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it return to the surface, stay at the depth to which it isd pushed or sink?

2007-05-06 12:33:31 · 4 answers · asked by Justin M 1 in Science & Mathematics Physics

4 answers

Assuming that "barely able to float" means that it still IS able to float, then it's density is still less than water's and it will return to the surface after being pushed below the surface.

However, if by "barely able to float" the questions means that the weighted balloon's density is the same as water's, then it will not return to the surface of the water; it will just continue to float at the depth that it was pushed to.

2007-05-06 12:41:39 · answer #1 · answered by JaniesTiredShoes 3 · 0 1

If the depth of the balloon is the only changed variable then the balloon would float upwards to the original depth at which it was introduced. It won't sink because the air in the balloon is still lighter than the water around and above it. The weight is what keeps it suspended at its initial depth. As you push the balloon down deeper water pressure increases and this causes the air pressure inside the balloon to increase. It's still the same amount of air so the density of it increases and the balloons volume of space decreases. It's buoyancy will actually increase the deeper you take it below the water and you will have to exert more force the deeper you try to submerge it. You could actually experiment with this yourself in a pool or at a swimming hole (a place where the water is relatively calm) if weather permits. Blow up a balloon and tie a basket, toss it in the water and record it's initial position with the weight of the basket on it. Add gravel, rocks, pebbles to the container until the balloon sinks to about six inches below the water and maintains its position. Push the balloon down several feet and see what happens. To lazy to try it then ask a science teacher to do it in class...

2016-05-17 06:11:59 · answer #2 · answered by ? 3 · 0 0

Presumably this is a regular novelty balloon. So when you push it under, the pressure will make the volume of the balloon decrease, so its density will increase, so the buoyancy will decrease, and it will sink. Submarines are not flexible like balloons.

2007-05-06 14:36:42 · answer #3 · answered by sojsail 7 · 1 0

It has all to do with displacement. If the baloon in negative, positive or neutral regarding displacement then it will act accordingly.
Negative....will sink
Neutral.......remain in position
positive.....return to the surface.
A submarine uses this principle to dive, stay underwater and return to the surface. It has ballast tanks which it floods and this gives it negative displacement and it can dive. The submarine trims its ballast at Neutral and it is able to stay level and use its planer (wings) to "fly" underwater. When it wants to surface it blows its ballast, (forces water out of the tanks) causing positive displacement (bouyancy) and it can return to the surface and float. A submarine uses torpedoes to punch holes in other ships, turning their displacement from positive to negative and sinking them. But submarines rarely attack baloons :-)

2007-05-06 13:36:41 · answer #4 · answered by WYNNER01 5 · 0 1

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