2 points Determine the density of a cylindrical solid with a mass of 16.9 g , a diameter of 7.3 mm , and a length of 19.3 mm in \( \mathrm{g} / \mathrm{cm}^{3} \). \( \mathrm{Pi}=3.14 \) ROUND ANSWER TO ONE DECIMAL PLACE. 20.9

To determine the density of the cylindrical solid, we can use the formula for density: \[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \] First, we need to calculate the volume of the cylinder using the formula: \[ \text{Volume} = \pi r^2 h \] where: - \( r \) is the radius of the cylinder, - \( h \) is the height (or length) of the cylinder. Given: - Mass = 16.9 g - Diameter = 7.3 mm, so the radius \( r = \frac{7.3 \, \text{mm}}{2} = 3.65 \, \text{mm} \) - Length \( h = 19.3 \, \text{mm} \) First, we convert the radius and height from millimeters to centimeters, since we want the density in \( \mathrm{g/cm^3} \): \[ r = 3.65 \, \text{mm} = 0.365 \, \text{cm} \] \[ h = 19.3 \, \text{mm} = 1.93 \, \text{cm} \] Now we can calculate the volume: \[ \text{Volume} = \pi r^2 h = 3.14 \times (0.365 \, \text{cm})^2 \times (1.93 \, \text{cm}) \] Calculating \( r^2 \): \[ (0.365 \, \text{cm})^2 = 0.133225 \, \text{cm}^2 \] Now substituting back into the volume formula: \[ \text{Volume} = 3.14 \times 0.133225 \, \text{cm}^2 \times 1.93 \, \text{cm} \] Calculating the volume: \[ \text{Volume} \approx 3.14 \times 0.133225 \times 1.93 \approx 0.785 \, \text{cm}^3 \] Now we can calculate the density: \[ \text{Density} = \frac{16.9 \, \text{g}}{0.785 \, \text{cm}^3} \approx 21.5 \, \text{g/cm}^3 \] Rounding to one decimal place, the density of the cylindrical solid is: \[ \boxed{21.5} \, \text{g/cm}^3 \]