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

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

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

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

whpalmer4
 2 years 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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