Density is a fundamental concept in physics and is defined as the mass of an object divided by its volume. The formula for density is represented by the Greek letter rho (ρ), and can be expressed mathematically as:
$$\rho = \frac{m}{V}$$
where m is the mass and V is the volume. In the International System of Units (SI), density is measured in kilograms per cubic meter (kg/m³). Understanding how to manipulate this equation is crucial for solving density-related problems, which often involve various geometric shapes such as cubes, spheres, and cylinders.
To illustrate the application of density, consider the average density of the Earth, which is approximately 55100 kg/m³. To find the mass of the Earth, we can use the density formula. First, we need to calculate the volume of the Earth, which can be modeled as a sphere. The volume of a sphere is given by the formula:
$$V = \frac{4}{3} \pi r^3$$
where r is the radius of the sphere. For Earth, the radius is given as 3960 miles. Since this measurement is not in SI units, we must convert it to meters. The conversion factor is:
1 mile ≈ 1609 meters
Thus, the radius in meters is calculated as:
$$r = 3960 \text{ miles} \times \frac{1609 \text{ meters}}{1 \text{ mile}} \approx 6.37 \times 10^6 \text{ meters}$$
Now that we have the radius in the correct units, we can calculate the volume of the Earth:
$$V = \frac{4}{3} \pi (6.37 \times 10^6 \text{ m})^3 \approx 1.08 \times 10^{21} \text{ m}^3$$
With both the density and volume known, we can now find the mass of the Earth by rearranging the density equation to solve for mass:
$$m = \rho \times V$$
Substituting the known values:
$$m = 55100 \text{ kg/m}^3 \times 1.08 \times 10^{21} \text{ m}^3 \approx 5.94 \times 10^{24} \text{ kg}$$
This calculation yields an approximate mass of the Earth, which is consistent with known scientific values, although slight variations may occur due to rounding. Mastering these calculations and unit conversions is essential for solving density problems effectively.
15 Terms
