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 36855 by Tinkutara last updated on 06/Jun/18

Commented by Tinkutara last updated on 06/Jun/18

Answered by ajfour last updated on 11/Jun/18

(τ_A /τ_B ) = (6/(√(11))) .

$$\frac{\tau_{{A}} }{\tau_{{B}} }\:=\:\frac{\mathrm{6}}{\sqrt{\mathrm{11}}}\:. \\ $$

Commented by Tinkutara last updated on 11/Jun/18

How?

Commented by ajfour last updated on 11/Jun/18

Commented by ajfour last updated on 11/Jun/18

(τ_A /τ_B )=(((D/(v+u))+(D/(v−u)))/((2D)/(√(v^2 −u^2 )))) with v=1.2u=((6u)/5) v+u=((11u)/5) ; v−u=(u/5) ; (√(v^2 −u^2 )) = ((√(11))/5) (τ_A /τ_B ) = (((1/(11))+1)/(2/(√(11)))) = (6/(√(11))) .

$$\frac{\tau_{{A}} }{\tau_{{B}} }=\frac{\frac{{D}}{{v}+{u}}+\frac{{D}}{{v}−{u}}}{\frac{\mathrm{2}{D}}{\sqrt{{v}^{\mathrm{2}} −{u}^{\mathrm{2}} }}}\: \\ $$$${with}\:{v}=\mathrm{1}.\mathrm{2}{u}=\frac{\mathrm{6}{u}}{\mathrm{5}} \\ $$$${v}+{u}=\frac{\mathrm{11}{u}}{\mathrm{5}}\:\:;\:\:{v}−{u}=\frac{{u}}{\mathrm{5}}\:;\: \\ $$$$\sqrt{{v}^{\mathrm{2}} −{u}^{\mathrm{2}} }\:=\:\frac{\sqrt{\mathrm{11}}}{\mathrm{5}} \\ $$$$\frac{\tau_{{A}} }{\tau_{{B}} }\:=\:\frac{\frac{\mathrm{1}}{\mathrm{11}}+\mathrm{1}}{\frac{\mathrm{2}}{\sqrt{\mathrm{11}}}}\:=\:\frac{\mathrm{6}}{\sqrt{\mathrm{11}}}\:. \\ $$

Commented by Tinkutara last updated on 14/Jun/18

Thanks Sir!

Terms of Service

Privacy Policy

Contact: info@tinkutara.com