## 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

1. IsTim Group Title

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

2. whpalmer4 Group Title

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?

3. IsTim Group Title

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

4. IsTim Group Title

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

5. whpalmer4 Group Title

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

6. IsTim Group Title

In J/S

7. whpalmer4 Group Title

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

8. whpalmer4 Group Title

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

9. IsTim Group Title

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

10. IsTim Group Title

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

11. IsTim Group Title

Man, can not realize I forgot this so fast.

12. whpalmer4 Group Title

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