Question

Water has a density of 1.0 g/cm3 and about 25 percent of the block is below water. What is a good estimate of the density of the block? A 0.25 g/cm3 B 0.50 g/cm3

C 0.75 g/cm3 D 1.0 g/cm3

Answer 1

0.25 g/cm^3

Step-by-step explanation:

Answer 2

A 0.25 g/cm3

Step-by-step explanation:

At equilibrium, the buoyant force acting on the block (B, upward) is equal to its weight (W, downward):

`B=W\\\rho_w V_(imm) g =  \rho_b V g`

where

`\rho_w` is the water density

`V_(imm)` is the part of the volume of the block immersed in the water

g is the acceleration of gravity

`\rho_b` is the density of the block

V is the volume of the block

Re-arranging the equation,

`\rho_b = (V_(imm))/(V)\rho_w`

where we know:

`(V_(imm))/(V)=0.25`, since the fraction of volume immersed is 25%

`\rho_w = 1.0 g/cm^3`

Substituting,

`\rho_b = (0.25)(1.0)=0.25 g/cm^3`