Water pressure? Where do you get the conversion factor of 32 from?
97996
The normal equation for water pressure is P=rho*g*h\n\nwhere g is the acceleration of gravity, rho is density, or\nmass per unit volume and h is the depth of the water. \n\nSince you want to compute in avoirdupois units for some incredible reason, you are going to have to deal with some confusion. The density of water in avoirdupois units is actually m/L^3 where m is in slugs, and L is in feet. \n\nHowever, we know that the weight of a cubic foot of water is 62 lbs. A lb is a unit of force, and F=mg, so \nm=F/g=\n62 lbs/(32 ft/sec/sec)=\n62/32 slugs. \n\nThen in your equation, \n\nP=rho*g*h\n=(F/g)/(L^3)*g*h\n=(F/L^3)*g*h/g\n\nNow substituting, F/L^3=62 lb/ft^3\ng=32 ft/sec/sec\nLet h=50 feet, say\n\nthen P=(62 lb/ft^3)*50ft\n=62*50lb/ft^2\n=3100 lb/ft^2\n\nI do not understand how you put h in meters and somehow ended up with an answer per square inch.\n\nIf you want to convert this to square inches, then\n\n3100 lb/ft^2 *(1ft/12in)^2=3100/144 lb/in^2\n\nIf you want to have h in meters, I will leave that as an exercise to the reader.
344788
hey man, dont worry ill help. first of all our given:\nh=280m\nsg=1.03\ngc=9.7m/s^2\n\nreq'd Pressure(absolute)\n----------------------------------------------\nsol'n:\nfor DENSITY of subs:\nformula\nSpecific Gravity=density of subs/density of water\ndensity of subs=1.03 x 1000(density for water)\ndensity of subs=1030 kg/m^3\n\nFOR PRESSURE GAGE:\nP(gage)=Vh\nP(gage)=density x gravitational constant x height\nP(gage)=1030 x 9.7 x 280\nP(gage)=2,797,480 Pa\n\nFOR PRESSURE ABSOLUTE:\nP(abs)=P(gage) - P(atm)\n\nwhy minus? bec. the submarine is pressurized, meaning its vacuumed\n\nP(abs)=2,797,480 Pa - 1atm(101.325x10^3 Pa)\nP(abs)=2,696,155 Pa\n\nconvert Pa to Kpa\n=2,696,480 /1000\n=2,696.155 Kpa\n\nconvert to bar\n=2696.155kPa x (1.01325 bar / 101.325 kPa)\n=26.96155 bar\n\ndont mind the others, they dont even know how to answer this simple question. They just want to earn points!
902702
I plugged it into my calculator and got 9.6585....\n\nFor these, for a given x, just plug it into the calculator.  Step by step, this is what it means:\n\n\n14.7e^(-.21*2)=\n14.7e(-.42)=\n9.6585.....\n\nYour process is correct, but the answer does not seem to be.  Your answer seems unlikely since e^-.21x will be 1/e^(.21x) will be way smaller than 8 (120/14.7) for x=2.\n\nYour equation represents exponential decay.  It means that the atmospheric pressure is decreasing at an increasingly large rate.  Why decreasing?  Because the exponent in -.  There is no way it is 120 because the maximum value is when x=0, in which case the pressure would be 14.7.
985939
Let's see....\n1 liter=1000 ml\n1000ml=4 cups water\n1 cup=8 oz\n 32 oz=1 liter\n32x3=96 oz\n\nSpace out the water. Try dividing it into three 1 liter bottles and carry one with you while at work, school,...Good luck cuz that's alot of water....
1051954
All you need to do is know the density of water- not too difficult to find! THe psi necessary to support a column then must match the weight of that column. So, if a column is 1inch in area and Hinches high, if you know D the density of water per cubic inch in pounds, the height of the column in inches times this value is the psi needed.\n\nPressure psi=H*D\n\nThis also works for metric, just use cms and kgs as the base sizes.
456039
C = 5(F - 32)/9\nK = P/2.2\n\nThe easiest place for all conversions is google. It has a built in calculator and converter that automatically gives you the results. Simply type in something like:\n\n30 degrees centigrade in fahrenheit\nor\n50 pounds in kilograms\nor even\n50 lb in kg
751939
you can either do the math, look it up on a chart or go here:\n\nhttp://leebrewery.com/hydromet.htm\n\nand use the java-calculator.
1099854
4.44 ºC\n\nAn easy way to find this out is do do a google search.\n\nGoogle has a buildt in calculator/conversion machine.\n\nhttp://www.google.com/search?q=40+F+in+C&sourceid=mozilla-search&start=0&start=0&ie=utf-8&oe=utf-8&client=firefox-a&rls=org.mozilla:en-US:official
456275
9.81 m/s² or 32.2 ft/s²
780627
WATER BOIL AT:  27*C at 26.739 (pressure(torr)  thru 30*C at 31.824\n pressure(torr)
363092
Well I'm not going to do the calculation for you but you can.\n\nDo a web search and you'll find dozens of sites to help you if you don't have a fluid flow handbook.
474374
Standard atmospheric pressure is 101 325 Pa = 101.325 kPa = 1013.25 hPa = 1013.25 mbar = 760 Torr. This definition is used for pneumatic fluid power (ISO R554), and in the aerospace (ISO 2533) and petroleum (ISO 5024) industries.
63901
Not sure what kind of conversions, to cover them all visit:\nhttp://www.onlineconversion.com/\nShould be able to help you out.
845517
One way is to use the Clausius-Clayperon Equation.  See the link below and use the boiling point of water for one of the pressure and temperature pairs.  \n\nP.S.  It's equation 22 in the linked document.
708158
well aside from that you said the vapor pressure is 18.6 and put 18.9 in your calculations, it looks great! Besides, even with that difference, it doesn't change the answer at all, its still .877 atm with both values.
61304
Thy this website out: http://www.convertit.com/Go/ConvertIt/Measurement/Converter.ASP
912617
P=P(0)*exp(-1.1e-4*H)\n\nP(0) is the pressure at sea level. Note it's temperature dependent and it assumes a stable Boltzmann type distribution.