Originally Posted By: paul
I didnt botch it up , I worked it exactly as you posted it.
meaning that I need to check all of your formulas plus your math.

F=(10132kPa - 101kPa)* 0.00064516m^2


You did not work it exactly as posted. Had you done that you'd have done the calcs using kilopascals, as indicaterd in what I wrote. You didn't, despite the fact kPa was typed out.

Ergo, my math is right, ignoring the PSI vs ATA foopah...

Originally Posted By: paul

it can be fast or slow but faster is better in this case.

In your case the speed of the gas will always be the same - the speed of sound.

Originally Posted By: paul
do you deny that the air mass moving inside the pipe would move the pipe?


It would. But that change in momentum would be countered by an equal, but opposite momentum due to the momentum of the tank the air was contained in. Thus total change in your pipes momentum would be zero. It may rock back and forth, but it isn't going to go anywhere.

Its that later half you always miss - anytime you give something momentum in a closed system, you get an equal but opposite momentum produced. It doesn't matter which of those you use as a reference point - in the end, the net change in momentum will always be zero.

Bryan
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