1. **State the problem:** A metal cube with side length 10.0 cm weighs 29.4 N in air and 19.6 N when submerged in a fluid. We need to find the density of the fluid. 2. **Relevant formulas:** - Weight in air: $W_{air} = mg$ - Apparent weight in fluid: $W_{fluid} = W_{air} - F_b$ - Buoyant force: $F_b = \rho_{fluid} V g$ Where $\rho_{fluid}$ is the fluid density, $V$ is the volume of the cube, and $g$ is acceleration due to gravity. 3. **Calculate volume of the cube:** Side length $= 10.0$ cm $= 0.10$ m $$V = (0.10)^3 = 0.001 \text{ m}^3$$ 4. **Calculate buoyant force:** $$F_b = W_{air} - W_{fluid} = 29.4 - 19.6 = 9.8 \text{ N}$$ 5. **Calculate fluid density:** Using $F_b = \rho_{fluid} V g$, solve for $\rho_{fluid}$: $$\rho_{fluid} = \frac{F_b}{V g}$$ Assuming $g = 9.8$ m/s$^2$: $$\rho_{fluid} = \frac{9.8}{0.001 \times 9.8} = \frac{9.8}{0.0098} = 1000 \text{ kg/m}^3$$ **Final answer:** The density of the fluid is $1000$ kg/m$^3$.