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A coin of mass 3.68 grams is placed on a record turning at 33 1/3 rpm (revolutions per minute = angular frequency). What is the period of the record? Include units in your answer. Answer must be in 3 significant digits.

Sagot :

33 1/3 rpm can be written as follow;

[tex]33\frac{1}{3}\text{rpm}=\frac{100}{3}rpm[/tex]

The period is given by:

[tex]T=\frac{2\pi}{\omega}[/tex]

where w = 100/3 rpm is the angular frequency. Convert 100/3 rpm to rev/s, as follow:

[tex]\frac{100}{3}\frac{\text{rev}}{\min}\cdot\frac{1\min }{60s}=0.555\frac{rev}{s}[/tex]

By replacing into the formula for the period T you obtain:

[tex]T=\frac{2\pi}{0.555\frac{\text{rev}}{s}}\approx11.3s[/tex]

Hence, the period is approximately 11.3s

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