# Archimedes principle. The weight of the bouy is given, so the weight of water displaced by the submerged part of the bouy equals that. Figure the volume that weight of water displaces, then figure the height of a .33m diameter column of water will have of that volume.

A large buoy used to support an oceanographic research instrument is made of a cylindrical, hollow iron tank. The tank is 2.1 m in height and 0.33 m in diameter. The total mass of the buoy and the research instrument is about 120 kg. The buoy must float so that one end is above the water to support a radio transmitter. Assuming that the mass of the buoy is evenly distributed, how much of the buoy will be above the waterline when it is floating?

8 months ago

## To determine how much of the buoy will be above the waterline when it is floating, we can apply Archimedes' principle. According to this principle, the buoyant force exerted on an object immersed in a fluid is equal to the weight of the fluid displaced by the object.

Here's how you can approach this problem:

1. Find the weight of the buoy:
Since the total mass of the buoy and the research instrument is given as 120 kg, you can determine the weight using the formula: weight = mass Ã— acceleration due to gravity.
weight = 120 kg Ã— 9.8 m/sÂ²
weight = 1176 N

2. Calculate the volume of water displaced by the submerged part of the buoy:
The buoy is in the shape of a cylindrical, hollow iron tank. To calculate its volume, use the formula for the volume of a cylinder:
volume = Ï€ Ã— rÂ² Ã— h
where r is the radius and h is the height.
Since the diameter is provided (0.33 m), the radius can be found by dividing the diameter by 2:
r = 0.33 m Ã· 2
r = 0.165 m
Now, substitute the radius and height values into the volume formula:
volume = Ï€ Ã— (0.165 m)Â² Ã— 2.1 m
volume = 0.553 mÂ³

3. Determine the height of a column of water:
To find the height of a column of water with a diameter of 0.33 m that has the same volume as the water displaced, divide the volume by the cross-sectional area of the column:
height = volume / area
To find the area, use the formula for the area of a circle:
area = Ï€ Ã— rÂ²
area = Ï€ Ã— (0.165 m)Â²
Now substitute the values into the height formula:
height = 0.553 mÂ³ / (Ï€ Ã— (0.165 m)Â²)
height â‰ˆ 6.59 m

Therefore, when the buoy is floating, approximately 6.59 meters of it will be above the waterline to support the radio transmitter.

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