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.

Question

istim · Answer

This is part of a physics question, but I thought it was simple algebra, and deserved to be let free here.

whpalmer4 · Answer

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?

istim · Answer

I think this is extremely over-complicating the original intent of this thread...but...yes?

istim · Answer

Simply asking how to cancel out "s" in this situation.

whpalmer4 · Answer

Where are you canceling the s?  Maybe with more context I can give you a better answer.

istim · Answer

In J/S

whpalmer4 · Answer

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

whpalmer4 · Answer

Remember, dividing by s is equivalent to multiplying by 1/s.

istim · Answer

well, you know, the same well you can "remove" distance (m) in speed (m/s)  to get time (t) by

istim · Answer

t=v/d?
t=d/v?

istim · Answer

Man, can not realize I forgot this so fast.

whpalmer4 · Answer

d = vt

m = (m/s)*(s)
m = m

But there's no cancellation in J/s...