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-0.01 rad/s2 2.0m Point P is on the rim of a wheel of radius 2.0 m. At time t = 0, the wheel is at rest, and P is on the x-aaa. The wheel undergoes a uniform angular acceleration of g.01 r9d/a? about the center O. In Figure 9.1, the linear speed of P, when it reaches the y-axis, is closest to: O A. 0.71 m's O B. 0.35 m/s OC.0.49 m/s O D.0.24 m's O E. 0.18 m/s

-0.01 rad/s2 2.0m Point P is on the rim of a wheel of radius 2.0 m. At time t = 0, the wheel is at rest, and P is on the x-aaa. The wheel undergoes a uniform angular acceleration of g.01 r9d/a? about the center O. In Figure 9.1, the linear speed of P, when it reaches the y-axis, is closest to: O A. 0.71 m's O B. 0.35 m/s OC.0.49 m/s O D.0.24 m's O E. 0.18 m/s

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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**Figure 9.1 Description:** The image shows a diagram of a wheel with point \( P \) on its rim. The wheel has a radius of 2.0 meters, and point \( P \) is initially aligned with the x-axis. An angular acceleration \( \alpha = 0.01 \, \text{rad/s}^2 \) is applied about the center \( O \). **Problem Statement:** Point \( P \) is on the rim of a wheel with a radius of 2.0 m. At time \( t = 0 \), the wheel is at rest, and \( P \) is on the x-axis. The wheel undergoes a uniform angular acceleration of \( 0.01 \, \text{rad/s}^2 \) about the center \( O \). In Figure 9.1, the linear speed of \( P \), when it reaches the y-axis, is closest to: - A. 0.71 m/s - B. 0.35 m/s - C. 0.49 m/s - D. 0.24 m/s - E. 0.18 m/s **Explanation of Diagram:** - The wheel is represented as a circle with a center labeled \( O \). - Point \( P \) is marked on the rim of the circle, located along the radius initially aligned with the x-axis. - The diagram indicates that the wheel rotates counter-clockwise, with the angular acceleration vector (\( \alpha = 0.01 \, \text{rad/s}^2 \)) shown as an arrow around the circle. This educational resource illustrates the relationship between angular acceleration and linear speed as point \( P \) moves along the circumference of the wheel.
Transcribed Image Text:**Figure 9.1 Description:** The image shows a diagram of a wheel with point \( P \) on its rim. The wheel has a radius of 2.0 meters, and point \( P \) is initially aligned with the x-axis. An angular acceleration \( \alpha = 0.01 \, \text{rad/s}^2 \) is applied about the center \( O \). **Problem Statement:** Point \( P \) is on the rim of a wheel with a radius of 2.0 m. At time \( t = 0 \), the wheel is at rest, and \( P \) is on the x-axis. The wheel undergoes a uniform angular acceleration of \( 0.01 \, \text{rad/s}^2 \) about the center \( O \). In Figure 9.1, the linear speed of \( P \), when it reaches the y-axis, is closest to: - A. 0.71 m/s - B. 0.35 m/s - C. 0.49 m/s - D. 0.24 m/s - E. 0.18 m/s **Explanation of Diagram:** - The wheel is represented as a circle with a center labeled \( O \). - Point \( P \) is marked on the rim of the circle, located along the radius initially aligned with the x-axis. - The diagram indicates that the wheel rotates counter-clockwise, with the angular acceleration vector (\( \alpha = 0.01 \, \text{rad/s}^2 \)) shown as an arrow around the circle. This educational resource illustrates the relationship between angular acceleration and linear speed as point \( P \) moves along the circumference of the wheel.
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