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Question Number 155919 by horpy4 last updated on 05/Oct/21
Answered by physicstutes last updated on 05/Oct/21
F×V=|ijk3uu2u+22u−3uu−2|=i(u3−2u2+3u2−6u)−j(3u2−6u−2u2−4u)+k(−9u2−2u3)F×V=(u3+u2−6u)i+(−u2+10u)j+(−9u2−2u3)k∫20F×Vdu=[(u44+u22−3u2)i+(−u33+5u2)j+(−3u3−12u4)k]02
Commented by horpy4 last updated on 05/Oct/21
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