Vector components: 1. (30'sqrt3)) m/s West, 30 m/s South
Explanation:
A vector can be resolved into its horizontal and vertical component by using trigonometry.
In particular, the horizontal component of a vector is given by
[tex]v_x = v cos \theta[/tex]
where
v is the magnitude of the vector
[tex]\theta[/tex] is its angle measured counterclockwise from the positive x-direction
And the vertical component is given by
[tex]v_y=v sin \theta[/tex]
In this problem, the magnitude of the vector is
[tex]v=60 m/s[/tex]
And the angle is
[tex]\theta=210^{\circ}[/tex]
Therefore, its  components are:
[tex]v_x=(60)(cos 210^{\circ})=60\cdot (-\frac{\sqrt{3}}{2})=-30\sqrt{3} m/s[/tex]
[tex]v_y=(60)(sin 210^{\circ})=60\cdot (-\frac{1}{2})=-30 m/s[/tex]
And both the components are negative: this means that the vector lies in the 3rd quadrant, so the x-component is towards west, while the y-component is towards south. So the correct answer is
1. (30'sqrt3)) m/s West, 30 m/s South
Learn more about vector components:
brainly.com/question/2678571
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