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what is the magnitude of bouyant force that a 2.1 kg solid gold ball would expereince when fully submerged

User Predictor
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Final answer:

The magnitude of the buoyant force that a 2.1 kg solid gold ball would experience when fully submerged can be calculated using Archimedes' principle. It is approximately 2.0 x 10^3 N.

Step-by-step explanation:

The magnitude of the buoyant force that a 2.1 kg solid gold ball would experience when fully submerged can be calculated using Archimedes' principle. According to Archimedes' principle, the buoyant force is equal to the weight of the water displaced by an object.

In this case, the buoyant force can be calculated as:

Buoyant force = weight of water displaced

Given that the density of water is approximately 1,000 kg/m³ and the volume of the gold ball can be calculated using its mass and density, we can determine the weight of the water displaced. Finally, by multiplying this weight by the acceleration due to gravity (9.8 m/s²), we can find the magnitude of the buoyant force.

Therefore, the magnitude of the buoyant force that a 2.1 kg solid gold ball would experience when fully submerged is approximately 2.0 x 103 N.

User Jschnurr
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