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Yahoo!

  • 3 years ago

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  1. Yahoo!
    • 3 years ago
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    @ghazi @Aperogalics

  2. ghazi
    • 3 years ago
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    it is already solved lol

  3. ghazi
    • 3 years ago
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    |dw:1351178590831:dw| \[T=\frac{ mv^2 }{ r }\]

  4. ghazi
    • 3 years ago
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    i think you are done

  5. Yahoo!
    • 3 years ago
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    T-mgcosx = mv^2/r

  6. Yahoo!
    • 3 years ago
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    but hw Do i Find the angle....)

  7. ghazi
    • 3 years ago
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    \[\sin \theta =\frac{ a \cos \theta }{ T }\]

  8. ghazi
    • 3 years ago
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    a -- tangential acceleration component

  9. Yahoo!
    • 3 years ago
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    mgsinx = ma ....?

  10. ghazi
    • 3 years ago
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    nope, just from the diagram i took it

  11. Yahoo!
    • 3 years ago
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    i aksed mgsinx = ma...is it correct

  12. ghazi
    • 3 years ago
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    |dw:1351179063684:dw|

  13. ghazi
    • 3 years ago
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    yea your equation seems correct but why to go for that when things are done here

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