Question

1.) A thin sheet of gold foil has an area of 3.12 cm^2 and weighs 6.5 mg. how thick is the sheet?
Rho(Gold)= 19300 kg/m^3

2.) An irregular piece of metal weighs 10 g in air & 8.0 g when submerged in H2O.
a.) Determine the volume of the metal & its density
b.) If the same piece of metal weighs 8.5 g when immersed in a particular oil, what is the
density of the oil?

Answer 1

Answer on Question #73254 - Physics / Other

1) A thin sheet of gold foil has an area of $A = 3.12 \, \text{cm}^2$ and weighs $m = 6.5 \, \text{mg}$. How thick is the sheet? ($\rho = 19300 \, \text{kg/m}^3$)

Solution:

The weight of gold foil $m = \rho V = \rho Ah$. Thus, the thick of foil is

$$h = \frac{m}{\rho A} = \frac{6.5 \times 10^{-3}}{19300 \times 3.12 \times 10^{-4}} = 0.001 \, \text{m} = 1 \, \text{mm}$$

Answer: 1 mm

2) An irregular piece of metal weighs $m = 10 \, \text{g}$ in air & $m' = 8.0 \, \text{g}$ when submerged in $\mathrm{H}_2\mathrm{O}$.

a.) Determine the volume of the metal & its density

b.) If the same piece of metal weighs $m'' = 8.5 \, \text{g}$ when immersed in a particular oil, what is the density of the oil?

Solution:

The weight of piece of metal in $\mathrm{H}_2\mathrm{O}$

$$m' g = m g - \rho_w g V$$

Thus, the volume of a piece of metal

$$V = \frac{m - m'}{\rho_w} = \frac{10 - 8.0}{1.0} = 2.0 \, \text{cm}^3$$

The density of a piece of metal

$$\rho_{\text{metal}} = \frac{m}{V} = \frac{10}{2} = 5.0 \, \text{g/cm}^3$$

The weight of piece of metal in oil

$$m'' g = m g - \rho_{\text{oil}} g V$$

So

$$\rho_{\text{oil}} = \frac{m - m''}{V} = \frac{10 - 8.5}{2.0} = 0.75 \, \text{g/cm}^3$$

Answers:

$V = 2.0 \, \text{cm}^3$

$\rho_{\text{metal}} = 5.0 \, \frac{\text{g}}{\text{cm}^3}$

$\rho_{\text{oil}} = 0.75 \, \text{g/cm}^3$

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