Question and Answers Forum

All Questions      Topic List

Mechanics Questions

Previous in All Question      Next in All Question      

Previous in Mechanics      Next in Mechanics      

Question Number 120237 by aurpeyz last updated on 30/Oct/20

Commented by Dwaipayan Shikari last updated on 30/Oct/20

△P=∣m.v_1 −mv_2 ∣=m(√(2g))(2+(√3))  v_1 =(√(2g.4))  v_2 =(√(2g.3))  ((△P)/(△t))=F_(ext) =((m(√(2g))(2+(√3)))/(0.01))  a=(F_(ext) /m)=100(√(2g)) (2+(√3))  (upward)

$$\bigtriangleup{P}=\mid{m}.{v}_{\mathrm{1}} −{mv}_{\mathrm{2}} \mid={m}\sqrt{\mathrm{2}{g}}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right) \\ $$$${v}_{\mathrm{1}} =\sqrt{\mathrm{2}{g}.\mathrm{4}} \\ $$$${v}_{\mathrm{2}} =\sqrt{\mathrm{2}{g}.\mathrm{3}} \\ $$$$\frac{\bigtriangleup{P}}{\bigtriangleup{t}}={F}_{{ext}} =\frac{{m}\sqrt{\mathrm{2}{g}}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)}{\mathrm{0}.\mathrm{01}} \\ $$$${a}=\frac{{F}_{{ext}} }{{m}}=\mathrm{100}\sqrt{\mathrm{2}{g}}\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\:\:\left({upward}\right) \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com