Question #7e5b1

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Answer 1 · 2017-06-05T00:10:45

Here's how I would do it.

(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

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