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vA=0i+0j
vB=i+root of 3 j
rC=2i+0j
position vector of center of mass M
=(MrA+MrB+MrC)(M+M+M)
=(rA+rB+rC)^1/3
=(3i+root of 3j)^1/3
=i+0.577j
so,the answer is i+0.577j
vB=i+root of 3 j
rC=2i+0j
position vector of center of mass M
=(MrA+MrB+MrC)(M+M+M)
=(rA+rB+rC)^1/3
=(3i+root of 3j)^1/3
=i+0.577j
so,the answer is i+0.577j
sayalilokhande711:
two balls each of mass m are placed on the vertices A&B of an equilateral triangle ABC of side 1m . a ball of this system from vertex A(located at origin)is
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theortical question hey
perpendicular thoery apply koro
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hi sayalil
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