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A point on a spinning record experiences a centripetal acceleration of .75 cm/s^2 . How far is the point from center of the record if the record spins at 0.5 revolutions per second?

Answer :

ANSWER:

0.076 cm

STEP-BY-STEP EXPLANATION:

Given:

a = 0.75 cm/s^2

angular velocity = 0.5 rps

[tex]\begin{gathered} \omega=0.5\cdot2\pi\text{ rad/s} \\ \omega=\pi\text{ rad/s} \end{gathered}[/tex]

If the distance of record from the center be r, then as we know:

[tex]\begin{gathered} a=\omega^2\cdot r \\ \text{ we solve for r} \\ r=\frac{a}{\omega^2} \\ r=\frac{0.75}{(3.14)^2} \\ r=0.076cm \end{gathered}[/tex]

Therefore, this distance would be 0.076 centimeters

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