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Let's solve each part of this problem.

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### Given Data:
- Diameter of turntable, \( d = 0.750 \, \text{m} \)
- Radius of turntable, \( r = \frac{d}{2} = 0.375 \, \text{m} \)
- Initial angular velocity, \( \omega_0 = 0.250 \, \text{rev/s} \)
- Angular acceleration, \( \alpha = 0.900 \, \text{rev/s}^2 \)
- Time, \( t = 0.200 \, \text{s} \)

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### Solution Steps:

#### a) Compute the angular velocity of the turntable after \( t = 0.200 \, \text{s} \).

The angular velocity \( \omega \) after time \( t \) can be found using the equation:
\[
\omega = \omega_0 + \alpha t
\]
Substitute the given values:
\[
\omega = 0.250 + (0.900)(0.200) = 0.250 + 0.180 = 0.430 \, \text{rev/s}
\]

#### b) Through how many revolutions has the turntable spun in this time interval?

The angular displacement \( \theta \) (in revolutions) can be found using:
\[
\theta = \omega_0 t + \frac{1}{2} \alpha t^2
\]
Substitute the values:
\[
\theta = (0.250)(0.200) + \frac{1}{2}(0.900)(0.200)^2
\]
\[
\theta = 0.050 + \frac{1}{2}(0.900)(0.040)
\]
\[
\theta = 0.050 + 0.018 = 0.068 \, \text{revolutions}
\]

#### c) What is the tangential speed of a point on the rim of the turntable at \( t = 0.200 \, \text{s} \)?

The tangential speed \( v \) is related to the angular velocity by:
\[
v = \omega \times 2\pi r
\]
Convert \( \omega = 0.430 \, \text{rev/s} \) to radians per second:
\[
\omega = 0.430 \times 2\pi \approx 2.701 \, \text{rad/s}
\]
Now, calculate \( v \):
\[
v = 2.701 \times 0.375 \approx 1.013 \, \text{m/s}
\]

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### Summary of Answers:
- **(a)** Angular velocity after \( 0.200 \, \text{s} \): \( 0.430 \, \text{rev/s} \)
- **(b)** Total revolutions in \( 0.200 \, \text{s} \): \( 0.068 \, \text{revolutions} \)
- **(c)** Tangential speed at \( t = 0.200 \, \text{s} \): \( 1.013 \, \text{m/s} \)

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Would you like more details on any step, or have questions? 

### Additional Related Questions:
1. How would the solution change if the angular acceleration were negative?
2. How do we convert from revolutions per second to radians per second?
3. What would be the centripetal acceleration of a point on the rim at \( t = 0.200 \, \text{s} \)?
4. How can we calculate the angular displacement in radians instead of revolutions?
5. If the angular velocity were initially zero, how many revolutions would the turntable complete in \( 0.200 \, \text{s} \)?

### Tip:
To find the tangential speed, always remember to convert angular velocity to radians per second when using meters for radius.

An electric turntable 0.750 [m] in diameter is rotating about a fixed axis with an initial angular velocity of 0.250 [rev/s] and a constant angular acceleration of 0.900 [rev/s^2].

(a) Compute the angular velocity of the turntable after 0.200 [s].
(b) Through how many revolutions has the turntable spun in this time interval?
(c) What is the tangential speed of a point on the rim of the turntable at t = 0.200 [s]?

Explore the angular kinematics of a rotating turntable with a diameter of 0.750 m, initial angular velocity of 0.250 rev/s, and angular acceleration of 0.900 rev/s². Calculate the angular velocity, total revolutions, and tangential speed at t = 0.200 s with step-by-step solutions.

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