How far up in the sky can a balloon filled with helium go?
94588
It depends on various factors but most important would be pressure. As the balloon ascends into the upper atmosphere pressure is slowly reduced as the air gets thinner farther up the balloon goes, Pressure within the balloon where the helium is held soon expands out as the air molecules in the atmosphere reduces and at one point it will burst as the helium expands past the stretchability limit of the balloon. So it depends on how strong the balloon fiber is, how much helium is stored within the balloon, how thin is the surrounding atmosphere compared to pressure within the balloon and also if the balloon meets any obstruction on its way up.
19534
ok, so lets calculate this.  a balloon with a foot diameter has a volume of .7854 cubic feet. .5 cubic feet converts into 14.158423356 liters.  The density of helium is .1785 g/L (grams per liter) and the density of air is usually approximately 1.25 g/L.  The difference between these densities is 1.07g/L.  The formula for density is d=m/V (density equals mass over volume)  so we set d=m/V, 1.07=m/14.158423356.  the 1.07 was the difference in density(how much less dense helium is than air) and the V i said(14.158423356) is how many Liters i said was in .5 cubic feet before.  it has to be in liters since the density measurement i gave you was in grams per liters, so for the formula to work we need to work in liters and receive an answer in grams.  ok, so solving for m in that density formula, we get 15.95g.  so each balloon can hold about 16g.  but i looked up the average weight of a regular party balloon and it's 2g, so each balloon can hold 14 more grams.  converting grams to pounds, 14 grams equals .030864717 pounds.  so each balloon can hold .030864717 pounds.  now if we divide your weight by the amount of weight each balloon can hold, we will get the number of balloons you need to support your weight. 160/.030864717 = 5183.9 so 5,184 balloons. \n\nyou would need 5,184 balloons to equal your weight.  this only makes you weightless, so the force of these balloons would basically perfectly counteract the force of gravity, so you would need to jump to push off and you wouldnt be pulled back down. if you wanted the balloons to actually lift you, you would need more, and depending on how many more, you would have a greater acceleration upward.  this is also assuming that each balloon is perfectly one foot in diameter, so the number of balloons may fluctuate a bit.\n\neven so, might i suggest you use a balloon with a larger diameter?\n\nif you want to find out for sure with specific balloons, you can always get like 50 balloons and some weights.  then you can measure how much weight the 50 balloons hold. then you set up a proportion saying that 50 balloons divided by the weight you measure it can hold(in pounds) equals X balloons divided by 160 lbs. then you can solve more precisely for the specific balloons you have.
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
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
190543
That depends upon the location of the air and the mass of the balloon compared to its volume.\n\nAt extreme altitudes the air is less dense than the helium (if it is at room temperature) and a helium filled balloon will sink.  This is also true if the mass of the balloon plus the mass of the helium is more than the buoyancy provided by the volume of the balloon.\n\nUnder “normal” circumstances, with a typical latex or Mylar balloon filled with helium, the difference between the mass of the helium in the volume of the balloon and a similar volume of air (which is denser) is what provides the “lift.”  This is the same reason that most wood floats on water.
611090
a bit over 29,000 feet high
140469
The helium balloon will continue to travel forward until it collides with the right side of the van that is now moving towards the balloon.  Once the side of the van and the balloon meet, the balloon will travel to the left.\n\nFrom the perspective of someone in the van, the balloon will APPEAR to move to the right.
310208
The balloon rises in atmosphere and remains intact till the point when its internal pressure is lower than the atmospheric pressure. At higher altitude where the atmospheric pressure is lower and the pressure inside the balloon is higher, it bursts as the rubber cannot withstand the internal pressure.
678477
9.8 x .020 will give you the gravitational force between the balloon and the earth.  The electrostatic force between the balloon and the ceiling could be different but must be at least 9.8 x .020 (otherwise the balloon would fall).  If the electrostatic force was larger than the gravitational force, there would be a normal force from the ceiling pushing down on the balloon (otherwise the balloon would try to calapse into the ceiling).
404977
Hahahahahahaha, how amusing the previous anwers are.\n\nThe balloon would continue to rise in the atmosphere until it reached a height where the pressure inside the balloon was equal to the pressure outside the balloon...then it would remain at the same height, moving in the wind, maybe slightly up and down due to thermals...until it burst !
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.
240987
put the opening of the balloon on the spigot of the helium tank.  they are usually made of rubber.  when you tip the spigot to one side, helium comes out.
459122
the side length is 2\nso the altitude should be 3^(1/2)\ngood luck
707890
If you drop a vertical line h representing the balloon's altitude, you have two right triangles.  Let the base of the 83 degree triangle be x.  Now here's where you run into a question and if the book shows a diagram, it might answer that question.  Are the stations looking toward a balloon in between them, or are they both looking in the same direction at a balloon very much farther away, and very much higher?  If it's between them, the base of the 57 degree triangle will be 15 km-x, but if it's on the same side of them both, that base will be 15 km+x.  Do you know the trick to remembering how the sides relate to the trig functions?  Ooh--aah--oh ah--ah oh--ha--ho.  You have to sound this out and it makes a distinctive but stupid sounding little chant.  For the sides opposite an angle, adjacent to the angle and the hypotenuse in a right triangle, sin=o/h, cos=a/h, tan=o/a, cot=a/o, sec=h/a, csc=h/o.  In this case we want the ratio of the height to the base, the opposite side being the height and the adjacent side being the base, so we want the tangents of the angles.  The altitude, opposite side to the angles, will be equal for both triangles.  Call it y.  x will be the base for the 83 degree triangle in both cases, but for the 57 degree triangle the base will be (15 km-x) if the balloon is between the stations or (15 km+x) if it's to the same side of both stations.  Tan 83=h/x and tan 57=h/(15-x) (between) or tan 57=h/(15+x) (same side).  Setting the heights of the triangles equal since the balloon can only be at one height: x tan 83=(15-x) tan 57 if between, x tan 83=(15+x) tan 57 if to the same side.  Solve for x.  Then plug x back into the earlier equations to get y, the altitude.  Work out the answer on your own, but I'll give you this: if the balloon is between the stations, x is between 2 and 3 and y is between 15 and 20.  If the balloon is to one side of both the stations, x is between 3 and 4 and y is between 25 and 30.
378440
carnival baloons reach a specific hieght at which they stop verical movement and start to float around losing gas and eventually coming down to earth. i dont know why hey stop movung up but my guess is that they stop when the thickness of air at that altitude is less than the density of helium.
351878
They pop. The higher the altitude, the lower the atmospheric pressure outside the balloon. But the same amount of gas is still inside the balloon, and it wants to be the same pressure as the air outside the balloon. Thus, the gas inside expands and pops your balloon.
686605
Helium is lighter than air and cause the ball to float away.\n\nCarbon Dioxide is heavier than air so the ball would not bounce so well and it would be harder to kick.\n\nOxygen is lighter than air, but the difference is so small you wouldn't notice, no more that a ball being slightly over/under inflated.\n\nSince air is the cheapest, at a cost of nil, I suggest you put that in you balls.