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Posts posted by Backes
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Hey, I bet that many of you know Torr, or mm/hg. It is defined as 1/760 of the standard atmospheric pressure. Below is an image of the setup to determine the value. My idea was to simply look for a formula to calculate the height for any fluid, not only Hg.
It sounds not that hard, and is in fact pretty easy, BUT... My results are somehow not accurate.
[latex]P_{ext}= P_{int}
\Leftrightarrow P_{ext} = \frac{m \times g}{\pi \times r^2}
\Leftrightarrow P_{ext} = \frac{\rho \times h \times \pi \times r^2 \times g}{\pi \times r^2}
\Leftrightarrow P_{ext} = \rho \times h \times g
\Leftrightarrow h = \frac{P_{ext}}{\rho \times g}[/latex](please ingore the <br>, linebreaks are somehow interpreted wrong... I did not post them )
With [latex]\rho[/latex] the density of the fluid.
When I know take those values: g = 9.81 N/kg and the density as 13.534 g/cm^3 (wikipedia) at 20°C and as outside pressure 1atm, which is 1013.25 hPA I get as a result 763,17mm. Did I miss something in the calculation? I don't think so... So where did I took a "wrong" value? Do I need to take the density of Hg at 0°C? Or is my g not accurate?
Thank you!
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Is Multiplication in the Real Set a group
in Linear Algebra and Group Theory
Posted
Hey, I started to learn what groups are... I asked myself, is the multiplication in R a group?
My counterargument is, that there is no identity element, because 1*e = e*1 = e, expect for e=0. And the second one is, that there is no symmetric element, because: a*b = b * a = e, where b = 1/a and e = the no identity element. This works, except for a=0...
Am I wrong in my argumentation, or is the multiplication in the real set no group?
Thank you!