1. **Problem statement:** We have two spheres cut from uniform rock. Sphere 1 has radius $r_1 = 4.50$ cm. The mass of sphere 2 is five times the mass of sphere 1. We want to find the radius $r_2$ of sphere 2. 2. **Key concept:** The mass of a sphere made of uniform material is proportional to its volume. The volume of a sphere is given by $$V = \frac{4}{3}\pi r^3$$ 3. Let $m_1$, $m_2$ be masses of spheres 1 and 2 respectively. Since the rock is uniform, density $\rho$ is the same: $$m_1 = \rho V_1 = \rho \frac{4}{3}\pi r_1^3$$ $$m_2 = \rho V_2 = \rho \frac{4}{3}\pi r_2^3$$ 4. Given mass relation: $$m_2 = 5 m_1$$ Substitute masses: $$\rho \frac{4}{3}\pi r_2^3 = 5 \left( \rho \frac{4}{3}\pi r_1^3 \right)$$ 5. Cancel common factors $\rho$, $\frac{4}{3}\pi$: $$r_2^3 = 5 r_1^3$$ 6. Take cube root on both sides: $$r_2 = \sqrt[3]{5} \times r_1$$ 7. Calculate numerical value: $$r_2 = \sqrt[3]{5} \times 4.50 \approx 1.71 \times 4.50 = 7.70$$ cm **Final answer:** The radius of the second sphere is approximately $7.70$ cm.