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A vinyl record is played by rotating the record so that an approximately circular groove in the vinyl slides under a stylus. Bumps in the groove run into the stylus, causing it to oscillate. The equipment converts those oscillations to electrical signals and then to sound. Suppose that a record turns at the rate of 33 1/3 rev/min, the groove being played is at a radius of 10.0 cm, and the bumps in the groove are uniformly separated by 1.75 mm. At what rate (hits per second) do the bumps hit the stylus?

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Kazeemsodikisola

Answer:

Explanation:

Given that,

Angular velocity w=33⅓ rev/min

Let convert to rad/sec

1 Revolution=2πrad

1 minute =60 seconds

Then,

33⅓rev/min= 100/3 rev/mins × 2πrad/1 rev × 1min/60sec

Therefore, w=3.491rad/sec

Also, given that radius of grove is 10cm

Then, r=0.1m

The distance between each bump is 1.75mm

Also, d=1.75/1000=0.00175m

The linear velocity(v) of the record is given as

v=wr

v=3.491×0.1

v=0.3491 m/s

So, the number hit strike by the bump is given as

N= v/d

Since v=0.3491 and d=0.00175m

Then,

N=0.3491/0.00175

N=199.49hits/sec

The rate at which the bumps hit the stylus is 199.5hits/secs

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