Question

A solid metallic having a volume of .3m^3 is completely submeged in water. The weight of the cube when submerged in water is 52300 N. Determine the kind of material the cube is made of. (density of gold= 19300 kg/m^3 , density of silver= 18200 kg/m^3 , density of copper= 17200 kg/m^3

Answer 1

Question#18741

A solid metallic having a volume of .3m^3 is completely submerged in water. The weight of the cube when submerged in water is 52300 N. Determine the kind of material the cube is made of. (density of gold= 19300 kg/m^3, density of silver= 18200 kg/m^3, density of copper= 17200 kg/m^3

Solution:

Let:

$$V = 0.3 \, \text{m}^3$$ $$P_{\text{in water}} = 52300$$ $$\rho_{Au} = 19300 \, \text{kg/m}^3, \rho_{Ag} = 18200 \, \text{kg/m}^3, \rho_{Cu} = 17200 \, \text{kg/m}^3$$

$\rho - ?$

The weight of metallic in water is:

$$P_{\text{in water}} = mg - F_A, \text{ where: } m \text{ - mass of metallic}, g = 9,8, F_A - \text{buoyant force (Archimedes' force)}$$ $$F_A = \rho_{\text{water}} g V$$ $$m = \frac{P_{\text{in water}} + F_A}{g} = \frac{P_{\text{in water}} + \rho_{\text{water}} g V}{g} = \frac{P_{\text{in water}}}{g} + \rho_{\text{water}} V$$

Such as: $m = \rho V, \rho = \frac{m}{V}$

$$\rho = \frac{\frac{P_{\text{in water}}}{g} + \rho_{\text{water}} V}{V} = \frac{P_{\text{in water}}}{g V} + \rho_{\text{water}}$$ $$\rho = \frac{52300}{9.8 \times 0.3} + 1000 = 18789.12 \, \text{kg/m}^3$$

Answer: the density of metallic is middle from gold and silver, maybe it is an alloy.