Answer
10.9 Hz
Work Step by Step
Find the reactance of a capacitor with equation 21–12b, $X_C=\frac{1}{2 \pi f C}$, then solve for the frequency.
$f=\frac{1}{2 \pi X_C C}=\frac{1}{2 \pi (6.10\times10^3\Omega) (2.40\times10^{-6}F)}=10.9Hz$
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