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An automobile engine slows down from 4500 rpm to 1600 rpm in 6.0 s. (a) calculate its angular acceleration, assumed constant. 3036.87 incorrect: your answer is incorrect. rad/s2 (b) calculate the total number of revolutions the engine makes in this time. 8700 incorrect: your answer is incorrect. rev

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skyluke89
1) The angular acceleration of the engine is given by:
[tex]\alpha = \frac{\omega _f - \omega _i}{\Delta t} [/tex]
where [tex]\omega _f [/tex] and [tex]\omega _i[/tex] are the final and initial angular velocity of the engine, while [tex]\Delta t[/tex] is the time interval considered.
We need first to convert the velocities from rpm (revolutions per minute) into rad/s. Keeping in mind that 
[tex]1 rev=2\pi rad[/tex]
[tex]1min = 60 s[/tex]
The factor of conversion is
[tex]1 \frac{rev}{min} = \frac{2 \pi rad}{60 s} [/tex]
So, the two velocities become
[tex]\omega _i = 4500 rpm \cdot ( \frac{2 \pi rad}{60 s} )=471 rad/s[/tex]
[tex]\omega _f = 1600 rpm \cdot ( \frac{2 \pi rad}{60 s} )=167 rad/s[/tex]
And using [tex]\Delta t=6.0 s[/tex] we can find the angular acceleration of  the engine:
[tex]\alpha = \frac{167 rad/s - 471 rad/s}{6.0 s}=-50.7 rad/s^2 [/tex]
where the negative sign means the engine is decelerating.

2) The total number of revolutions is given by the law of angular motion:
[tex]S(\Delta t)= \omega _i \Delta t + \frac{1}{2} \alpha (\Delta t)^2 = (471 rad/s)(6.0 s) + \frac{1}{2}(-50.7 rad/s^2)(6.0 s)=[/tex]
[tex]=1915 rad [/tex]
And keeping in mind that [tex]1 rev=2 \pi rad[/tex], the number of revolutions is
[tex]1915 rad \cdot \frac{1}{2 \pi}=305 rev [/tex]
onyebuchinnaji

The angular acceleration of the automobile is 50.62 rad/s².

The total number of revolutions within the given time is 290 revolutions.

Angular acceleration of automobile

The angular acceleration of the automobile is calculated as follows;

[tex]\alpha = \frac{\omega _f - \omega _i}{t} \\\\[/tex]

ωf = 1600 rpm = 167.57 rad/s

ωi = 4500 rpm = 471.3 rad/s

[tex]\alpha = \frac{167.57 - 471.3}{6} \\\\\alpha = -50.62 \ rad/s^2[/tex]

Total number of revolutions in 6 s

N = (4500 rpm - 1600 rpm)

N = 2,900 rpm

N = 2,900 x (6/60)

N = 290 rev

Learn more about angular speed here: https://brainly.com/question/6860269