air-filled baloon is weighted (added a weight at the bottom) so it sink in water. near the surface the balloon has a certain volume. at the bottom, is the balloon smaller or larger or same in size? as the weighted balloon sinks, what happens to its density? what happens to its buoyant force?
2006-12-17 09:14:49 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
As the balloon sinks, the deeper water has a higher pressure. The pressure of the air in the balloon has to become higher to balance the water pressure. For that to happen, the ballon's volume needs to decrease. (Same air in the balloon, same temperature, less volume means more pressure.)
Density = Mass/volume. Mass won't change, but volume does.
Buoyant force: if the volume of the water displaced decreases, the buoyant force decreases.
2006-12-17 14:58:57 · answer #1 · answered by sojsail 7 · 0⤊ 0⤋
You really should do your own homework.
The balloon is full of air, and it must remain in pressure equilibrium as it sinks. How can you increase the pressure of a fixed amount of air? What must happen to the volume?
2006-12-17 09:20:24 · answer #2 · answered by cosmo 7 · 0⤊ 0⤋
i'm reading your project as pointing out that: The mass OF THE HELIUM is a million kg and The mass of it relatively is field is negligible. If it floats, then the mass to volume ratio (density) is the comparable for the two the air and the balloon. a million.29 kg /a million m^3 = a million kg / x m^3 x = a million/a million.29 x ? 0.7752 .
2016-10-15 03:23:20 · answer #3 · answered by ? 4 · 0⤊ 0⤋