# Question 7e5b1

Jun 5, 2017

Here's how I would do it.

#### Explanation:

You get the radius of the sphere from the volume of the displaced water.

$V = {\text{30 cm"^3 – "20 cm"^3 = "10 cm}}^{3}$

The formula for the volume $V$ of a sphere is

color(blue)(bar(ul(|color(white)(a/a) V = 4/3πr^3color(white)(a/a)|)))" "

where $r$ is the radius.

We can rearrange this formula to get

r = root3((3V)/(4π))

r = root3(("3 × 10 cm"^3)/(4π)) = root3("2.39 cm"^3) = "1.3 cm"

(b) Density of metal block

You get the density ρ of the metal block from its mass $m$ and volume $V$.

color(blue)(bar(ul(|color(white)(a/a)ρ = m/Vcolor(white)(a/a)|)))" "

Assume that $m = \text{39 g}$.

$V = {\text{35 cm"^3 - "30 cm"^3 = "5 cm}}^{3}$

ρ = m/V= "39 g"/"5 cm"^3 = "7.8 g/cm"^3

(c) Density of wood

You apparently removed the metal sphere for this experiment.

The original volume was ${\text{20 cm}}^{3}$.

The metal block had a volume of ${\text{5 cm}}^{3}$, so

${\text{Volume of water + metal = 25 cm}}^{3}$.

When you submerged the wooden block,

${\text{Volume of water + metal + wood = 40 cm}}^{3}$.

∴The volume of the wooden block is

$V = {\text{40 cm"^3 - "25 cm"^3 = "15 cm}}^{3}$

Assume that the mass of the wood is 7.5 g. Then

ρ = m/V = "7.5 g"/"15 cm"^3 = "0.50 g/cm"^3#