### Tire Rotation and Speed Calculation**Problem Overview:**A car tire has a diameter of 55.0 cm. The car is traveling at a speed of 24.0 m/s.---**Part A: Tire's Rotation Frequency****Question:**What is the tire's rotation frequency in rpm (revolutions per minute)?**Explanation:**1. **Formula to Use:** \[ \text{Rotation frequency (rpm)} = \left( \frac{\text{linear speed (m/s)}}{\text{circumference of the tire (m)}} \right) \times 60 \]2. **Circumference Calculation:** Given diameter = 55.0 cm = 0.55 m; \[ \text{Circumference} = \pi \times 0.55 \]3. **Substitute the Values:** - Linear speed = 24.0 m/s - Calculate and substitute to find the rotation frequency.4. **User Response:** - The user attempted an answer: 706 rpm. - Feedback: Incorrect; the user has three attempts remaining.---**Part B: Speed at the Top Edge of the Tire****Question:**What is the speed of a point at the top edge of the tire?**Explanation:**1. **Concept:** - The top edge of a rotating tire moves at the speed of the car plus the speed of the tire's rotation (relative to the ground). - Speed at the top = 2 x car speed2. **Calculation:** - Given car speed = 24.0 m/s - Speed at the top = 48.0 m/s3. **User Response:** - The user entered: 48.0 m/s. - Feedback: Correct.--- This exercise requires understanding the relationship between linear speed, rotational speed, and how they relate to tire mechanics in physics. ## Problem Statement**Part C**What is the speed of a point at the bottom edge of the tire?Express your answer as an integer and include the appropriate units.### Answer Submission- Input fields for: - **Value** - **Units**#### Options Available:- Submit your answer- Request Answer### Additional Resources:- [Provide Feedback]- [Terms of Use](#)- [Privacy Policy](#)- [Permissions](#)---**Copyright © 2022 Pearson Education Inc. All rights reserved.**

Concept: Given: Diameter of tire (d)=55 cmIn metersDiameter (d)=0.55 mSpeed of car (v)=24.0 m/s To determine the Rotational frequency we will use the formula Rotational frequency=VelocityCircumference of the tireSolution: Part A The rotational Frequency of the tire will be, According to the Formula of Rotational frequency Rotational frequency=VelocityCircumference of the tireHere,Circumference of the tire=2πd2Circumference of the tire=2π0.55 m2Circumference of the tire=1.727 mThus,Rotational frequency=24 m/s1.727Rotational frequency=13.8969 rpsIn rpmRotational frequency=13.8969×60Rotational frequency=833.814 rpmRotational frequency=834 rpm Part B: The speed of a point at the top edge of the tire will be Twice the speed of the car that is, vtop=2vhere,v= 24 m/sSubstituting the valuesvtop=2×24vtop=48 m/s Part C: The speed of a point at the bottom edge of the tire will be, The point at the bottom edge is stationary due to the static frictional force. vbottom=0 m/s While the bottom point of the tyre remains fixed as it remains in contact with the road due to which it experiences static frictional force, while the top portion of the point travels at a speed twice as fast as the car.

Science

Advanced Physics

A car tire is 55.0 cm in diameter. The car is traveling at a speed of 24.0 m/s. Part A What is the tire's rotation frequency, in rpm? Express your answer to three significant figures and include the approp [5] ΑΣΦ 706 Submit Previous Answers Request Answer X Incorrect; Try Again; 3 attempts remaining Part B 48.0 B ? What is the speed of a point at the top edge of the tire? Express your answer to three significant figures and include the appropria m S Submit Previous Answers rpm

A car tire is 55.0 cm in diameter. The car is traveling at a speed of 24.0 m/s. Part A What is the tire's rotation frequency, in rpm? Express your answer to three significant figures and include the approp [5] ΑΣΦ 706 Submit Previous Answers Request Answer X Incorrect; Try Again; 3 attempts remaining Part B 48.0 B ? What is the speed of a point at the top edge of the tire? Express your answer to three significant figures and include the appropria m S Submit Previous Answers rpm