To find the magnitude of the vector $\mathbf{A}_2$ represented by the components $A_{x2} = -6.70 \, \text{m}$ and $A_{y2} = -3.00 \, \text{m}$, you can use the following formula for the magnitude of a vector in two dimensions:

$|\mathbf{A}_2| = \sqrt{A_{x2}^2 + A_{y2}^2}$

Substituting the given values:

$|\mathbf{A}_2| = \sqrt{(-6.70)^2 + (-3.00)^2}$

Let's calculate that:

$|\mathbf{A}_2| = \sqrt{44.89 + 9.00} = \sqrt{53.89}$

$|\mathbf{A}_2| \approx 7.34 \, \text{m}$

So, the magnitude of the vector $\mathbf{A}_2$ is approximately 7.34 meters.

Would you like to know more details or have any questions?

Here are 5 related questions for further exploration:

1. How would the magnitude change if $A_{y2}$ were positive?
2. What is the direction (angle) of the vector $\mathbf{A}_2$ relative to the positive x-axis?
3. How would you find the magnitude of a 3-dimensional vector?
4. Can a vector's magnitude be negative? Why or why not?
5. How would the magnitude change if both components were doubled?

Tip: Always square the components before summing when calculating the magnitude of a vector.