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Mathematics 8 Online
OpenStudy (anonymous):

given: p=aV^-3/2, p1=2000Pa and V1= 1x10^-2 m^3 find: a

OpenStudy (anonymous):

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 :/

OpenStudy (anonymous):

\[\large p = a(V)^{- \frac{3}{2}}\] Is it??

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

happy new year by the way :)

OpenStudy (anonymous):

Happy New Year to you and to your Family @bronzegoddess

OpenStudy (anonymous):

So you got: \[\large p(V)^{\frac{3}{2}} = a\]

OpenStudy (anonymous):

thx, yes

OpenStudy (anonymous):

So: \[\huge a = 2000 \times (10^{-2})^{\frac{3}{2}}\]

OpenStudy (anonymous):

Now look carefully: \[\huge (x^b)^a = x^{b \times a}\]

OpenStudy (anonymous):

i see

OpenStudy (anonymous):

So: \[\large (10^{-2})^{\frac{3}{2}} = ??\]

OpenStudy (anonymous):

:) 10^-6/2=10^-3

OpenStudy (anonymous):

but 2000 x 10^-3 = 2...

OpenStudy (anonymous):

Try to figure out, anna is calling me..

OpenStudy (anonymous):

What is the units of a given?/ What is units given in answer??

OpenStudy (anonymous):

joules

OpenStudy (anonymous):

i thought of converting into liters for the volume but that is not possible because joules are newton meters..

OpenStudy (anonymous):

Do you think it is right? \[Pa \times m^3 = Joules\]

OpenStudy (anonymous):

In simpler notations, how can we write joules??

OpenStudy (anonymous):

i dont understand..

OpenStudy (anonymous):

If your teacher is not understanding then how can the student..!!

OpenStudy (anonymous):

pascal=kg/ms^2

OpenStudy (anonymous):

Oh okay.. Similarly, can you say something about Joules??

OpenStudy (anonymous):

therefore Pa xm^3= kg/m

OpenStudy (anonymous):

kg/ms^2 i mean

OpenStudy (anonymous):

wait, the units are write.. Pa x m^3= kg/ms^2 xm^3= kgm^2/s^2

OpenStudy (anonymous):

Pascal is : \[\frac{kg}{ms^2}\] Is this right??

OpenStudy (anonymous):

the units are right*

OpenStudy (anonymous):

can someone just tell me why i am getting 2 and not 2x10^-3...

OpenStudy (anonymous):

\[Pa = \frac{N}{m^2}\]

OpenStudy (anonymous):

So: \[Joule = N \cdot m = Force \times Distance \; \; \; (Right)\]

OpenStudy (anonymous):

I think there is something wrong with your question.. Can you scan and upload the page??

OpenStudy (anonymous):

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 : \[\large V = 1 \times 10^{-6} m^3\]

OpenStudy (anonymous):

@waterineyes

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