A gasoline-driven lawnmower has a blade that extends out 1.5 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. 733 rpm 5,254 rpm 7,006 rpm 21,019 rpm
A gasoline-driven lawnmower has a blade that extends out 1.5 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. 733 rpm 5,254 rpm 7,006 rpm 21,019 rpm
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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Transcribed Image Text:**Question:**
A gasoline-driven lawnmower has a blade that extends out 1.5 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.
**Answer Options:**
1. O 733 rpm
2. O 5,254 rpm
3. O 7,006 rpm
4. O 21,019 rpm
---
**Explanation:**
To find the angular velocity, we can use the following formulas and steps:
1. **Determine the circumference of the circle traced by the blade tip:**
The radius \( r \) = 1.5 ft.
The circumference \( C \) = \( 2 \pi r \) = \( 2 \times 3.14 \times 1.5 \) = 9.42 ft.
2. **Calculate the number of revolutions per second:**
The speed of the blade tip \( v \) = 1,100 ft/sec.
The number of revolutions per second \( \text{rev/sec} \) = \( \frac{v}{C} \) = \( \frac{1,100}{9.42} \) ≈ 116.74 rev/sec.
3. **Convert to revolutions per minute (rpm):**
\( 116.74 \) rev/sec × 60 sec/min = 7,004.4 rpm.
After rounding the answer to the nearest rpm, we obtain 7,006 rpm.
Therefore, the correct answer is:
- \( \boxed{7,006 \; \text{rpm}} \).
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