Converting kg/m3 to atm

[emericana]

More specifically, in EK 1001 Physics, prob 548...

in this thread p will == density

"If the values of p1, p2, p3, and p4 are 500kg/m3, 100kg/m3, 2000kg/m3, and 4000kg/m3, respectively, then what is the approximate pressure at p3? "

If needed, I will submit the pic which they are referring to so let me know and i will upload it.


Essentially the answer says to :

add up p1gh1+ p2gh2+ p3gh3 + 1atm

The answer choices are all in atm.

When you multiply the values together for p1Vg etc you get huge numbers.

For example

p1Vg

(500kg/m3)(10m)(10m/s2) = 50,000

How is this converted to atm?

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[TXKnight]

More specifically, in EK 1001 Physics, prob 548...

in this thread p will == density

"If the values of p1, p2, p3, and p4 are 500kg/m3, 100kg/m3, 2000kg/m3, and 4000kg/m3, respectively, then what is the approximate pressure at p3? "

If needed, I will submit the pic which they are referring to so let me know and i will upload it.


Essentially the answer says to :

add up p1gh1+ p2gh2+ p3gh3 + 1atm

The answer choices are all in atm.

When you multiply the values together for p1Vg etc you get huge numbers.

For example

p1Vg

(500kg/m3)(10m)(10m/s2) = 50,000

How is this converted to atm?

I am failing to understand the question correctly I guess.
What are your options?
seems to me they're talking about Total Pressure=gauge P+ATM Pressure? so you just add them up to that point plus atmospheric pressure?
1 atm=760 torr
1 torr=1mmhg=13.5g/cm3....density of mercury?
don't think that's correct though
sorry...more specific?

 

[emericana]

I am failing to understand the question correctly I guess.
What are your options?
seems to me they're talking about Total Pressure=gauge P+ATM Pressure? so you just add them up to that point plus atmospheric pressure?
1 atm=760 torr
1 torr=1mmhg=13.5g/cm3....density of mercury?
don't think that's correct though
sorry...more specific?

Question 548

How did they get answer B from this calculation?

 

[WhiteWashed]

Question 548

How did they get answer B from this calculation?

Friend, the answer to your question is C. the rho*g*h calculation will yield pressure in pascals. assume that 10^5 pascals=1atm and add the fluid pressure to atmospheric pressure...will equal 2.5atm

 

[TXKnight]

Friend, the answer to your question is C. the rho*g*h calculation will yield pressure in pascals. assume that 10^5 pascals=1atm and add the fluid pressure to atmospheric pressure...will equal 2.5atm

sounds about right i think. haven't made the calc yet but units sound good.
1pa=N/m2
and you got 150000 (kg/m3) * (m/s2) * (m)=Newtons

 

[emericana]

Friend, the answer to your question is C. the rho*g*h calculation will yield pressure in pascals. assume that 10^5 pascals=1atm and add the fluid pressure to atmospheric pressure...will equal 2.5atm

Ahh great thanks. I didnt understand pascals but now I do. Thanks so much.