Given that $\vec{A}$ has a magnitude of $8.0 \, \text{m}$ and is directed along the negative $y$-axis (at $270^\circ$ or $-90^\circ$ with respect to the positive $x$-axis):

• $A_x = 8.0 \times \cos(270^\circ) = 0 \, \text{m}$
• $A_y = 8.0 \times \sin(270^\circ) = -8.0 \, \text{m}$

So, the components are:

$A_x = 0 \, \text{m}, \quad A_y = -8.0 \, \text{m}$

Would you like any further details or have additional questions?

Here are some related questions:

1. How would you find the components of a vector if the angle was 45° instead of 90°?
2. What is the significance of the negative sign in $A_y$?
3. How can the direction of a vector be verified using its components?
4. How does changing the magnitude affect the components of a vector?
5. How would the calculation change if the vector was not aligned with one of the axes?

Tip: Always ensure that the angle used in the trigonometric functions is correctly referenced from the positive $x$-axis, adjusting for any necessary quadrant changes.