Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 57551 by necx1 last updated on 07/Apr/19

Commented by necx1 last updated on 07/Apr/19

$$pleasehelpafriend.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 08/Apr/19

$$Tcos\theta ={mg}_{moon}$$$$Tsin\theta ={mw}^{2}r$$$$tan\theta =\frac{w^{2}r}{g_{moon}}$$$$r=\frac{tan\theta \times g_{moon}}{w^2}$$$$sin\theta =\frac{r}{l}$$$$l=\frac{r}{sin\theta }=\frac{tan\theta \times g_{moon}}{w^{2}sin\theta }=\frac{g_m}{w^{2}cos\theta }=\frac{10}{6}\times \frac{1}{2^2}\times 2$$$$l=\frac{5}{6}=0.8333$$$$$$$$l=83.33cm$$$$$$

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

$$Thanks$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com