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 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.
 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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IsTim
 one year ago
Best ResponseYou've already chosen the best response.0This is part of a physics question, but I thought it was simple algebra, and deserved to be let free here.

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.0joules are kg m^2/s^2 \[J/s= \frac{{kg}*m^2}{s^2}*\frac{1}{s}=\frac{{kg}*m^2}{s^3}\] Right?

IsTim
 one year ago
Best ResponseYou've already chosen the best response.0I think this is extremely overcomplicating the original intent of this thread...but...yes?

IsTim
 one year ago
Best ResponseYou've already chosen the best response.0Simply asking how to cancel out "s" in this situation.

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.0Where are you canceling the s? Maybe with more context I can give you a better answer.

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.0They don't cancel out. J/s = watts, and watts are kg m^2 / s^3, just like I showed above.

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.0Remember, dividing by s is equivalent to multiplying by 1/s.

IsTim
 one year ago
Best ResponseYou've already chosen the best response.0well, you know, the same well you can "remove" distance (m) in speed (m/s) to get time (t) by

IsTim
 one year ago
Best ResponseYou've already chosen the best response.0Man, can not realize I forgot this so fast.

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.0d = vt m = (m/s)*(s) m = m But there's no cancellation in J/s...
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