1. **State the problem**: We have two spheres made of the same uniform rock. The first sphere has a radius of $4.50$ cm. The mass of the second sphere is five times the mass of the first. 2. **Recall the formula for the volume of a sphere**: The volume $V$ of a sphere with radius $r$ is given by the formula $$V=\frac{4}{3}\pi r^3.$$ Since the rock is uniform, mass is proportional to volume. 3. **Set up the relationship between masses and radii**: Let the radius of the second sphere be $r_2$. Since mass is proportional to volume, we have: $$\frac{\text{mass of second sphere}}{\text{mass of first sphere}}=\frac{V_2}{V_1}=\frac{\frac{4}{3}\pi r_2^3}{\frac{4}{3}\pi (4.50)^3}=\frac{r_2^3}{(4.50)^3}.$$ Given the mass of the second sphere is five times the first, we write: $$5=\frac{r_2^3}{(4.50)^3}.$$ 4. **Solve for $r_2$**: $$r_2^3=5 \times (4.50)^3.$$ Calculate $(4.50)^3$: $$4.50^3=4.50 \times 4.50 \times 4.50=91.125.$$ Therefore, $$r_2^3=5 \times 91.125=455.625.$$ Take the cube root: $$r_2=\sqrt[3]{455.625}.$$ 5. **Calculate the cube root**: $$r_2\approx 7.7\text{ cm}.$$ **Final answer**: The radius of the second sphere is approximately $7.7$ cm.