2007-03-26 10:21:25 · 2 answers · asked by AxSr89 2 in Science & Mathematics ➔ Earth Sciences & Geology
If the balloon was blow up fast, there would be no difference.
The only change would be the water cooling or heating the air in the balloon. The cold water would shrink the diameter (higher density), the hot water would expand it (lower density).
If you pulled both out, and they had the exact same volume, then there would be no difference in density, because the air's density came from and was controlled by you.
2007-03-26 10:32:24 · answer #1 · answered by QFL 24-7 6 · 0⤊ 0⤋
Cold water.
The hot water would cause the air to expand and the tension in the balloon wall to decrease. Both changes mean the balloon will increase in volume while its mass stays the same.
2007-03-26 17:27:14 · answer #2 · answered by novangelis 7 · 0⤊ 0⤋