- Gold has a mass of 50 Kg , occupying a volume of \( 10 \mathrm{~cm}^{\wedge} 3 \) - Calculate the density

Given: - Mass: \( m = 50 \, \mathrm{kg} \) - Volume: \( V = 10 \, \mathrm{cm}^{3} \) **Step 1. Write the definition of density.** The density \( \rho \) is defined by \[ \rho = \frac{m}{V}. \] **Step 2. Convert the volume from cubic centimeters to cubic meters.** Since \[ 1 \, \mathrm{cm}^{3} = 10^{-6} \, \mathrm{m}^{3}, \] the volume in cubic meters is \[ V = 10 \, \mathrm{cm}^{3} = 10 \times 10^{-6} \, \mathrm{m}^{3} = 10^{-5} \, \mathrm{m}^{3}. \] **Step 3. Calculate the density using consistent units.** Substitute the values into the density formula: \[ \rho = \frac{50 \, \mathrm{kg}}{10^{-5} \, \mathrm{m}^{3}}. \] Perform the division: \[ \rho = 50 \times 10^{5} \, \mathrm{kg/m}^{3} = 5 \times 10^{6} \, \mathrm{kg/m}^{3}. \] **Final Answer:** The density of gold is \[ \boxed{5 \times 10^{6} \, \mathrm{kg/m}^{3}}. \]