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The angle that the net force makes with respect to the x- axis, is determined as 55.3⁰.
The angle that the net force makes with respect to the +x axis, is determined as follows;
F = qvBsinθ
Fy = qv(By)sinθ
Fx = qv(Bx)sinθ
The angle that the net force makes with respect to the +x axis;
tanθ = Fy/Fx
tanθ = qv(By)sinθ / qv(Bx)sinθ
tanθ = By/Bx
tanθ = 0.065/0.045
tanθ = 1.444
θ = tan⁻¹(1.444)
θ = 55.3⁰
Thus, the angle that the net force makes with respect to the x- axis, is determined as 55.3⁰.
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