Question

Pixel A is located 80 km from the Jacksonville radar. How many seconds did it take for the radar pulse to travel to the target at Pixel A and return to the radar? (It will be a very small answer!)

C= 3 × 10^8 m s−1
R= 80,000
R= CT/2

Answer 1

`C=3* 10^8~(m)/(s)\hspace{5em}R=80000\implies R=8* 10^4 \\\\[-0.35em] ~\dotfill\\\\ R=\cfrac{CT}{2}\implies 2R=CT\implies \cfrac{2R}{C}=T\implies \cfrac{2(8* 10^4)}{3* 10^8}=T \\\\\\ \cfrac{16* 10^4}{3* 10^8}=T \implies \cfrac{16}{3}*\cfrac{10^4}{10^8}=T\implies \cfrac{16}{3}*\cfrac{1}{10^8\cdot 10^(-4)}=T \\\\\\ \cfrac{16}{3}*\cfrac{1}{10^(8-4)}=T \implies \cfrac{16}{3* 10^4}=T\implies \cfrac{16}{30000}=T\implies \cfrac{1}{1875}=T`