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a fan rotating with an initial angular velocity of 1000 rev/min is switched off. in 2.0 seconds, the angular velocity decreases to 200 rev/min.
assuming the angular acceleration is constant, how many revolutions does the blade undergo during this time?
1200
20
10
100

Explanation:

Step1: Convert time to minutes

The time is 2.0 seconds. To convert to minutes, use \( t = \frac{2.0}{60} \) min \( = \frac{1}{30} \) min.

Step2: Use the angular displacement formula for constant acceleration

The formula for angular displacement \( \theta \) (in revolutions) with constant angular acceleration is \( \theta=\frac{\omega_i + \omega_f}{2}\times t \), where \( \omega_i = 1000 \) rev/min, \( \omega_f = 200 \) rev/min, and \( t=\frac{1}{30} \) min.
Substitute the values: \( \theta=\frac{1000 + 200}{2}\times\frac{1}{30} \)
First, calculate the average angular velocity: \( \frac{1000 + 200}{2}=600 \) rev/min.
Then, multiply by time: \( 600\times\frac{1}{30}=20 \) rev? Wait, no, wait. Wait, 2 seconds is \( \frac{2}{60}=\frac{1}{30} \) minutes? Wait, no, wait, 2 seconds is \( \frac{2}{60}= \frac{1}{30} \approx 0.0333 \) minutes. But wait, maybe I made a mistake in time conversion. Wait, let's re - evaluate. Wait, the initial and final angular velocities are in rev/min, and time is in seconds. Let's convert time to minutes: \( t = 2.0\ s=\frac{2.0}{60}\ min=\frac{1}{30}\ min\approx0.0333\ min \). But the formula \( \theta=\frac{\omega_i+\omega_f}{2}\times t \) (where \( \theta \) is in revolutions, \( \omega \) in rev/min, \( t \) in min) gives \( \theta=\frac{1000 + 200}{2}\times\frac{2}{60} \). Wait, I see my mistake. The time is 2 seconds, so \( t=\frac{2}{60}\ min=\frac{1}{30}\ min \)? No, wait, 2 seconds is \( \frac{2}{60}=\frac{1}{30}\ min \), but when I use the formula \( \theta=\frac{\omega_i+\omega_f}{2}\times t \), \( \omega \) is in rev per minute, \( t \) in minutes. So \( \omega_i = 1000 \) rev/min, \( \omega_f = 200 \) rev/min, \( t=\frac{2}{60}\ min=\frac{1}{30}\ min \). Then \( \theta=\frac{1000 + 200}{2}\times\frac{2}{60} \). Wait, the 2 in the numerator and 60 in the denominator: \( \frac{1200}{2}\times\frac{2}{60}=600\times\frac{2}{60}=20 \)? Wait, no, \( \frac{1200}{2}=600 \), \( 600\times\frac{2}{60}=20 \)? Wait, no, \( \frac{2}{60}=\frac{1}{30} \), \( 600\times\frac{1}{30}=20 \). Wait, but that seems high? Wait, no, 1000 rev/min is 1000 revolutions per 60 seconds, so per second it's \( \frac{1000}{60}\approx16.67 \) rev/s, and 200 rev/min is \( \frac{200}{60}\approx3.33 \) rev/s. The average angular velocity is \( \frac{16.67 + 3.33}{2}=10 \) rev/s. Then in 2 seconds, the number of revolutions is \( 10\times2 = 20 \) revolutions. Ah, that's a better way. Convert \( \omega_i \) and \( \omega_f \) to rev/s. \( \omega_i=1000\ rev/min=\frac{1000}{60}\ rev/s=\frac{50}{3}\ rev/s \), \( \omega_f = 200\ rev/min=\frac{200}{60}=\frac{10}{3}\ rev/s \). The average angular velocity \( \omega_{avg}=\frac{\omega_i+\omega_f}{2}=\frac{\frac{50}{3}+\frac{10}{3}}{2}=\frac{\frac{60}{3}}{2}=\frac{20}{2} = 10\ rev/s \). Then the number of revolutions \( \theta=\omega_{avg}\times t \), where \( t = 2\ s \). So \( \theta=10\ rev/s\times2\ s = 20 \) revolutions.

Answer:

20