1.

How can the we find out the result in magnitude of p and q are equal and the angles between them is tete

Answer»

ong>Answer:

R=P=[P2+Q2+2PQcosθ]1/2

R=P=[P2+Q2+2PQcosθ]1/2Q2+2PQcosθ=0

R=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 α

R=P=[P2+Q2+2PQcosθ]1/2Q2+2PQcosθ=0Q+2Pcosθ=0......{1}let the angle between resultant of 2P and Q be αtanα=Q+2Pcosθ2Psinθ

R=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 1

R=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

R=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



Discussion

No Comment Found

Related InterviewSolutions