given: p=aV^-3/2, p1=2000Pa and V1= 1x10^-2 m^3 find: a
ok so i tried to work it out by filling in but my answer is wrong: a=p1v1^3/2 = (2000)(1x10^-2)^3/2, my answer is 2 but my book says 2x10^-3 :/
\[\large p = a(V)^{- \frac{3}{2}}\] Is it??
yes
happy new year by the way :)
Happy New Year to you and to your Family @bronzegoddess
So you got: \[\large p(V)^{\frac{3}{2}} = a\]
thx, yes
So: \[\huge a = 2000 \times (10^{-2})^{\frac{3}{2}}\]
Now look carefully: \[\huge (x^b)^a = x^{b \times a}\]
i see
So: \[\large (10^{-2})^{\frac{3}{2}} = ??\]
:) 10^-6/2=10^-3
but 2000 x 10^-3 = 2...
Try to figure out, anna is calling me..
What is the units of a given?/ What is units given in answer??
joules
i thought of converting into liters for the volume but that is not possible because joules are newton meters..
Do you think it is right? \[Pa \times m^3 = Joules\]
In simpler notations, how can we write joules??
i dont understand..
If your teacher is not understanding then how can the student..!!
pascal=kg/ms^2
Oh okay.. Similarly, can you say something about Joules??
therefore Pa xm^3= kg/m
kg/ms^2 i mean
wait, the units are write.. Pa x m^3= kg/ms^2 xm^3= kgm^2/s^2
Pascal is : \[\frac{kg}{ms^2}\] Is this right??
the units are right*
can someone just tell me why i am getting 2 and not 2x10^-3...
\[Pa = \frac{N}{m^2}\]
So: \[Joule = N \cdot m = Force \times Distance \; \; \; (Right)\]
I think there is something wrong with your question.. Can you scan and upload the page??
In my sense, you might be given that Volume is given to you in 10^{-2} per metre sense, so that Volume in m^3 will be : \[\large V = 1 \times 10^{-6} m^3\]
@waterineyes
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