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SNIIX18 Group Title

Imagine that you are swinging back and forth a chain with a locket on the end. Assuming that you are on Earth so that gravity is constant at 9.81 meters/second², and you know that the chain is 43 centimeters long, what is the period of this simple pendulum? A. 13 seconds B. 1.3 seconds C. 3.5 seconds D. 0.76 seconds

  • one year ago
  • one year ago

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  1. souvik Group Title
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    \[T=2\pi \sqrt{L/g}\]

    • one year ago
  2. theEric Group Title
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    http://hyperphysics.phy-astr.gsu.edu/hbase/pend.html

    • one year ago
  3. theEric Group Title
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    That link backs up souvik's response. It might also help you find other information.

    • one year ago
  4. SNIIX18 Group Title
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    Thank you both :) @theEric that link was very helpful Grazie

    • one year ago
  5. theEric Group Title
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    You're welcome! And I guess that equation only works for a "simple pendulum."

    • one year ago
  6. souvik Group Title
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    any problem? @SNIIX18

    • one year ago
  7. SNIIX18 Group Title
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    yahh i cant wrap my finger around this problem its kind of difficult

    • one year ago
  8. souvik Group Title
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    this question?

    • one year ago
  9. SNIIX18 Group Title
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    yes it's kind of difficult for me

    • one year ago
  10. souvik Group Title
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    what happened?

    • one year ago
  11. SNIIX18 Group Title
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    i just dont understand

    • one year ago
  12. souvik Group Title
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    |dw:1372793064271:dw|

    • one year ago
  13. souvik Group Title
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    |dw:1372793264146:dw|

    • one year ago
  14. souvik Group Title
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    i divide the force mg into two component one is mgcos(theta) and mssin(theta) any question?

    • one year ago
  15. SNIIX18 Group Title
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    no i understand know i was doing everything wrong Thank you for the help

    • one year ago
  16. souvik Group Title
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    ok....

    • one year ago
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