Objective: Calculate the density of a metallic cube and express it in the format $X \times 10^3 \text{ kg m}^{-3}$, adhering to significant figure rules.

Density Calculation Steps

1. Convert Units and Calculate Volume:
   • Given mass $m = 5.580 \text{ kg}$. This has 4 significant figures.
   • Given side length $s = 9.0 \text{ cm}$. This has 2 significant figures.
   • Convert side length to meters: $s = 9.0 \text{ cm} = 0.090 \text{ m}$.
   • Calculate the volume $V$: $V = s^3 = (0.090 \text{ m})^3 = 0.000729 \text{ m}^3$.
   • Since the side length measurement ($9.0 \text{ m}$) has 2 significant figures, the calculated volume should be rounded to 2 significant figures: $V \approx 0.00073 \text{ m}^3$ or $7.3 \times 10^{-4} \text{ m}^3$.
2. Calculate Density:
   • Use the density formula: $\rho = \frac{m}{V}$.
   • Substitute the values: $\rho = \frac{5.580 \text{ kg}}{0.00073 \text{ m}^3}$.
   • Perform the division: $\rho \approx 7643.83... \text{ kg m}^{-3}$.
   • The result must be rounded to 2 significant figures, as determined by the least precise measurement (side length).
   • Rounding $7643.83...$ to 2 significant figures: The first two digits are 7 and 6. The next digit is 4, which is less than 5, so we keep the '6'.
   • Therefore, $\rho \approx 7600 \text{ kg m}^{-3}$.
3. Determine the value of X:
   • Express the calculated density in the required format: $\rho = X \times 10^3 \text{ kg m}^{-3}$.
   • $7600 \text{ kg m}^{-3}$ can be written as $7.6 \times 1000 \text{ kg m}^{-3}$, which is $7.6 \times 10^3 \text{ kg m}^{-3}$.
   • By comparison, $X = 7.6$.