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Margaret [11]
3 years ago
7

The old-fashioned record player had two speeds: one for long-playing albums (LP) and one for singles. The turntable speed for LP

s was 33 1/3 rpm (revolutions per minute). This was equivalent to 100 revolutions every three minutes. The speed of a single was 45 rpm. Suppose it took 4.0 seconds for the turntable to ‘speed up’ from 33 1/3 rpm to 45 rpm when a switch was pressed to change the speed. What would be this angular acceleration in rad/sec2?
Physics
1 answer:
eimsori [14]3 years ago
3 0

Answer:

\alpha = 0.305 rad/s^2

Explanation:

initial frequency of revolution is given as

f_1 = 100/3 rpm

now initial angular speed is

\omega_i = 2\pi f

\omega_i = 2\pi(\frac{100}{3\times 60})

\omega_i = 3.49 rad/s

Similarly final angular speed is given as

\omega_f = 2\pi f_2

\omega_f = 2\pi(\frac{45}{60})

\omega_f = 4.71 rad/s

Now angular acceleration is given as

\alpha = \frac{\omega_f - \omega_i}{\Delta t}

\alpha = \frac{4.71 - 3.49}{4}

\alpha = 0.305 rad/s^2

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