Problem 92 - Physics

Question

An irregularly shaped chunk of concrete has a hollow spherical cavity inside. The mass of the chunk is \(33 \mathrm{~kg}\), and the volume enclosed by the outside surface of the chunk is \(0.025 \mathrm{~m}^{3}\). What is the radius of the spherical cavity?

Step-by-Step Solution

Answer

The radius of the spherical cavity is approximately 0.134 meters.

Step 1: Understand the Problem

We have an irregularly shaped piece of concrete with an internal hollow, spherical cavity. The total mass of the chunk is given as 33 kg, and the total external volume of the chunk is 0.025 m³. We need to find the radius of the spherical cavity.

Step 2: Determine Density of Concrete

First, estimate the density of concrete if not explicitly provided. A standard average density for concrete is approximately \(2400 \text{ kg/m}^3\).

Step 3: Calculate Mass of Full Chunk Without Cavity

Using the formula \(\text{mass} = \text{density} \times \text{volume}\), compute the mass the chunk would have if it did not have a cavity. So, \(\text{mass} = 2400 \times 0.025 = 60 \text{ kg}\).

Step 4: Calculate Mass of Missing Cavity Material

Subtract the actual mass of the concrete chunk from the calculated mass if there were no cavity. This will give us the mass that is missing due to the spherical cavity: \(60 - 33 = 27 \text{ kg}\).

Step 5: Calculate Volume of the Cavity

Use the calculated mass of the cavity and the density of concrete to find the volume of the cavity. Use the formula \(\text{volume} = \frac{\text{mass}}{\text{density}}\). Thus, \(\text{volume} = \frac{27}{2400} = 0.01125 \text{ m}^3\).

Step 6: Calculate Radius of the Cavity

Since the cavity is spherical, use the volume formula for a sphere, \(V = \frac{4}{3} \pi r^3\), to find the radius \(r\). Rearrange to solve for \(r\): \(r = \left(\frac{3V}{4\pi}\right)^{\frac{1}{3}}\). Substitute the volume found: \(r = \left(\frac{3 \times 0.01125}{4\pi}\right)^{1/3}\), resulting in \(r \approx 0.134 \text{ m}\).

Key Concepts

Density of concrete

Density is a fundamental property of materials that describes how much mass is contained in a given volume. For concrete, a typical density value is approximately 2400 kg/m³. This value can slightly vary depending on the type of concrete and its composition. Generally:

• Higher density means the material is heavier for the same size.
• Concrete's density influences its strength, durability, and load-bearing capacity.
• It's crucial not only in construction but in applications like this physics problem, where knowing density allows us to calculate missing mass or volumes.

In physical problems involving concrete, density often acts as a key to unlock unknowns, linking mass and volume through straightforward calculations.

Volume of sphere calculation

Finding the volume of a sphere is an essential skill in geometry and physics. For a sphere with radius \(r\), the formula to calculate its volume \(V\) is:\[V = \frac{4}{3} \pi r^3\]Understanding this formula requires recognizing that:

• The factors \(\frac{4}{3}\) and \(\pi\) are constants that adjust the shape (sphere) and size within a three-dimensional space.
• The radius is cubed, showing that as the radius increases, the volume increases rapidly.
• In practical problems like the concrete chunk, using this formula helps determine the size of spherical spaces within larger structures.

By solving the formula for \(r\), you can find the radius when the volume is known, which is a key step in analyzing hollow spaces like cavities.

Mathematical formulas in physics

Physics relies heavily on mathematical formulas to describe and analyze real-world phenomena. These formulas provide a bridge between abstract concepts and measurable properties. In the exercise provided:

• We use the formula \( \text{mass} = \text{density} \times \text{volume} \) to connect mass, density, and volume, highlighting their interdependence.
• The sphere's volume formula transforms geometric principles into usable data for volume and radius calculations.
• This exercise showcases the importance of formulas in translating knowledge of one property (like mass) into another (like volume) using constants and relationships.

The consistent application of such formulas is what enables problem solving in physics, ensuring precise predictions and analysis.

Concrete properties

Concrete is a ubiquitous construction material known for its strength and versatility. Its properties extend beyond mere weight and density:

• It is a composite material, typically made from a mix of cement, aggregates (like sand and gravel), and water.
• Concrete's properties can vary: high-strength types are denser, while lighter versions might use lightweight aggregates.
• Thermal properties make it a good insulator, and it sets and hardens over time, achieving its full strength weeks after pouring.

In physics exercises, understanding these properties helps contextualize why a cavity might exist (for weight reduction) and how different formulas apply in the context of a given scenario.

Physics education

Physics education aims to develop a deep understanding of natural phenomena through critical thinking and problem solving. Exercises like the one involving the concrete chunk are designed to enhance learning by:

• Encouraging students to apply density, volume, and mass relationships in a practical context.
• Building skills in interpreting mathematical formulas to uncover unknowns.
• Fostering a methodical approach to problem solving that combines conceptual knowledge with logical reasoning.

Such problems not only improve comprehension of specific topics like density and volume but also instill a broader capacity for analytical thinking vital in academics and beyond.