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IsTim Group Title

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.

  • one year ago
  • one year ago

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  1. IsTim Group Title
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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.

    • one year ago
  2. whpalmer4 Group Title
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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?

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

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

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

    • one year ago
  6. IsTim Group Title
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    In J/S

    • one year ago
  7. whpalmer4 Group Title
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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.

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

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

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

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

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

    • one year ago
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