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

Answer:

A 0.25 g/cm3

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,

$\rho_b = \frac{V_{imm}}{V}\rho_w$

where we know:

$\frac{V_{imm}}{V}=0.25$, since the fraction of volume immersed is 25%

$\rho_w = 1.0 g/cm^3$

Substituting,

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

Answer 2

Answer:

0.25 g/cm^3

Explanation: