Answer
(a) $0.75\%$
(b) $69\%$
Work Step by Step
(a) We know that
$f=1-\frac{v^2}{2c^2[\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}]-1}$
We plug in the known values to obtain:
$f=1-\frac{(0.10c)^2}{2c^2[\frac{1}{\sqrt{1-(\frac{0.10c}{c})^2}}-1]}$
This simplifies to:
$f=0.0075=0.75\%$
(b) We know that
$f=1-\frac{v^2}{2c^2[\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}]-1}$
We plug in the known values to obtain:
$f=1-\frac{(0.90c)^2}{2c^2[\frac{1}{\sqrt{1-(\frac{0.90c}{c})^2}}-1]}$
This simplifies to:
$f=0.687=68.7\%$
$\implies f=69\%$
