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How can the we find out the result in magnitude of p and q are equal and the angles between them is tete |
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Answer» ong>Answer: R=P=[P2+Q2+2PQcosθ]1/2R=P=[P2+Q2+2PQcosθ]1/2Q2+2PQcosθ=0R=P=[P2+Q2+2PQcosθ]1/2Q2+2PQcosθ=0Q+2Pcosθ=0......{1}R=P=[P2+Q2+2PQcosθ]1/2Q2+2PQcosθ=0Q+2Pcosθ=0......{1}LET the angle between resultant of 2P and Q be αR=P=[P2+Q2+2PQcosθ]1/2Q2+2PQcosθ=0Q+2Pcosθ=0......{1}let the angle between resultant of 2P and Q be αtanα=Q+2Pcosθ2PsinθR=P=[P2+Q2+2PQcosθ]1/2Q2+2PQcosθ=0Q+2Pcosθ=0......{1}let the angle between resultant of 2P and Q be αtanα=Q+2Pcosθ2Psinθfrom EQUATION 1R=P=[P2+Q2+2PQcosθ]1/2Q2+2PQcosθ=0Q+2Pcosθ=0......{1}let the angle between resultant of 2P and Q be αtanα=Q+2Pcosθ2Psinθfrom equation 1tanα=∞→α=90R=P=[P2+Q2+2PQcosθ]1/2Q2+2PQcosθ=0Q+2Pcosθ=0......{1}let the angle between resultant of 2P and Q be αtanα=Q+2Pcosθ2Psinθfrom equation 1tanα=∞→α=90 |
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