how many atmospheres in 2175 psi, approximately?
339441
1 atm of pressure is about 14.7 psi,\nso 2175 psi would be about 148 atm.
874775
ignor the atmospheric changes, it would be 4.8s if I'm not mistaken
865811
30.5 m of water.\n\nConvert units to desired units.\n\n1 atm is 14.?? psi is about 32 ft of water.\n\nLook in reference for more detailed unit conversions.\n\nOr use P= &#92;rho g H.
665192
I believe in the US we use millimeters of mercury and in other places they use millibars.\n\nAccording to my conversion chart one atomsphere is 76 centimeters of mercury and one bar is 0.9869 atmospheres.\n\none atomsphere is 1.0333 kgs/ sq cm or 14.7 pounds per square inch.
1145988
It is filled with air.  It has a circumference of 27-28 inches with a weight of 14-16 ounces.  It should have the air pressure of 8.5 to 15.6 PSI.
933654
Mass of moon\n-------------------- ------------------- =\nMass of earth's atmosphere\n\n73 800 000 000 000 000 000\n-------------------- ------------------- =\n5 700 000 000 000 000\n\n≈13000 (nearest thousand)
265928
1.27 X 1022 KG (0.0021 Earths)
1153522
The quantities you have listed are in fact related via the ideal gas law.  So, assuming you aren't in huge temperature or pressure extremes, you can use the ideal gas law with decent results.\n\nIdeal Gas Law\nPV = nRT\n\nwhere P = pressure (N/m^2)\nV = volume (m^3)\nn = number of moles of gas\nR = universal gas constant = 8.3145 J/(mol*K)\nT = temp (in Kelvins)\n\nSo, by looking at the equation you can not convert air pressure directly to volume without also knowing the temperature and number of moles of gas.
304941
As per ASTM requirement\n\nAt 3 days - 1740 psi\nAt 7 days - 2760 psi
548733
Pressure increases with depth ,each 10 meters 1 atmosphere\nto calculate the pressure for a certain depth use the formula.\nPa= (depth/10)+1\nwhere Pa is the pressure\nExample at 20 meters\nPa= (20/10)+1\nPa= 2+1= 3 Atmospheres
686168
That depends on the depth. For each 27.68" of liquid water, there is an increase of 1 psi. This would be affected slightly by anything disolved in the water, such as salt in the oceans. Water vapor in the atmosphere is also sometimes included in the term hydrosphere. Pressure at the Earth's surface is figured as 14.7 psi in common usage, which accounts for the difference in psi(a)
268223
Temperature is 59 degrees F\nbarometric pressure is 29.92 inches of mercury\nstill air
838967
I think the pressure is actually 2.3E5 Pa.  That would make more sense.
105912
1 atm. = 760.0 mM Hg = 760.0 Torr = 14.70 psi = 1.013 Bars = 1033 cM water = 101.325 kPa.\n\n745 mM Hg = 745/760 = 0.983 or so atmospheres.\n\nA great conversion program can be found at:
492280
The equation for this is (PV)/(nT).  Pressure must be in kPa, volume must be in L, and temperature must be in Kelvin.\n\nTherefore, 572 cm3 = 0.572 L, 725 mm = 96.635 kPa, and 130 C = 403 K.\n\nIn the first equation, n = 0.137 moles\n\nIf you input it into the second equation, where:\n240 C = 513 K\n720 mm = 97.168 kPa\n\nYou get V = 0.724 L, or 724 cm3.\n\nOf course, these are all approximations due to rounding.  If you want, you can try it with a calculator, but make sure the units are correct.  Also, 101.3 kPa = 760 mm.\n\nThis is used in basic chemistry, which I just took.
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.
315199
It is .433 psi per foot of height or .0361 psi per inch to three significant figures