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a hollow plastic sphere is held below the surface of a freshwater lake by a cord anchored to the bottom of the lake. the sphere has a volume of 0.650 m3 and tension in the cord is 1120 n. (a) calculate the buoyant force exerted by the water on the sphere. (b) what is the mass of the sphere? (c) the cord breaks and the sphere rises to the surface. when the sphere comes to rest, what fraction of its volume is submerged?

User Ben Wilson
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Final answer:

(a) The buoyant force exerted by the water on the sphere is 6,370 N.

(b) The mass of the sphere is 114.29 kg.

(c) The fraction of its volume is submerged is 1.

Step-by-step explanation:

(a) To calculate the buoyant force exerted by the water on the sphere, we can use Archimedes' principle, which states that the buoyant force is equal to the weight of the fluid displaced by the object.

In this case, the fluid is water. The volume of the sphere is 0.650 m³ and the density of water is approximately 1000 kg/m³. Therefore, the weight of the water displaced is:

Weight = Density × Volume = 1000 kg/m³ × 0.650 m³

= 650 kg

Since the weight of the water displaced is equal to the buoyant force, the buoyant force exerted by the water on the sphere is 650 kg × 9.8 m/s² (acceleration due to gravity) = 6,370 N.

(b) The mass of the sphere can be found using the equation:

Mass = Weight / Acceleration due to gravity = 1120 N / 9.8 m/s²

= 114.29 kg.

(c) When the cord breaks and the sphere rises to the surface, it becomes fully submerged. The fraction of its volume submerged is equal to the ratio of the volume of the water displaced to the volume of the sphere.

Volume of water displaced = Volume of sphere = 0.650 m³

Fraction of volume submerged = Volume of water displaced / Volume of sphere

= 0.650 m³ / 0.650 m³

= 1.