Question

The force,

Answer 1

$\qquad \qquad \textit{inverse proportional variation} \\\\ \textit{\underline{y} varies inversely with \underline{x}} ~\hspace{6em} \stackrel{\textit{constant of variation}}{y=\cfrac{\stackrel{\downarrow }{k}}{x}~\hfill } \\\\ \textit{\underline{x} varies inversely with }\underline{z^5} ~\hspace{5.5em} \stackrel{\textit{constant of variation}}{x=\cfrac{\stackrel{\downarrow }{k}}{z^5}~\hfill } \\\\[-0.35em] \rule{34em}{0.25pt}$

$\begin{array}{llll} \textit{"F" is inversely proportional}\\ \textit{to the square of "d"}\\ F=\cfrac{k}{d^2} \end{array}\qquad \textit{we also know that} \begin{cases} F=\stackrel{Newtons}{0.009}\\ d=\stackrel{meters}{2} \end{cases}$

$0.009=\cfrac{k}{(2)^2}\implies 0.009=\cfrac{k}{4}\implies 0.036=k~\hfill \boxed{F=\cfrac{0.036}{d^2}} \\\\\\ \textit{when F = 0.062, what is "d"?}~~~~~~0.062=\cfrac{0.036}{d^2} \\\\\\ d^2=\cfrac{0.036}{0.062}\implies d^2=\cfrac{18}{31}\implies d=\sqrt{\cfrac{18}{31}}\implies d\approx 0.76$