The problem in the image reads:

"The x component of vector $\vec{A}$ is -27.3 m and the y component is +52.5 m. (a) What is the magnitude of $\vec{A}$? (b) What is the angle between the direction of $\vec{A}$ and the positive direction of x?"

Let's solve both parts step-by-step.

Part (a): Magnitude of $\vec{A}$

The magnitude $|\vec{A}|$ of a vector is calculated using the Pythagorean theorem:

$|\vec{A}| = \sqrt{A_x^2 + A_y^2}$

Where:

• $A_x = -27.3 \, \text{m}$
• $A_y = 52.5 \, \text{m}$

Part (b): Angle between $\vec{A}$ and the positive x-direction

The angle $\theta$ is given by:

$\theta = \tan^{-1} \left( \frac{A_y}{A_x} \right)$

I will now calculate these values.### Solutions: (a) The magnitude of vector $\vec{A}$ is approximately $59.2 \, \text{m}$.

(b) The angle between the direction of $\vec{A}$ and the positive x-direction is approximately $117.5^\circ$.

Would you like more details or have any questions?

Here are 5 related questions to expand your understanding:

1. How do you find the components of a vector if the magnitude and direction are known?
2. What is the difference between scalar and vector quantities?
3. Can you calculate the resultant of two vectors using their components?
4. How do you find the direction of a vector given its components?
5. How does changing the angle of a vector affect its x and y components?

Tip: Always use the correct quadrant when finding the angle of a vector to avoid directional mistakes.