Question-57551

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

Tcosθ=mg_(moon) Tsinθ=mw^2 r tanθ=((w^2 r)/g_(moon) ) r=((tanθ×g_(moon) )/w^2 ) sinθ=(r/l) l=(r/(sinθ))=((tanθ×g_(moon) )/(w^2 sinθ))=(g_m /(w^2 cosθ))=((10)/6)×(1/2^2 )×2 l=(5/6)=0.8333 l=83.33cm

$${Tcos}\theta={mg}_{{moon}} \\ $$$${Tsin}\theta={mw}^{\mathrm{2}} {r} \\ $$$${tan}\theta=\frac{{w}^{\mathrm{2}} {r}}{{g}_{{moon}} } \\ $$$${r}=\frac{{tan}\theta×{g}_{{moon}} }{{w}^{\mathrm{2}} } \\ $$$${sin}\theta=\frac{{r}}{{l}} \\ $$$${l}=\frac{{r}}{{sin}\theta}=\frac{{tan}\theta×{g}_{{moon}} }{{w}^{\mathrm{2}} {sin}\theta}=\frac{{g}_{{m}} }{{w}^{\mathrm{2}} {cos}\theta}=\frac{\mathrm{10}}{\mathrm{6}}×\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }×\mathrm{2} \\ $$$${l}=\frac{\mathrm{5}}{\mathrm{6}}=\mathrm{0}.\mathrm{8333} \\ $$$$ \\ $$$${l}=\mathrm{83}.\mathrm{33}{cm} \\ $$$$ \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 07/Apr/19