Physics
18 Online
OpenStudy (anonymous):
please show ur Work >>>>
Dimension of a pressure ?
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OpenStudy (anonymous):
i need to see the steps of solving it plz :)
OpenStudy (ash2326):
Do you know the relation between Pressure and force?
OpenStudy (anonymous):
i know \[p = \frac{ F }{ a }\] So \[= \frac{ m a }{a }\]
OpenStudy (ash2326):
\[P=ma / A\]
\[a\ne A\]
can you tell me the difference between a and A?
OpenStudy (anonymous):
idk , u tell me :B
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OpenStudy (anonymous):
i just wanna know how we got L to the power -1 and T to the power -2
OpenStudy (anonymous):
how we got the negative powers ?
OpenStudy (ash2326):
a= acceleration= distance/ (time )^2= L/T^2=\(LT^{-2}\)
A= area= \(L^2\)
OpenStudy (anonymous):
oh yeah
OpenStudy (ash2326):
Can you try now?
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OpenStudy (anonymous):
gemme a minute
OpenStudy (ash2326):
sure :)
OpenStudy (anonymous):
= \[\frac{ M L T ^{-2} }{L T ^{-2} }\]
OpenStudy (anonymous):
is that right -0-
OpenStudy (ash2326):
Why you have \(Lt^{-2}\) in the denominator?
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OpenStudy (anonymous):
because pressure = \[\frac{ m a }{a }\] and \[a = l T ^{-2}\]
OpenStudy (anonymous):
u just said that >_>
OpenStudy (ash2326):
Read my post again,
OpenStudy (anonymous):
is p pressure Force over area ?
OpenStudy (ash2326):
Yes :)
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OpenStudy (anonymous):
i thought its force over acceleration -.-
OpenStudy (anonymous):
okie my bad
OpenStudy (ash2326):
No problem, you got it now?
OpenStudy (anonymous):
yeah -,- thanks again