Answer
a. $L(0.5c)=10\sqrt \frac{3}{4}=5\sqrt3,$
$L(0.75c)=10\sqrt \frac{7}{16}=\frac{5\sqrt7}{2},$
$L(0.9c)=10\sqrt \frac{19}{100}=\sqrt{19},$
b. The legth of an object decreases as its velocity Increases
Work Step by Step
$L(x)=10\sqrt {1-\frac{v^2}{c^2}}$
a. $L(0.5c)=10\sqrt {1-\frac{(300,000(0.5))^2}{(300,000^2)}}$,
$L(0.5c)=10\sqrt \frac{3}{4}=5\sqrt3,$
$L(0.75c)=10\sqrt {1-\frac{(300,000(0.75))^2}{(300,000^2)}}$,
$L(0.75c)=10\sqrt \frac{7}{16}=\frac{5\sqrt7}{2},$
$L(0.9c)=10\sqrt {1-\frac{(300,000(0.9))^2}{(300,000^2)}}$,
$L(0.9c)=10\sqrt \frac{19}{100}=\sqrt{19},$
b. The legth of an object decreases as its velocity Increases
