Problem 14 - Applied Physics

Question

Find the volume of a spherical water tank with radius \(8.00 \mathrm{~m}\).

Step-by-Step Solution

Answer

The volume of the tank is approximately 2144.59 m³.

Step 1: Identify the formula for the volume of a sphere

The volume of a sphere can be calculated using the formula: \[ V = \frac{4}{3} \pi r^3 \] where \( V \) represents the volume and \( r \) is the radius of the sphere.

Step 2: Substitute the radius of the water tank into the formula

The radius of the spherical water tank is given as \( r = 8.00 \mathrm{~m} \). Substitute this value into the formula to get the volume expression: \[ V = \frac{4}{3} \pi (8.00)^3 \]

Step 3: Calculate the cube of the radius

Calculate \( (8.00)^3 \): \[ (8.00)^3 = 512 \] So the expression becomes \( V = \frac{4}{3} \pi \times 512 \).

Step 4: Complete the multiplication

Multiply \( 512 \) by \( \frac{4}{3} \): \[ \frac{4}{3} \times 512 = \frac{2048}{3} \] Now the expression is \( V = \frac{2048}{3} \pi \).

Step 5: Estimate the value of \( \pi \) and compute the volume

Assume that \( \pi \approx 3.14 \). Then, substitute \( \pi \) to approximate the volume: \[ V = \frac{2048}{3} \times 3.14 \approx 2144.59 \] m³.

Step 6: Round the volume to a reasonable precision

For practical purposes, round the calculated volume to two decimal places: \( V \approx 2144.59 \mathrm{~m}^3 \).

Key Concepts

Spherical water tank

Spherical water tanks are commonly used in various industries due to their efficient use of materials and space. These tanks hold liquids in a sphere shape, ensuring that pressure is evenly distributed along the tank walls. This uniformity minimizes the risk of pressure-induced damage and allows the tank to withstand various environmental conditions.

The design of a spherical tank is especially optimal for storing large quantities of fluids, as the sphere has the smallest surface area for a given volume compared to other shapes. This minimizes the heat exchange with the surroundings, making it ideal for temperature-sensitive liquids.

• Efficient space usage
• Uniform pressure distribution
• Temperature regulation benefits

Understanding how to calculate the volume of such a tank is essential for determining its capacity and ensuring its contents are properly managed.

Radius

The radius is a crucial component of a sphere, defined as the distance from the center of the sphere to any point on its surface. It is a straight line measurement, which is vital for calculating other key metrics like volume and surface area.

In this particular exercise, the radius of the spherical water tank is given as 8 meters. This measurement allows us to proceed with calculating the volume by substituting the value into the appropriate formula.

• Distance from center to surface
• Essential for other calculations
• Unit-specific (meters in this case)

By understanding the role of the radius in geometric calculations, students can more easily approach problems involving spheres and other circular shapes.

Mathematical formula

Mathematical formulas are among the most powerful tools in mathematics, enabling us to calculate unknown values based on known quantities. The formula to find the volume of a sphere is particularly important for dealing with three-dimensional shapes.

The formula used here is: \[ V = \frac{4}{3} \pi r^3 \] where \( V \) is the volume and \( r \) is the radius. The constant \( \pi \) represents the ratio of the circumference to the diameter of a circle, approximately equal to 3.14159.

• Defines relationships between different quantities
• Involves constants like \( \pi \)
• Provides a method to find unknown values

Mastery of formula-based calculations allows for more efficient problem-solving and a deeper understanding of geometry.

Volume calculation

Volume calculation is the process of determining the amount of space occupied by a three-dimensional object. For a sphere, this involves using the volume formula and the specific measurements of the object.

In our exercise, we were tasked with finding the volume of a spherical water tank with a radius of 8 meters: \[ V = \frac{4}{3} \pi (8)^3 \] After finding \( 512 \) from \( (8)^3 \), you multiply by \( \frac{4}{3} \), resulting in \( \frac{2048}{3} \pi \). Approximating \( \pi \) as 3.14, the volume becomes roughly \( 2144.59 \) m³.

• Requires radius substitution
• Utilizes cube of radius
• Involves estimating \( \pi \) value

Understanding the completion of these calculations ensures accurate solutions to real-world problems involving spherical volumes.