91Ó°ÊÓ

To determine the volume of a cylinder, a special formula is required: \[ V = \pi r^2 h \] This formula multiplies the area of the base (a circle) by the height of the cylinder. The base's area is given as \( \pi r^2 \), with \( r \) being the radius, and \( h \) stands for the cylinder's height.
Given a cylinder with a radius of 1.22 meters and a height of 3.71 meters, we simply need to plug these values into the formula. Here’s a breakdown of the calculation:

• Find \( r^2 \): \( 1.22^2 = 1.4884 \)
• Multiply this by the height: \( 1.4884 \times 3.71 = 5.523564 \)
• Finally, multiply by \( \pi \): \( 5.523564 \times \pi \approx 17.353 \text{ m}^3 \)

Thus, the volume of the cylinder is approximately 17.353 cubic meters. This volume is essential in understanding how much space the cylinder occupies, which allows us to look further into mass-volume relationships.

The mass-volume relationship is crucial in calculating density, which helps describe how much mass is present in a given volume. In simple terms, density shows how tightly matter is packed together in a specific space.
The formula for density is: \[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \] So, whenever you know the mass and volume of a substance, you can calculate its density, providing valuable insight into its physical properties. In our case, we have an industrial solvent with a mass of 14300 kilograms and a calculated volume of 17.353 cubic meters. By dividing the mass by the volume, the result is: \[ \text{Density} = \frac{14300}{17.353} \approx 823.92 \text{ kg/m}^3 \] This numerical value, 823.92 kg/m³, indicates how much of the solvent's mass is contained per unit volume.

Understanding the density of industrial solvents is important for applications across various industries. Solvents with higher density generally have more mass in a given volume compared to those with lower density. Such characteristic properties affect not only storage and transportation but also the performance of the solvents in chemical processes.
In the current problem, we calculated the density of an industrial solvent as approximately 823.92 kg/m³. This figure enables engineers and chemists to understand how the solvent behaves in different scenarios. For example, this property informs choice when selecting the appropriate solvent for dissolving materials based on density compatibility. Moreover, it's a key factor in designing equipment for processing or storing the solvent.
Overall, by understanding and calculating such densities, industries ensure efficient and effective use of materials while adhering to standards and safety measures.