### Current Attempt in ProgressA ceiling fan is turned on and a net torque of 2.9 N·m is applied to the blades. The blades have a total moment of inertia of 0.23 kg·m². What is the angular acceleration of the blades?---**Answer:**To calculate the angular acceleration (\(\alpha\)) of the blades, use the formula:\[ \tau = I \cdot \alpha \]Where:- \(\tau\) is the net torque- \(I\) is the moment of inertia- \(\alpha\) is the angular accelerationGiven:- \(\tau = 2.9 \, \text{N·m}\)- \(I = 0.23 \, \text{kg·m}^2\)Rearrange the formula to solve for the angular acceleration:\[ \alpha = \frac{\tau}{I} \]Substitute the given values:\[ \alpha = \frac{2.9 \, \text{N·m}}{0.23 \, \text{kg·m}^2} \]\[ \alpha = 12.61 \, \text{rad/s}^2 \]Thus, the angular acceleration of the blades is \(12.61 \, \text{rad/s}^2\). To enter this solution in the given fields:- Number: `12.61`- Units: `rad/s²`

Given: The torque on the blade of ceilling fan is 2.9 N.m. The moment of the inertial of the blades…

Current Attempt in Progress A ceiling fan is turned on and a net torque of 2.9 N-m applied to the blades. The blades have a total moment of inertia of 0.23 kg-m2. What is the angular acceleration of the blades? Number Units

Current Attempt in Progress A ceiling fan is turned on and a net torque of 2.9 N-m applied to the blades. The blades have a total moment of inertia of 0.23 kg-m2. What is the angular acceleration of the blades? Number Units

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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Question

### Current Attempt in Progress

A ceiling fan is turned on and a net torque of 2.9 N·m is applied to the blades. The blades have a total moment of inertia of 0.23 kg·m². What is the angular acceleration of the blades?

---

**Answer:**

To calculate the angular acceleration (\(\alpha\)) of the blades, use the formula:

\[ \tau = I \cdot \alpha \]

Where:
- \(\tau\) is the net torque
- \(I\) is the moment of inertia
- \(\alpha\) is the angular acceleration

Given:
- \(\tau = 2.9 \, \text{N·m}\)
- \(I = 0.23 \, \text{kg·m}^2\)

Rearrange the formula to solve for the angular acceleration:

\[ \alpha = \frac{\tau}{I} \]

Substitute the given values:

\[ \alpha = \frac{2.9 \, \text{N·m}}{0.23 \, \text{kg·m}^2} \]

\[ \alpha = 12.61 \, \text{rad/s}^2 \]

Thus, the angular acceleration of the blades is \(12.61 \, \text{rad/s}^2\).

To enter this solution in the given fields:
- Number: `12.61`
- Units: `rad/s²`

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