A balloon filled with air has a string with a weight attached to it. It is placed in water. The weight is heavy enough to keep the balloon two feet below the surface of the water. The balloon stays two feet below the surface, without moving. It then is pushed down 18 inches. What happens to the whole system
Answers
Answer:
The system (Balloon + string + weight) would stay where you push it to.
In this case 2 feet + 18 inches = 3 1/2 below the surface of the water.
Explanation:
According to Archimedes Principle The buoyant force (upthrust) experienced by an object in water is equal to the weight of the water it displaces. In this case it's the density of water multiplied by the volume of the system.
If the system floats 2 feet under the water and doesn't sink or rise it is neutrally buoyant. Which means the average density (mass / volume)of the system is the same as water. Therefore, when it is pushed further down it will simply rest in this new location.
It is tempting to think that pushing something further into water will increase the upthrust, but this is just when the object is partially in the water. In that case pushing it down displaces more and more water so the upthrust increases. Once the object is fully immersed the upthrust is constant regardless of how far down in the water.