Question Number 57551 by necx1 last updated on 07/Apr/19
Commented by necx1 last updated on 07/Apr/19
$${please}\:{help}\:{a}\:{friend}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 08/Apr/19
$${Tcos}\theta={mg}_{{moon}} \\ $$$${Tsin}\theta={mw}^{\mathrm{2}} {r} \\ $$$${tan}\theta=\frac{{w}^{\mathrm{2}} {r}}{{g}_{{moon}} } \\ $$$${r}=\frac{{tan}\theta×{g}_{{moon}} }{{w}^{\mathrm{2}} } \\ $$$${sin}\theta=\frac{{r}}{{l}} \\ $$$${l}=\frac{{r}}{{sin}\theta}=\frac{{tan}\theta×{g}_{{moon}} }{{w}^{\mathrm{2}} {sin}\theta}=\frac{{g}_{{m}} }{{w}^{\mathrm{2}} {cos}\theta}=\frac{\mathrm{10}}{\mathrm{6}}×\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }×\mathrm{2} \\ $$$${l}=\frac{\mathrm{5}}{\mathrm{6}}=\mathrm{0}.\mathrm{8333} \\ $$$$ \\ $$$${l}=\mathrm{83}.\mathrm{33}{cm} \\ $$$$ \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 07/Apr/19
Commented by tanmay.chaudhury50@gmail.com last updated on 07/Apr/19
Commented by necx1 last updated on 08/Apr/19
$${Thanks} \\ $$