91Ó°ÊÓ

Understanding how to calculate surface area is crucial in many physics and engineering problems.
The surface area of any object is the total area covered by its outer surface. For instance, calculating the surface area of a cube is straightforward. If each edge of the cube measures 1 meter, then the surface area is determined by the formula:
\[ A_{\text{cube}} = 6 \times \text{side}^2 \]
In our exercise, substituting the edge length, we get: \[ A_{\text{cube}} = 6 \times 1^2 = 6 \text{ m}^2 \]
For spherical objects like sand grains, we use a different formula:
\[ A_{\text{sphere}} = 4 \times \text{pi} \times r^2 \]
Here, 'r' represents the radius of the sphere. Given a radius of 50 \mu \text{m} (50 x 10^{-6} meters), the calculation would be:
\[ A_{\text{sphere}} = 4 \times \text{pi} \times (50 \times 10^{-6})^2 = 10 \text{pi} \times 10^{-9} \text{ m}^2 \]
Thus, understanding these formulas lets you compute the surface area of both simple and complex shapes.

Calculating the volume of a sphere is another key concept in physics.
The volume determines how much space an object occupies. For a sphere, the volume is given by:
\[ V_{\text{sphere}} = \frac{4}{3} \times \text{pi} \times r^3 \]
Using the given radius of 50 \mu \text{m} or 50 x 10^{-6} meters, we perform the following calculation:
\[ V_{\text{sphere}} = \frac{4}{3} \times \text{pi} \times (50 \times 10^{-6})^3 = \frac{500 \times \text{pi}}{3} \times 10^{-18} \text{ m}^3 \]
Understanding this formula is essential when dealing with problems involving spherical objects. It helps in determining how many such objects can fill a certain volume or the material required to create a spherical object.

Mass and density are fundamental concepts that frequently appear in physics problems.
The mass of an object indicates how much matter it contains, while density is the measure of how much mass is contained in a given volume. The relationship between mass, volume, and density is given by: \[ m = \rho \times V \]
Here, 'm' is the mass, '\rho' is the density, and 'V' is the volume.
In the exercise, the density of silicon dioxide is provided as \[ \rho = 2600 \text{ kg/m}^3 \]
To find the total mass of sand grains, you first calculate the total volume of the grains using the number of grains and the volume of one grain:
\[ V_{\text{total \text{grains}}} = 1.91 \times 10^8 \times \frac{500 \text{pi}}{3} \times 10^{-18} \text{ m}^3 \]
This gives approximately 10 \text{ m}^3.
Then, the total mass can be calculated by:
\[ m = 2600 \times 10 = 26 \text{ kg} \]
Understanding how to calculate mass from density and volume is widely applicable in science and engineering, from calculating the mass of materials to determining buoyancy and other properties.