My question is: "One atmosphere of pressure is equal to 101,325 Pa. If the density of water is 998 kg/m3, what is the necessary depth to reach 2 atm of pressure relative to the surface"
Now the hydrostatic equation is p=-wh where p is change in density, w is specific weight, and h is change in altitude. Now, to get w, it is simply w=(998)(9.81)=9790.38. So we now have p=-(9790.38)h. Now it is (101325*2)=-(9790.38)h. Multiply and divide and we get -20.69889014=h.
So my two questions to you is: Did I do this right? and, Would I make my answer POSITIVE 20.698?
If I did something wrong please tell me.
Now the hydrostatic equation is p=-wh where p is change in density, w is specific weight, and h is change in altitude. Now, to get w, it is simply w=(998)(9.81)=9790.38. So we now have p=-(9790.38)h. Now it is (101325*2)=-(9790.38)h. Multiply and divide and we get -20.69889014=h.
So my two questions to you is: Did I do this right? and, Would I make my answer POSITIVE 20.698?
If I did something wrong please tell me.
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I didn't take the time to verify your reasoning and calculations, but 1 atmospheres = 10.3 meters of water, so 2 = 20.6 meters. Your answer is certainly in the ballpark. The negative in your answer means that you have to go DOWN to get that pressure.