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blondinia
2 months ago
15

A resultant vector is 8.00 units long and makes an angle of 43.0 degrees measured ������� – ��������� with respect to the positi

ve � − ����. What are the magnitude and angle (measured ������� – ��������� with respect to the positive � − ����) of the equilibrant vector? Please show all steps in your calculations

Physics
1 answer:
Maru [3.3K]2 months ago
4 0

Answer:

223 degrees

Explanation:

We have the following information:

Magnitude of the resultant vector = 8 units

The resultant vector makes a counterclockwise angle with the positive x-axis

\theta=43^{\circ}

Our goal is to determine the magnitude and angle of the equilibriant vector.

We understand that the equilibrium vector has the same magnitude but is oriented in the opposite direction of the given vector.

Thus, the equilibrium vector's magnitude = 8 units

x-component of a vector=v_x=vcos\theta

Where v = Magnitude of vector

Using the formula:

x-component of the resultant vector=v_x=8cos43=5.85

y-component of the resultant vector=v_y=vsin\theta=8sin43=5.46

x-component of the equilibrium vector=v_x=-5.85

y-component of the equilibrium vector=-v_y=-5.46

Because the equilibrium vector lies in the third quadrant

\theta=tan^{-1}(\frac{v_x}{v_y})=tan^{-1}(\frac{-5.46}{-5.85})=43^{\circ}

The angle \theta' is in the third quadrant

In the third quadrant, the angle =\theta'+180^{\circ}

The angle of the equilibrium vector measured from the positive x-axis in a counterclockwise direction is 180+43=223 degrees

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1 month ago
Question 1
Keith_Richards [3271]

Answer:

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