A gasoline-driven lawnmower has a blade that extends out 2.0 ft from its center. The tip of the blade is traveling at the speed of sound, which is 1,100 feet/sec. What is the angular velocity of the tip of the blade? Express your answer in revolutions per minute (rpm). Use 3.14 as an approximation for and round your answer to the nearest rpm. 33,000 rpm 7,006 rpm 5,254 rpm 550 rpm 10:09 PM 6/18/2023
A gasoline-driven lawnmower has a blade that extends out 2.0 ft from its center. The tip of the blade is traveling at the speed of sound, which is 1,100 feet/sec. What is the angular velocity of the tip of the blade? Express your answer in revolutions per minute (rpm). Use 3.14 as an approximation for and round your answer to the nearest rpm. 33,000 rpm 7,006 rpm 5,254 rpm 550 rpm 10:09 PM 6/18/2023
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
![### Problem Statement
A gasoline-driven lawnmower has a blade that extends out 2.0 ft from its center. The tip of the blade is traveling at the speed of sound, which is 1,100 feet/sec. What is the angular velocity of the tip of the blade? Express your answer in revolutions per minute (rpm). Use 3.14 as an approximation for π and round your answer to the nearest rpm.
### Given Options for the Answer
- 33,000 rpm
- 7,006 rpm
- 5,254 rpm
- 550 rpm
### Detailed Solution
1. **Identify the given values:**
- Radius \( r \) = 2.0 ft
- Speed of the tip \( v \) = 1,100 feet/sec
- \( \pi \) ≈ 3.14
2. **Angular velocity (ω)** is given by the formula:
\[
\omega = \frac{v}{r}
\]
\[
\omega = \frac{1,100 \text{ feet/sec}}{2.0 \text{ ft}} = 550 \text{ radians/sec}
\]
3. **Convert angular velocity to revolutions per second (rps):**
\[
\omega_{rps} = \frac{\omega}{2\pi} = \frac{550}{2 \times 3.14} \approx 87.58 \text{ rps}
\]
4. **Convert rps to revolutions per minute (rpm):**
\[
\omega_{rpm} = 87.58 \text{ rps} \times 60 \text{ seconds/minute} \approx 5,255 \text{ rpm}
\]
### Conclusion
The angular velocity of the tip of the blade is approximately **5,254 rpm**. Hence, the correct answer is:
- 5,254 rpm
### Visual Explanation
There are no graphs or diagrams in the provided text. This transcription was done to make the content suitable for an Educational website by breaking down the steps and including all math formulas and calculations for better understanding.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe9fc99ab-b3a9-4577-a5d7-bccfea12d417%2Fc1370c74-488b-408e-83cc-e5fb4245ea36%2Fqsmmoht_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
A gasoline-driven lawnmower has a blade that extends out 2.0 ft from its center. The tip of the blade is traveling at the speed of sound, which is 1,100 feet/sec. What is the angular velocity of the tip of the blade? Express your answer in revolutions per minute (rpm). Use 3.14 as an approximation for π and round your answer to the nearest rpm.
### Given Options for the Answer
- 33,000 rpm
- 7,006 rpm
- 5,254 rpm
- 550 rpm
### Detailed Solution
1. **Identify the given values:**
- Radius \( r \) = 2.0 ft
- Speed of the tip \( v \) = 1,100 feet/sec
- \( \pi \) ≈ 3.14
2. **Angular velocity (ω)** is given by the formula:
\[
\omega = \frac{v}{r}
\]
\[
\omega = \frac{1,100 \text{ feet/sec}}{2.0 \text{ ft}} = 550 \text{ radians/sec}
\]
3. **Convert angular velocity to revolutions per second (rps):**
\[
\omega_{rps} = \frac{\omega}{2\pi} = \frac{550}{2 \times 3.14} \approx 87.58 \text{ rps}
\]
4. **Convert rps to revolutions per minute (rpm):**
\[
\omega_{rpm} = 87.58 \text{ rps} \times 60 \text{ seconds/minute} \approx 5,255 \text{ rpm}
\]
### Conclusion
The angular velocity of the tip of the blade is approximately **5,254 rpm**. Hence, the correct answer is:
- 5,254 rpm
### Visual Explanation
There are no graphs or diagrams in the provided text. This transcription was done to make the content suitable for an Educational website by breaking down the steps and including all math formulas and calculations for better understanding.
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