The magnitude of a vector G pointed 36.9° clockwise from the positive y-axis, with an x-component of 3, is 5.
The magnitude of the vector is given by:
[tex] \overline{G} = \sqrt{G_{x}^{2} + G_{y}^{2}} [/tex]
Where:
[tex] G_{x}[/tex]: is the x-component of the vector G = 3
[tex]G_{y}[/tex]: is the y-component of the vector G
We need to calculate the y-component of the vector. We can use the following trigonometric function:
[tex] tan(\theta) = \frac{G_{x}}{G_{y}} [/tex]
Where:
θ: is the angle = 36.9° (clockwise from the positive y-axis)
Hence, the y-component of the vector is:
[tex] G_{y} = \frac{G_{x}}{tan(\theta)} = \frac{3}{tan(36.9)} = 4 [/tex]
Now, the magnitude of the vector is:
[tex] \overline{G} = \sqrt{G_{x}^{2} + G_{y}^{2}} = \sqrt{(3)^{2} + (4)^{2}} = 5 [/tex]
Therefore, the magnitude of the vector G is 5.
You can find another example of the calculation of vector's magnitude here: https://brainly.com/question/13134973?referrer=searchResults
I hope it helps you!
