The y-component of vector vec{A} is +3.0 m. | vedclass

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Question

(A) Let the magnitude of vector $\vec{A}$ be $A$. Given that the $y$-component of vector $\vec{A}$ is $A_y = 3.0 \ m$. The angle $	heta = 30^{\circ}$

Vedclass · Accepted Answer

$2 \sqrt{3} \ m$

Vedclass · Answer

(A) Let the magnitude of vector $\vec{A}$ be $A$. Given that the $y$-component of vector $\vec{A}$ is $A_y = 3.0 \ m$. The angle $	heta = 30^{\circ}$ is measured counterclockwise from the positive $y$-axis. Using the definition of the component of a vector, the $y$-component is given by $A_y = A \cos(	heta)$. Substituting the given values: $3.0 = A \cos(30^{\circ})$. Since $\cos(30^{\circ}) = \frac{\sqrt{3}}{2}$, we have $3.0 = A 	imes \frac{\sqrt{3}}{2}$. Solving for $A$: $A = \frac{3.0 	imes 2}{\sqrt{3}} = \frac{6}{\sqrt{3}} = 2\sqrt{3} \ m$.

The $y$-component of vector $\vec{A}$ is $+3.0 \ m$. If $\vec{A}$ makes an angle of $30^{\circ}$ counterclockwise from the positive $y$-axis, the magnitude of $\vec{A}$ is (assume $\vec{A}$ is in the $x-y$ plane).

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