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## IsTim 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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1. IsTim

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

2. whpalmer4

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

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

4. IsTim

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

5. whpalmer4

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

6. IsTim

In J/S

7. whpalmer4

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

8. whpalmer4

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

9. IsTim

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

10. IsTim

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

11. IsTim

Man, can not realize I forgot this so fast.

12. whpalmer4

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

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