$x_L=2fL=2(1050)(00*{10}^{-6})=\mathrm{.}6597$

$x_C=\frac{1}{2fC}=\frac{1}{2(1050)(5\mathrm{.}03)}=3,01*{10}^{-5}$

$x_L=x_L=\mathrm{.}6597*3,4\mathrm{.}10-5$

$($f$)$ If the input signal is a sinusoidal voltage with an amplitude of $90\mathrm{.}0$ mV $($at the AM frequency above$),$ what is the amplitude of the output voltage?

$($g$)$ If the knob on the capacitor is now turned so that $=120,$ what is the frequency the radio is now tuned to$?$