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the period T (in seconds) of a pendulum is given by t = 2 pi √(L/32), where else stands for the length in feet of the pendulum. if pi equals 3.14, and the period is 12.56 seconds, what is the length?

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The period T of a pendulum is given by:


T=2\pi*\sqrt[]{(L)/(32ft/s^2)}

Where T is the period (s)

L the length (ft)

π is 3.14

A value of T=12.56 seconds is given. Let's find the length then:


\begin{gathered} \text{Divide both sides by 2}\pi \\ (12.56s)/(2\pi)=\frac{2\pi*\sqrt[]{(L)/(32ft/s^2)}}{2\pi} \\ \text{Simplify} \\ (12.56s)/(2(3.14))=\sqrt[]{(L)/(32ft/s^2)} \\ 2s=\sqrt[]{(L)/(32ft/s^2)} \\ \text{Apply square to both sides} \\ (2s)^2=\sqrt[]{(L)/(32ft/s^2)}^2 \\ 4s^2=(L)/(32ft/s^2) \\ \text{Multiply both sides by 32}ft/s^2 \\ 4s^2*32ft/s^2=(L)/(32ft/s^2)*32ft/s^2 \\ \text{Simplify} \\ 128ft=L \\ \text{And reorder terms} \\ L=128ft \end{gathered}

Thus, the length of a pendulum with a period of T=12.56s is equal to 128ft

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