Gauge Pressure problem (algebra based)

[guynameddanny]

Homework Statement

What is the gauge pressure at a depth of 100 m. in water?

Homework Equations

Pgauge = rho*g*h (or the pressure above atmospheric pressure)

The Attempt at a Solution

I missed a day of class (and on a summer class, that's a LOT of material). I am working problems that are assigned by the teacher over the internet. I'm sure these change of problems are things I can figure out, the problem I am having is understanding the equation for this problem. I can't seem to find where it talks about it in my book, I'm thinking because the teacher used either alternative symbols (because of limited keyboard characters), or he just likes to be different from the book.

I'm really just looking for help understanding the equation, what are the meanings of the different variables? I take it g*h is gravity*height (depth I presume), I'm not sure what "rho" would be though.

Edit: My theory is that "rho" is density of the liquid, not sure why he would use rho instead of something that makes sense like d.

To solve:

1000 kg/m^3 * 9.8 m/s^2 * 100 m = 980 kPa

 

[PhanthomJay]

'rho' (the greek letter $$\rho$$) is the commonly used symbol to represent density (mass per unit volume). The letter 'd' is often used for 'distance'; after a while, you run out of letters to use...

 

[tomcue]

The equation for gauge pressure is Pgauge = rho*g*h, where rho is the density of the liquid, g is the acceleration due to gravity, and h is the depth of the liquid.

In this problem, the gauge pressure at a depth of 100 m in water is being asked. To solve, we need to know the density of water, which is 1000 kg/m^3, the acceleration due to gravity, which is 9.8 m/s^2, and the depth, which is 100 m.

Substituting these values into the equation, we get:

Pgauge = (1000 kg/m^3) * (9.8 m/s^2) * (100 m) = 980,000 Pa

This is the gauge pressure at a depth of 100 m in water. It is important to note that this is the pressure above atmospheric pressure, so if we want to find the total pressure, we would need to add the atmospheric pressure to this value.