Physics question (helium balloon)?
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
976120
Helium molecules are very small and slowly leak through the balloon wall.\n\nAir contains a mixture of gaseous molecules, primarily Nitrogen and Oxygen.  These are much larger than Helium.
822839
The net force is the volume of the balloon times the density difference between air and hydrogen (∆d) times g. From archimedes principle: the bouyant force is dair*V*g, the weight of hydrogen is dhyd*V*g, so the net force is the difference, or (dair - dhyd)*V*g =∆d*V*g.   The volume of the baloon is (4/3)πr^3.\n\nr = cuberoot[3*F/(4π * ∆d * g)]\n\nThis comes out 4.89 m
243047
If you wanted to have the effect of being on the mooon (with 1/6 gravity), you would need enoug Helium to balance out 5/6 of your weight.\n\nA 200 pound man has a mass of 90.9 kiligrams.  It takes about .7 kg of Helium to lift 1 kg.  So to cancel out 5/6 of the force of gravity acting on you, you would need about 53 kg of Helium.\nOne mole of Helium has a mass of 4.00 grams.  1 mole of Helium at STP has a volume of about 22.4 Liters.  Thus, you would need about 296800 Liters (946.4 cubic feet) of Helium, this is quite a lot.
417118
The experiment which you have suggested (although incompletely) is a classic method of verifying Bernoulli's equation, and it consists of hanging two balls in the air by strings, reasonably close to each other and blowing air through the gap between them. \n\nSound as the logic behind the experiment may be, it is in fact quite simple to perform the experiment incorrectly. One way in which this can be done is by blowing directly at the balls, which has the undesirable effect of pushing them in a haphazard direction. Hence, while blowing, keep in mind that you do not let your blows land on the surfaces of the balls. The air should strictly pass through the gap. \n \nThe other reason could be heavy balls. \n\nYou are best off using pingpong balls, with their surfaces spaced about 5-7 centimeters ( 2-3 inches) apart.
108625
I guess the helium ballon will lift the same mass, not more. \nThis is because although the mass will weight less (1/6), also the ballon will have less lift force because this depends on the volume x density x g(moon), and g(moon) is 1/6 also, so the less weight counteracts with less lift force.
10432
The balloon is filled with hot air by the burner.  The hot air inside the balloon rises because it is lighter than cold (or cooler air).  assuming you mean't a hot air balloon.\n\nNobody is sure about helium ballons because we always suck the helium out so we can talk weird.
282829
This is similar to asking why gravity doesn't affect boats. It does, but buoyancy keeps them above water. Similarly, if you push a hollow rubber ball down to the bottom of a pail of water and simply release it, it will pop back up to the surface.\nThe helium balloon is just being pushed up by the relatively higher air pressure under it. See the sources for a detailed calculation and a diagram of the pressure situation.
315960
Properties of Gases \n\n1.  Indefinite shape -- you can squash an air-filled  balloon to make circus animals\n2.  May expand -- balloons get bigger when temperature gets warmer\n3.  May compress -- can put lots of air into a car tire\n4.  Have low densities -- balloons filled with air will float on water\n5.  Diffuse uniformly -- put a pizza on the front seat of the car and the people in the back seat get hungry pretty quickly -- that pizza vapors travel to fill the car! \n\n11.2  Atmospheric Pressure \n\nEven though we can't see the air around us, we know it's there and those molecules bouncing around exert a force on our bodies.  Fortunately, the force is the same on all sides of us, so it equalizes out and we don't notice that we're pushing against the air as we move our bodies. \n\nWe use a barometer to measure pressure.  We can use lots of different units depending on the system we're using.  Normal atmospheric pressure at sea level is defined as "1 atmosphere of pressure";  we call this by the special name "standard pressure".  The equivalent measures in other  units for standard pressure are given in table 11.1;  when we talk about standard pressure in this course we will use units of atmospheres, mm Hg and torr.  You're probably familiar with the weatherperson talking about pressure in inches of Hg.  If you use 2.54 cm = 1 inch, you can do the conversion from inches of Hg to cm and then on to mm of Hg.  There's no magic here, it's just a metric/English conversion example. \n\nYou may be curious about how a barometer works.  Here's my interpretation.... \n\nLet's start with the diagram on page 295 which is a representation for the workings of a barometer. It helps to talk about how this setup is put together. One would hold the tube "right side up", that is, like a test tube and fill it "to the brim" with mercury. One then inverts the tube in a pool of mercury (taking appropriate safety precautions). Immediately some of the mercury will flow out of the tube. How much flows out depends on how much pressure the atmosphere exerts on the pool of mercury. If the atmospheric pressure is higher, there is more pressure on the pool of mercury and the material in the tube stays at a higher level. If the atmospheric pressure is lower on a given day, more mercury is allowed to flow out of the tube. If the atmospheric pressure increases again, more mercury is pushed pack up into the tube. It is important to recognize that a virtual vacuum (except for a few stray mercury molecules) exists above the liquid in the tube which exerts no pressure (has no influence) on the height of the material in the tube. The only influence on the height of the material, is the air pressure on the puddle of mercury at the bottom.\n\nThe number the weatherperson gives is the height of the column measured in inches or mm (of mercury). We use mercury because it is so dense. If we were to use a water column instead (mercury is 13+ times as dense as water), the column would have to be 13+ times as high.  \n\n  Do example exercise  11.1 and the self test.  You can make up unit factors using any pair of these units to do homework exercises 5 and 7. \n\n11.3  Variables affecting Gas Pressure \n\nThink of gas pressure as the effect of molecules hitting the walls of a container.  When the molecules hit harder, the pressure is increased.  The tire pressure in your car can be affected by 3 things:\n \n  \n\nVariable What we measure Predicted effect on pressure of the gas Why? Also true (follows logically); complete with increase or decrease \nvolume of the container how much air the tire can contain increase volume, decrease pressure more space for the molecules to move around, so don't hit walls of container so hard decrease pressure, (       ) volume \ntemperature of the gas temperature of the air in the tire increase temperature, increase pressure molecules moving faster, so hit
899593
Hello,\n\nIn this link you will have what you need:\n\nhttp://www.bsu.edu/web/fseec/pie/Lessons2/Physics/Potato%20Battery.doc\n\nI believe now will be OK.
