how many sq. feet of helium are required to lift one pound?
931678
The density of helium is 0.1786 g/L, and the density of air is 1.2 kg/m^3.  With proper unit conversion, the density difference is 0.064lb/ft^3.  Therefore it takes 15.7 ft^3 of helium to lift one pound.
355695
2000 lbs = 1 ton; .0005 tons = 1 lb
595563
Yes, it will weigh less. Fill ANY container with lighter-than-air gas, and it will weigh less.\n\nException: If the gas is under high pressure, such as in a pressurized tank, the container will actually weigh MORE.\n\nReason: Gas is compressible. If helium were half the weight of normal air, then helium under twice the pressure will weigh the SAME AS normal air under normal pressure. So if the pressure on the helium were increased, the container will weigh MORE.
536322
a tv program showed that approximately 2600 balloons filled with helium is needed to lift 20 kg child a few centimeters of the ground. (if i remember right, of course)
541177
A mole is 6.23 x 10^23 of something.  That number is Avogadro's number.  So a mole of Helium atoms is 6.23 x 10^23.  The weight listed on the periodic table is also the weight of a mole of atoms of a pure element in grams.\n\nThey are used because it gives us a nice round number to work with.  Some reactions require 2 of one molecule and 3 of another for example.  Well, 2 molecules and 3 molecules are hard to work with.  So instead, we could use 2 and 3 moles of the compounds and do the reaction, without having to worry about the molecular weights as much or densities or other physical properties.  Hope that makes sense.
293575
1 POUND PER SQUARE CENTIMETER
344744
I went to Nasa's website, which has an explanation about this. I calculated it out and it turned out to be 145 pounds, rounded to the nearest pound. I myself do not exactly understand the equation fully, but I do understand the concept. Hope this helps!
223066
foot pounds, 1 foot pound is the amount of force it takes to move a mass of 1 pound a distance of one foot. A newton is the equivalent measure in the metrc system.
530965
Pounds per square inch. Tensile Strength is expressed as the maximum load that a given material can withstand before failing. So the tensile strength of a high grade of steel might be 100,000 psi. but some steels are much lower. So if you are wanting to know what the maximum load is that you can place on the high grade of steel before it breaks, you will have to calculate what the smallest cross sectional area is that the load is being applied to. So if you have a rod that is .500" diameter the cross sectional area is 3.1416 X (Radius.250 squared)=.19635 square inches. So if 1 sq. in. will hold 100,000 Lbs. then you can multiply .19635 sq. in. by 100,000 psi to find pounds that .500" diameter rod will hold which is 19,635 Lbs. A .500" diameter bolt from the same material will hold less due to the reduced cross sectional area at the minor diameter of the threads on the bolt.  The diameter of a bolt always refers to it's nominal Major Diameter. The Minor Diameter has to be calculated, measured, or found in a table. In engineering practice a safety factor (increased cross sectional area) is added which may range from 200% to 1000% or more depending on the application. Material fatigue, vibratory loading, dynamic loading, corrosion, etc. and the possible dangers to people and property from under-engineering must always be taken into consideration in addition to this simple calculation for safety's sake.
315199
It is .433 psi per foot of height or .0361 psi per inch to three significant figures
367999
It weighs the amount that the air inside it weighs.
572696
density of pure water is 62.4 lbs/sq ft.\n\nSo, load = dAgh\n\nor A = 40000/(62.4*5*32) = 4.01 sq ft.
349993
1 Kip = 1 Kilo pound = 1,000 pounds
828487
Archimedes principle for buoyancy is:\nFb = (mass of the surrounding fluid)*(gravity)\nFb = buoyant force\nThe mass of the surrounding fluid is air and can then be calculated as:\nmass of surrounding fluid = (Density of Air)*(Volume of Balloon)\nThe buoyant force then can be written as\nFb = (Density of Air)*(Volume of Balloon)*(Gravity)\n\nThe opposing force to buoyancy is the gravity force and can be written for the helium balloon as:\nFg = (mass of helium + load of balloon)*(gravity)\nMass of Helium = (Density of Helium)*(Volume of Balloon)\nSo, then the equation looks like:\nFg = (Density of Helium)*(Volume of Balloon)*(gravity)\n+ (load of balloon)*(gravity)\n\nWhen a balloon is going up it has a larger buoyant force upward than the gravity force (Fg) downward.  When they are equal then the net force is zero and can be written as:\n\nFnet = Fb - Fg = 0\nor \nFb = Fg\nThe full equation then looks like:\n(Density of Air)*(Volume of Balloon)*(Gravity) = (Density of Helium)*(Volume of Balloon)*(gravity)\n+ (load of balloon)*(gravity)\n\nGravity cancels and the equation can then be algebraically arranged as: \n(Volume of Balloon) = (load of balloon)/(Density of Air - Density of Helium)\nor\nV= m / (Air density - Helium density)\n\nPlugging in the values that were stated in your problem we get:\n800kg/(1.29kg/m³ - 0.18kg/m³) = 721m³
1165094
It would depend on the amount of Helium within the truck.\n\nJust because Helium is lighter (less dense) than air does not mean that it does not have mass (and thus weight).\nFor a given volume of Helium and air at the same pressure, the Helium would have less mass.  However, if one pressurized the Helium and/or had enough moles of the gas, one could actually be increasing the overall weight of the car by adding the Helium.\nIn orer to lighten the car, the idea would be to displace the surrounding "fluid" (air) but at the same time, by adding as little additional mass as possibe.\n\nAny effects of Helium upon the overall weight of the vehicle will be totally negligible using any normal amount of Helium.  One would need a small blimp's worth of Helium to have any real effect.
463030
Conversion Calculator here\nhttp://education.yahoo.com/reference/weights_and_measures/index?p=volume
1063108
From one inch to 3/4  depending on length less than 500 feet and the demand for air at the station\n\n...\nno