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

whpalmer4Best ResponseYou've already chosen the best response.0
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

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

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

whpalmer4Best ResponseYou've already chosen the best response.0
Where are you canceling the s? Maybe with more context I can give you a better answer.
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.0
They don't cancel out. J/s = watts, and watts are kg m^2 / s^3, just like I showed above.
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.0
Remember, dividing by s is equivalent to multiplying by 1/s.
 one year ago

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

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

whpalmer4Best ResponseYou've already chosen the best response.0
d = vt m = (m/s)*(s) m = m But there's no cancellation in J/s...
 one year ago
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