    No, it's not. r is the "opposite" side of the right triangle:
    r= sin(26)*2.8
     
  4. Oct 7, 2003 #3

    HallsofIvy

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    "a= mg/(cos(ang))"

    No, that's wrong. For one thing, there is no "m" in the acceleration. For another thing, the acceleration you want, in order to calculate speed around the circle is the acceleration toward the center, that's directed along the opposite side of your triangle, not the hypotenuse. The formula you want is a/g= tan(ang) (a= opposite side, g= near side).

    You do want v2= r* a which is
    v2= (2.8 sin(26))(9.8 tan(26)).

    I get v= 2.42 m/s.
     
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