1. **Calculate the average density of the Earth.** The average density $\rho$ is defined as mass divided by volume: $$\rho = \frac{m}{V}$$ From the endpapers, the Earth's mass $m = 5.97 \times 10^{24}$ kg. The radius of the Earth $r = 6.37 \times 10^6$ m. The volume of a sphere is: $$V = \frac{4}{3}\pi r^3$$ Calculate volume: $$V = \frac{4}{3} \pi (6.37 \times 10^6)^3 = 1.08 \times 10^{21} \text{ m}^3$$ Calculate density: $$\rho = \frac{5.97 \times 10^{24}}{1.08 \times 10^{21}} \approx 5520 \text{ kg/m}^3$$ This value matches the Earth's average density listed in Table 14.1 which is approximately 5500 kg/m³. Typical granite density is about 2700 kg/m³, so Earth’s average density is roughly twice that of typical surface rock, indicating a denser core. 2. **Find the density of the standard kilogram (platinum-iridium cylinder).** Given dimensions: Height $h = 39.0$ mm = 0.0390 m, Diameter $d = 39.0$ mm = 0.0390 m, Radius $r = \frac{d}{2} = 0.0195$ m. Volume of cylinder: $$V = \pi r^2 h = \pi (0.0195)^2 (0.0390) \approx 4.67 \times 10^{-5} \text{ m}^3$$ Mass $m = 1$ kg (standard kilogram definition). Density: $$\rho = \frac{m}{V} = \frac{1}{4.67 \times 10^{-5}} \approx 2.14 \times 10^4 \text{ kg/m}^3$$ This density is consistent with platinum-iridium alloys. 3. **Calculate the mass of gold needed to cast the model car.** Given: Mass of iron model $m_{Fe} = 9.35$ kg, Density of iron $\rho_{Fe} = 7.87 \times 10^3$ kg/m³ (typical value), Density of gold $\rho_{Au} = 1.93 \times 10^4$ kg/m³. Volume of model (same for both metals): $$V = \frac{m_{Fe}}{\rho_{Fe}} = \frac{9.35}{7.87 \times 10^3} = 1.19 \times 10^{-3} \text{ m}^3$$ Mass of gold model: $$m_{Au} = \rho_{Au} V = 1.93 \times 10^4 \times 1.19 \times 10^{-3} = 23.0 \text{ kg}$$ 4. **Determine the density of a proton and compare it to lead's density.** Proton diameter $d = 2.4$ fm $= 2.4 \times 10^{-15}$ m, Radius $r = 1.2 \times 10^{-15}$ m, Mass $m = 1.67 \times 10^{-27}$ kg. Volume of proton (sphere): $$V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (1.2 \times 10^{-15})^3 = 7.24 \times 10^{-45} \text{ m}^3$$ Density: $$\rho = \frac{1.67 \times 10^{-27}}{7.24 \times 10^{-45}} = 2.31 \times 10^{17} \text{ kg/m}^3$$ Lead density from Table 14.1 is about $1.13 \times 10^4$ kg/m³. The proton’s density is enormously greater than lead’s density, illustrating the extreme compactness of atomic nuclei.