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For Part C, we need to find the \( x \)- and \( y \)-components of the vector \( \vec{a} = (13 \, \text{m/s}^2, \, 32^\circ \, \text{left of} \, -y\text{-axis}) \).

### Step-by-Step Solution:

1. **Interpret the Angle:**
   - The vector is \( 32^\circ \) left of the negative \( y \)-axis.
   - This means the angle with the positive \( x \)-axis is \( 180^\circ - 32^\circ = 148^\circ \).

2. **Calculate \( x \)-component:**
   \[
   a_x = a \cdot \cos \theta = 13 \cdot \cos(148^\circ)
   \]

3. **Calculate \( y \)-component:**
   \[
   a_y = a \cdot \sin \theta = 13 \cdot \sin(148^\circ)
   \]

4. **Evaluate \( \cos(148^\circ) \) and \( \sin(148^\circ) \):**
   - \( \cos(148^\circ) \approx -0.848 \)
   - \( \sin(148^\circ) \approx 0.530 \)

5. **Plug in the values:**
   - \( a_x = 13 \cdot (-0.848) = -11.024 \, \text{m/s}^2 \)
   - \( a_y = 13 \cdot 0.530 = 6.89 \, \text{m/s}^2 \)

6. **Final Components:**
   - \( \vec{a} = (-11.0, 6.9) \, \text{m/s}^2 \)

### Answer:
Enter the components as `-11.0, 6.9`.

Would you like any more details or have questions about this? Here are some related topics:

1. How to determine vector components given different angle conventions.
2. Using trigonometric functions for vectors in different quadrants.
3. Converting angles between different reference directions.
4. Understanding vector components and their physical interpretation.
5. Application of vector decomposition in physics problems.

**Tip:** Always check the reference direction and angle conventions before calculating vector components to ensure accuracy.

Find the x- and y-components of the vector \( \vec{a} = (13 \, \text{m/s}^2, \, 32^\circ \, \text{left of} \, -y\text{-axis}) \).

Learn how to find the x- and y-components of a vector given its magnitude and direction relative to the y-axis. This problem involves using trigonometric functions to decompose vectors and is suitable for high school students in Grades 10-12.

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