Physics 39 Online
OpenStudy (anonymous):

A swimming pool is 25 m long and 10 m wide. The bottom has an inclination; one of its sides is 1.3 m depth and the other side is 3 m depth. Find the pressure in each side (1.3 and 3 m) at the bottom. the pool has been filled with water (d=1000kg/m3)

OpenStudy (anonymous):

Okay, so you were ask to solve for the pressure on either side of the pool. The formula for pressure is pressure = density*gravity*height This is quite difficult since there is an inclination. I'm not sure if it will affect the pressure on both sides of the pool. Anyway, if you will solve it separately, just follow the formula. the density for water is 1g/cm^3 or 1000kg/m^3 gravity is 9.8m/s^2 and depths are: depth A = 1.3m depth B = 3 m ~~~~~~~~~~~~~ That's how I see it. But, if the inclination is needed to solve the problem, then you have to use trigonometric functions. You have to find the angle of inclination of the floor. Might as well seek help from others.

OpenStudy (anonymous):

Its all right man. I think i can figure it out. Thanks again

OpenStudy (anonymous):

im new with presure and this kind of units. Im already getting used :)

OpenStudy (anonymous):

That's nice. :) Good luck!

OpenStudy (anonymous):

1m3 = 1lit? Correct me if im wrong please

OpenStudy (anonymous):

That's right

OpenStudy (anonymous):

thanks

OpenStudy (anonymous):

how do i transform 1 20cm2 into m2?

OpenStudy (anonymous):

and cm3 into m3? can you show me a general procedure? Im really bad at remembering equivalences.

OpenStudy (anonymous):

1*1m^2. Is 100*100cm^2 So 1m^2 is 10000cm^2 same method for m^3 which is 1000000cm^3

OpenStudy (anonymous):

I just did it by working out a 1m by 1m box and it's easy to think of the dimensions in cm I don't memorise it but then you use that to then work out the proportion type thing

OpenStudy (anonymous):

I have to transform 1500 cm2 = 0.15m2 1.25cm3 = 1.25x10-6m3 correct?

OpenStudy (anonymous):

Yep

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

Can you correct this exercise for me? The base of a container is 1500 m2. It contains 110 litres of gas (d=1.25g/cm3). What pressure is exerted over the base? This is how i did A=1500 cm2 = 0.15m2 V=110lit =110m3 d=1.25g/cm3 =1250kg/m3 pr=Vdg/A = 8.98*10-6 Pa however the "correct" answer is 8.98*10-3Pa

OpenStudy (anonymous):

I found the answer! 1lit = 0.001 m3

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