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.

Learn how to calculate the components of a vector at an angle of 270 degrees using trigonometric functions. This problem covers vector magnitude, direction, and trigonometric calculations. Suitable for high school students.

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