Question

Question #7e5b1

Answer 1

Here's how I would do it.

Explanation:

(a) Radius of sphere

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

`V = "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 = "39 g"`.

`V = "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 `"20 cm"^3`.

The metal block had a volume of `"5 cm"^3`, so

`"Volume of water + metal = 25 cm"^3`.

When you submerged the wooden block,

`"Volume of water + metal + wood = 40 cm"^3`.

∴The volume of the wooden block is

`V = "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`