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Find the center of mass of a thin wire lying along the curve r(t) = 3+1 + 3tj + قN | ل قت نہ k, Osts 2 if the density is 8 =5
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Given! - r(t) = 3ti + 3ti + ( 2 % E ostea d=5vistt x = 3t y = 3t 2=(2) t da = 3 dy = 3 al at We know that Mass Į f (x, y, z)Myz My z = 116] sat st ristt. Vistt. dt. 2 Mxz = {(3t) (18 tt) It < (56 tº +3+²) de 2 [ 27 +²+ at 7 27X4+80= 116 Mxz 116 2 3/we know that X = Myz. m. 116 8190 95 58 95 S8 x = 58 95 y = Maez 116 58 95 Igo cl 58 95 Z = Moey 1088 E 1088 =217652 105x tgo

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