Velocity of the point D V₁ = 750w2j ej(105) AD = 750ej(105°) = 750x-0.2 xjx (cos 105° +j sin 105°) V=144.88+ j38.82

Elements Of Electromagnetics
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I want a way to calculate it on paper. Please, where did we get the number from (144.8+j38.8) ? very urgent (mechanical system course) i need explain
Velocity of the point D
V₁ = 750w2j ej(105)
AD = 750ej(105°)
= 750x-0.2 xjx (cos 105° +j sin 105°)
V=144.88+ j38.82
Transcribed Image Text:Velocity of the point D V₁ = 750w2j ej(105) AD = 750ej(105°) = 750x-0.2 xjx (cos 105° +j sin 105°) V=144.88+ j38.82
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