430468
First of all the distance is 20 METERS and not 20 METERS/SECOND. Also the unit of acceleration is m/sec*sec and not m/sec.\n\nNow, there are both horizontal and vertical speeds. Also the position is a vector that has both a horizontal factor and a vertical factor.\n\nTo find the horizontal speed:\nInitial speed: 0 m/sec.\nAcceleration: 5m/s^2\n=> Speed2 = Speed1 + Acc.* Time\n=> After 1 sec. speed= 0 + 1*5=5m/s\n=>After 2 secinds speed = 0 + 2*5= 10m/sec.\n\nTo find the vertical speed:\nInitial speed: 0 m/sec.\nAcceleration: -9.8m/s^2 [acceleration due to gravity]\n=> Speed2 = Speed1 + Acc.* Time\n=> After 1 sec. speed= 0 - 1*9.8=-9.8m/s\n=>After 2 seconds speed = 0 - 2*9.8= -19.6m/sec.\n\nTo find the horizontal position:\nDisplacement = (initial speed) * time + 0.5*acc.*time^2\n=> after 1 sec. displacement = 0 + 0.5*5 * 1 = 2.5 m\n=> position = 2.5 meters to the right of the initial horizontal position of the balloon\n\n=> after 2 seconds displacement = 0 + 0.5*5* 4 = 10 m\n=> position = 10 meters to the right of the initial horizontal position of the balloon\n\n\nTo find the vertical position:\nDisplacement = (initial speed) * time + 0.5*acc.*time^2\n=> after 1 sec. displacement = 0 + 0.5*-9.8 * 1 = -4.9 m\n=> position = 20[initial position] - 4.9 = 15.1 meters above the ground\n\n=> after 2 seconds displacement = 0 + 0.5*-9.8 * 4 = -19.6 m\n=> position = 20[initial position] - 19.6 = 0.4 meters above the ground
345743
i know!\n\nas temperature decreases, volume decreases. its part of the kinetic molecular theory.\n\nthe gas particles move slower because they lose energy as temperature decreases so they don't exert as much force on the walls of the balloon, so the balloon shrinks.
730471
Enrique_C is correct if there is no stretch in the balloon's material, like with a Mylar balloon.  With a latex balloon, it would take more helium, because it requires more pressure to stretch the balloon material.  This pressure will increase the density of the helium in the balloon, thus requiring more helium to fill the same space.  The smaller the balloon is when deflated, the more helium it will take to inflate it to 16ft.
379193
it has nothing to do with the air in the atmosphere its the air that fills the balloon up. when its hot the ballon goes up as it cools it goes down
485421
well u need to consider the formula of finding the wolume of a sphere. formula is 4/3*pi*r*r*r. so when the balloon is 2cm wide it has a radius,r, of 1cm. so v=kr*r*r. when the design is changed to a width of 10cm. the radius is now 5cm hence going under an increase of 5. the new radius is 5 times the first radius. so the nex volume=k5*5*5*r*r*r=125*k*r. so the new volume is 125*41=5125. (i have a query with the problem though since it did not follow the formula, 4/3*pi*r*r*r does not equal 41. usually a balloon even though is not a perfect circle is considered to follow this formula. if the balloon you are talking about is not a sphere shaped ballon then i would undersatnd why the formula does not apply and in that case the explanation i gave would be inacurate. check on that before you take my answer as right. i just gave you what i think is right based on what i understood of the problem. good luck.)
971254
v=sqrt(k*P/rho), where k is Adiabatic index, rho is density of helium\nfor monoatomic gases k=(3+2)/3=1.667\nrho=0.1786 g/L=0.1786kg/m^3 under standard conditions\nP=1atm=98 066 Pa\nso v=956.7m/s\n\nbut it's 970m/s from http://www.teachersparadise.com/ency/en/wikipedia/h/he/helium.html
629282
assuming the balloons are spherical and have no mass and the temperature is about 25 C then 418 balloon might just do it.