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dmitriy555 [2]
3 years ago
8

There is a small scratch on a disco record located 14.5 cm from the center. When played, the scratch makes the record skip 30 ti

mes each minute (i.e., the scratch passes under the record player needle 30 times per minute). What is the linear speed of the scratch as it turns
Physics
1 answer:
LuckyWell [14K]3 years ago
3 0

Answer:

The linear speed of scratch is 45.55 cm/dec

Explanation:

The scratch located from the center at radial distance = 14.5 cm.

Each minute the skip of scratch when disco record played  = 30 times

Since, the scratch on the disc start revolving when disc is played and in one minute the scratch revolve 30 times. Therefore, the revolution = 30 RPM

Below is the calculation of linear speed of the scratch.

Angular \ velocity, \ w \ = \frac{2 \pi  \times RPM}{60} \\

w  =  \frac{2 \pi  \times 30}{60} = \pi  \ rad/dec \\

Linear \ velocity = wr \\

r = radial \ distance \\

v = \pi \times 14.5 \\

v = 45.55 cm/dec

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Answer:

a)   m_v = m_s ((\frac{w_o}{w})² - 1) ,  b)  m_v = 1.07 10⁻¹⁴ g

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where k is the spring constant and m is the mass of the oscillator

let's apply this expression to our case,

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in the two expressions the constant k is the same and q as the one property of the silicon bar, let us equal

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