A small sphere of wood with a density $\rho$ = 0.40 g/cm³ is held at rest well under the surface of a pool of water.The magnitude of the initial acceleration of the sphere when it is released is A)15 m/s² B)9.8 m/s² C)33 m/s² D)23 m/s² E)3.4 m/s²

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Quizplus · Accepted Answer

The correct answer is A because the initial acceleration of the sphere is equal to the acceleration due to gravity minus the buoyant force. The buoyant force is equal to the weight of the water displaced by the sphere, which is equal to the volume of the sphere times the density of water. The density of water is 1 g/cm³, so the buoyant force is equal to the volume of the sphere times 1 g/cm³. The volume of the sphere is equal to 4/3πr³, where r is the radius of the sphere. The radius of the sphere is not given, but it can be found using the density of the sphere. The density of the sphere is equal to the mass of the sphere divided by the volume of the sphere. The mass of the sphere is not given, but it can be found using the density of the sphere and the volume of the sphere. The density of the sphere is 0.40 g/cm³, and the volume of the sphere is 4/3πr³. Therefore, the mass of the sphere is equal to 0.40 g/cm³ times 4/3πr³. The acceleration due to gravity is 9.8 m/s². Therefore, the initial acceleration of the sphere is equal to 9.8 m/s² minus the buoyant force, which is equal to 0.40 g/cm³ times 4/3πr³ times 1 g/cm³. The initial acceleration of the sphere is therefore equal to 9.8 m/s² minus 1.33 g/cm³ times r³. The radius of the sphere is not given, but it is assumed to be small. Therefore, the initial acceleration of the sphere is approximately equal to 9.8 m/s² minus 1.33 g/cm³ times r³, which is approximately equal to 15 m/s².

A Small Sphere of Wood with a Density ρ\rhoρ = 040 G/cm3 Is Held at Rest Well Under the Surface

A Small Sphere of Wood with a Density $\rho$ = 040 G/cm³ Is Held at Rest Well Under the Surface