Asphere completely submerged in water is tethered to the bottom with a string. the tension in the string is one-fourth the weight of the sphere.

800 kg/m³

Explanation:

I assume you want to find the density of the sphere?

Start with a free body diagram.  There are three forces acting on the sphere: gravity pulling the sphere down, buoyancy pushing the sphere up, and tension pulling the sphere down.

Applying Newton's second law:

∑F = ma

B - W - T = ma

Since the sphere isn't accelerating, a = 0.

B - W - T = 0

B = W + T

We know that the tension is one-fourth the weight:

B = W + W/4

B = 5/4 W

B = 5/4 mg

Buoyant force is defined as:

B = ρVg,

where ρ is the density of the fluid, V is the displaced volume, and g is acceleration of gravity.

ρVg = 5/4 mg

ρV = 5/4 m

The mass of the sphere is equal to its density times its volume.  Since the sphere is fully submerged, it's volume is the same as the volume of the displaced water.

ρV = 5/4 ρₓV

ρ = 5/4 ρₓ

ρₓ = 4/5 ρ

So the density of the sphere is 4/5 the density of the water.  Water's density is 1000 kg/m³, so:

ρₓ = 4/5 (1000 kg/m³)

ρₓ = 800 kg/m³

pretty sure it's false

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2