Problem 21 - Elementary Algebra

Question

For a fixed length of household copper wire, the relationship of the cross- sectional area, \(x\), and the resistance, \(y\), is an inverse variation. When the cross-sectional area is \(3.14 \times 10^{-6} \mathrm{~m}^{2}\), the resistance is \(5.4 \times 10^{-3} \mathrm{ohm}\). a. Find the constant of proportionality, \(k\). Use scientific notation. Include the units of measurement. b. Write an equation that represents this relationship. c. Find the resistance when the cross-sectional area is \(2.05 \times 10^{-6} \mathrm{~m}^{2}\). Round the mantissa to the nearest tenth.

Step-by-Step Solution

Answer

a. \( 1.70 \times 10^{-8} \text{ ohm} \text{ m}^2 \)b. \( y = \frac{1.70 \times 10^{-8}}{x} \)c. \( 8.3 \times 10^{-3} \text{ ohm} \)

Step 1 - Understand the Relationship

The problem states that resistance, \( y \), varies inversely with the cross-sectional area, \( x \). This means the relationship can be expressed as \( y = \frac{k}{x} \), where \( k \) is the constant of proportionality.

Step 2 - Use Given Values to Find \( k \)

Given that when \( x = 3.14 \times 10^{-6} \text{ m}^2 \), the resistance \( y = 5.4 \times 10^{-3} \text{ ohm} \), substitute these values into the equation \( y = \frac{k}{x} \) to find \( k \). \[ 5.4 \times 10^{-3} = \frac{k}{3.14 \times 10^{-6}} \] To isolate \( k \), multiply both sides by \( 3.14 \times 10^{-6} \): \[ k = (5.4 \times 10^{-3}) \times (3.14 \times 10^{-6}) \]

Step 3 - Calculate \( k \)

Perform the multiplication: \[ k = 5.4 \times 3.14 \times 10^{-9} \] Calculate the product: \[ k = 16.956 \times 10^{-9} \] Round the mantissa to three significant figures: \[ k = 1.70 \times 10^{-8} \text{ ohm} \text{ m}^2 \]

Step 4 - Write the Equation

Now that the constant of proportionality \( k \) is known, substitute it back into the inverse variation equation: \[ y = \frac{1.70 \times 10^{-8}}{x} \]

Step 5 - Calculate Resistance for New Cross-Sectional Area

Given a new cross-sectional area \( x = 2.05 \times 10^{-6} \text{ m}^2 \), substitute this into the equation to find the new resistance \( y \): \[ y = \frac{1.70 \times 10^{-8}}{2.05 \times 10^{-6}} \] Perform the division: \[ y = \frac{1.70}{2.05} \times 10^{-2} \] Round the mantissa to the nearest tenth: \[ y = 0.8 \times 10^{-2} = 8.3 \times 10^{-3} \text{ ohm} \]

Key Concepts

cross-sectional area

Cross-sectional area refers to the surface area of the section of an object that is perpendicular to its length. In the context of a wire, this means the area of the round cut through the wire. It's important because it affects how many electrons can flow through the wire at once. The larger the cross-sectional area, the lower the resistance. The formula to find this area for a cylindrical object (like a wire) is \[ A = \pi r^2 \] where \( r \) is the radius of the wire.

resistance

Resistance is a measure of how much an object opposes the flow of electric current. In a wire, it depends on the material, length, and cross-sectional area. In this problem, resistance varies inversely with the cross-sectional area. This means that as the cross-sectional area increases, the resistance decreases, and vice versa. The standard unit for resistance is the ohm (\( \Omega \) ).

constant of proportionality

The constant of proportionality, \( k \) in this problem, is a value that relates the resistance and the cross-sectional area of the wire through an inverse variation. The relationship is given by \[ y = \frac{k}{x} \] To find \( k \), multiply the given resistance by the given cross-sectional area. In this case: \[ k = (5.4 \times 10^{-3}) \times (3.14 \times 10^{-6}) = 1.70 \times 10^{-8} \text{ ohm } \text{ m}^2 \] This unchanging value connects different pairs of cross-sectional area and resistance.

scientific notation

Scientific notation is a way to express very large or very small numbers more conveniently. It represents numbers as a product of a number between 1 and 10 and a power of ten. For example, \[ 3.14 \times 10^{-6} \] represents 3.14 multiplied by ten raised to the negative sixth power. This helps to simplify calculations involving very small or very large figures. For instance, calculating the constant of proportionality \( k \) for this problem involves: \[ k = (5.4 \times 10^{-3}) \times (3.14 \times 10^{-6}) = 1.70 \times 10^{-8} \] Understanding scientific notation is crucial for accurately performing and interpreting these calculations.