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IsTim

  • one year ago

Simple dimensional analysis: To cancel out "s" in J/S, do I multiply or divide J/s by s? s in both cases are the same.

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  1. IsTim
    • one year ago
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    This is part of a physics question, but I thought it was simple algebra, and deserved to be let free here.

  2. whpalmer4
    • one year ago
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    joules are kg m^2/s^2 \[J/s= \frac{{kg}*m^2}{s^2}*\frac{1}{s}=\frac{{kg}*m^2}{s^3}\] Right?

  3. IsTim
    • one year ago
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    I think this is extremely over-complicating the original intent of this thread...but...yes?

  4. IsTim
    • one year ago
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    Simply asking how to cancel out "s" in this situation.

  5. whpalmer4
    • one year ago
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    Where are you canceling the s? Maybe with more context I can give you a better answer.

  6. IsTim
    • one year ago
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    In J/S

  7. whpalmer4
    • one year ago
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    They don't cancel out. J/s = watts, and watts are kg m^2 / s^3, just like I showed above.

  8. whpalmer4
    • one year ago
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    Remember, dividing by s is equivalent to multiplying by 1/s.

  9. IsTim
    • one year ago
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    well, you know, the same well you can "remove" distance (m) in speed (m/s) to get time (t) by

  10. IsTim
    • one year ago
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    t=v/d? t=d/v?

  11. IsTim
    • one year ago
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    Man, can not realize I forgot this so fast.

  12. whpalmer4
    • one year ago
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    d = vt m = (m/s)*(s) m = m But there's no cancellation in J/s...

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