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anonymous

  • 4 years ago

36.0kn/c=2(8.99x10^9)(Q/2.59)/.197 solve for Q

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  1. JamesJ
    • 4 years ago
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    The right hand side, RHS is \[ RHS = 2(8.99 \times 10^9)(Q/2.59)/.197 \] \[ = (3.524 \times 10^{10})Q \] \[ = \frac{36.0kn}{c} = LHS \] Now it should be easier for you.

  2. JamesJ
    • 4 years ago
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    Yes?

  3. JamesJ
    • 4 years ago
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    chan ... talk to me

  4. JamesJ
    • 4 years ago
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    ok ... hope that helped. find me or someone in Physics chat if you need more help

  5. anonymous
    • 4 years ago
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    i just didn't know how to deal with the Q/2.59

  6. JamesJ
    • 4 years ago
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    yes ... just put the 2.59 and the 0.197 in the denominator.

  7. anonymous
    • 4 years ago
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    so what did you get as a final answer.. just to compare and make sure i did it right

  8. JamesJ
    • 4 years ago
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    I didn't calculate the final answer, but it's obviously Q = (some constant, of order -9 or so)*(kn/c)

  9. anonymous
    • 4 years ago
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    thanks man.. some one that finally helped me and didn't start spittin out more complicated formulas

  10. JamesJ
    • 4 years ago
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    sure. Come back soon with some more Physics questions.

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