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Question Number 136593 by I want to learn more last updated on 23/Mar/21

Answered by mr W last updated on 24/Mar/21

Commented by otchereabdullai@gmail.com last updated on 25/Mar/21

World best!

Worldbest!

Commented by mr W last updated on 24/Mar/21

ω_(DB) =10 rad/s  v_B =150(√2)ω_(DB)  =1500(√2) mm/s  v_(Bx) =v_B cos 45°=1500 mm/s=1.5 m/s  v_(By) =v_B sin 45°=1500 mm/s=1.5 m/s  v_(Ax) =v_(Dx) =v_(Bx) =1.5 m/s  v_(Ay) =0  v_(Dy) =((150)/(150+200))×v_(By) =(3/7)×1.5=((4.5)/7) m/s  ⇒v_A =1.5 m/s  ⇒v_D =(√(1.5^2 +(((4.5)/7))^2 ))=1.642 m/s  ⇒ω_(AB) =(v_(By) /(150+200))=((1500)/(350))=((30)/7)=4.286 rad/s  ω_(CA) =(v_(Ax) /(150))=((1500)/(150))=10 rad/s

ωDB=10rad/svB=1502ωDB=15002mm/svBx=vBcos45°=1500mm/s=1.5m/svBy=vBsin45°=1500mm/s=1.5m/svAx=vDx=vBx=1.5m/svAy=0vDy=150150+200×vBy=37×1.5=4.57m/svA=1.5m/svD=1.52+(4.57)2=1.642m/sωAB=vBy150+200=1500350=307=4.286rad/sωCA=vAx150=1500150=10rad/s

Commented by I want to learn more last updated on 24/Mar/21

I really appreciste sir. God bless you.

Ireallyapprecistesir.Godblessyou.

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