Problem 12
Question
The radius, \(r,\) of a circle is proportional to the square root of the area, \(A\). Kinetic energy, \(K\), is proportional to the square of velocity, \(v .\)
Step-by-Step Solution
Verified Answer
Question: Explain the relationship between the radius of a circle and its area, and the relationship between kinetic energy and an object's velocity.
Answer: The radius of a circle is proportional to the square root of its area, represented by the equation \(r = k_1 \sqrt{A}\). This means that the radius increases or decreases with respect to the square root of the area, with \(k_1\) representing the proportionality constant.
On the other hand, the kinetic energy of an object is proportional to the square of its velocity, represented by the equation \(K = k_2 v^2\). This means that the kinetic energy increases or decreases with respect to the square of the object's velocity, with \(k_2\) representing the proportionality constant.
1Step 1: 1. Define the proportionality relationship
In the proportional relationship, we can represent it as \(r \propto \sqrt{A}\).
2Step 2: 2. Formulate the relationship with a constant of proportionality
We can introduce a constant of proportionality, \(k_1\), to express the relationship as an equation: \(r = k_1 \sqrt{A}\).
3Step 3: 3. Interpret the result
The equation \(r = k_1 \sqrt{A}\) tells us that the radius of the circle increases or decreases in relation to the square root of the area. The constant \(k_1\) indicates the proportion in which the radius changes in relation to the square root of the area.
Second Relationship:
Kinetic energy (K) is proportional to the square of velocity (v).
4Step 4: 4. Define the proportionality relationship
In the proportional relationship, we can represent it as \(K \propto v^2\).
5Step 5: 5. Formulate the relationship with a constant of proportionality
We can introduce a constant of proportionality, \(k_2\), to express the relationship as an equation: \(K = k_2 v^2\).
6Step 6: 6. Interpret the result
The equation \(K = k_2 v^2\) tells us that the kinetic energy of an object increases or decreases in relation to the square of its velocity. The constant \(k_2\) indicates the proportion in which the kinetic energy changes in relation to the square of velocity.
Key Concepts
Radius and AreaKinetic Energy and VelocityConstants of Proportionality
Radius and Area
Proportional relationships can help us understand how two quantities are connected. Here, the radius of a circle is linked to the square root of its area. This means if you know the area of a circle, you can determine its radius using this relationship.
Let's break this down further:
Let's break this down further:
- The relationship is expressed as \( r \propto \sqrt{A} \), meaning that the radius \( r \) is "proportional to" the square root of the area \( A \).
- To express this relationship with an equation, we use a constant, often represented by \( k_1 \), leading to \( r = k_1 \sqrt{A} \).
- This constant \( k_1 \) tells us the specific ratio or "proportion" of how the radius changes with the square root of the area.
Kinetic Energy and Velocity
Kinetic energy is a measure of an object's energy due to its motion. In this relationship, kinetic energy \( K \) is directly proportional to the square of the velocity \( v \).
Here's how this is presented in a more understandable way:
Here's how this is presented in a more understandable way:
- The basic relationship is presented as \( K \propto v^2 \), meaning kinetic energy changes with the square of the velocity.
- An equation form of this relationship can be represented as \( K = k_2 v^2 \), where \( k_2 \) is the constant of proportionality.
- \( k_2 \) determines exactly how kinetic energy will change for a particular increase in velocity.
Constants of Proportionality
In both referenced relationships—radius and area, and kinetic energy and velocity—the constant of proportionality plays a critical role. It serves as a bridge between the variables, providing a specific, measurable way to understand how changes in one quantity affect changes in another.
- The constant \( k_1 \) in \( r = k_1 \sqrt{A} \) helps determine how much the radius of a circle changes in response to changes in the area.
- Similarly, \( k_2 \) in \( K = k_2 v^2 \) tells us the precise relationship between velocity and kinetic energy.
- While the constants themselves are unique to each situation, they enable us to make precise calculations and predictions about proportional relationships in various scientific and mathematical contexts.
Other exercises in this chapter
Problem 11
In Exercises \(1-21,\) solve the equation for the variable. $$ \sqrt{3 x-2}+1=10 $$
View solution Problem 11
Write the expression as a constant times a power of a variable. Identify the coefficient and the exponent. $$ \frac{x^{4}}{4 \sqrt{x^{2}}}, x>0 $$
View solution Problem 12
In Exercises \(1-21,\) solve the equation for the variable. $$ (x+1)^{2}+4=29 $$
View solution Problem 12
Write the expression as a constant times a power of a variable. Identify the coefficient and the exponent. $$ \left(\frac{1}{5 \sqrt{x}}\right)^{3} $$
View solution