You haven't provided a specific question about the sphere. However, I can tell you that the decrease in the sphere's radius is due to the increased pressure at that depth in the sea. The pressure at a certain depth in a fluid is given by the equation P = ρgh, where ρ is the fluid density, g is the acceleration due to gravity, and h is the depth.

If you want to know by how much the volume of the sphere has decreased, you can calculate the volume of the sphere before and after the decrease in radius and then find the difference. The volume V of a sphere is given by the equation V = 4/3πr³, where r is the radius of the sphere.

So, the initial volume of the sphere is V1 = 4/3π(2 cm)³ = 33.51 cm³.

The final volume of the sphere is V2 = 4/3π(1.98 cm)³ = 32.67 cm³.

The decrease in volume is ΔV = V1 - V2 = 33.51 cm³ - 32.67 cm³ = 0.84 cm³.

If you have a different question about the sphere, please provide more details.

Question

You haven't provided a specific question about the sphere. However, I can tell you that the decrease in the sphere's radius is due to the increased pressure at that depth in the sea. The pressure at a certain depth in a fluid is given by the equation P = ρgh, where ρ is the fluid density, g is the acceleration due to gravity, and h is the depth.

If you want to know by how much the volume of the sphere has decreased, you can calculate the volume of the sphere before and after the decrease in radius and then find the difference. The volume V of a sphere is given by the equation V = 4/3πr³, where r is the radius of the sphere.

So, the initial volume of the sphere is V1 = 4/3π(2 cm)³ = 33.51 cm³.

The final volume of the sphere is V2 = 4/3π(1.98 cm)³ = 32.67 cm³.

The decrease in volume is ΔV = V1 - V2 = 33.51 cm³ - 32.67 cm³ = 0.84 cm³.

If you have a different question about the sphere, please provide more details.

Knowee AI · Accepted Answer

You haven't provided a specific question about the sphere. However, I can tell you that the decrease in the sphere's radius is due to the increased pressure at that depth in the sea. The pressure at a certain depth in a fluid is given by the equation P = ρgh, where ρ is the fluid density, g is the acceleration due to gravity, and h is the depth.

If you want to know by how much the volume of the sphere has decreased, you can calculate the volume of the sphere before and after the decrease in radius and then find the difference. The volume V of a sphere is given by the equation V = 4/3πr³, where r is the radius of the sphere.

So, the initial volume of the sphere is V1 = 4/3π(2 cm)³ = 33.51 cm³.

The final volume of the sphere is V2 = 4/3π(1.98 cm)³ = 32.67 cm³.

The decrease in volume is ΔV = V1 - V2 = 33.51 cm³ - 32.67 cm³ = 0.84 cm³.

If you have a different question about the sphere, please provide more details.

A small sphere of radius 2cm is taken to a depth of 500m under the sea. If the sphere shrinks to a radius of 1.98 cm, then

Question

Solution

Similar Questions

Upgrade your grade with Knowee