To solve for the magnitude of the resultant vector and the angle it makes with the positive x-axis, we can break the vectors $\vec{v}$ and $\vec{w}$ into their components, add the components to find the resultant vector, and then calculate the magnitude and direction of the resultant.

Step 1: Find the components of $\vec{v}$ and $\vec{w}$

1. Vector $\vec{v}$:
   • Magnitude = 5
   • Direction angle = 140° (measured from the positive x-axis)

The components of $\vec{v}$ are given by: $v_x = |\vec{v}| \cdot \cos(\theta_v) = 5 \cdot \cos(140^\circ)$ $v_y = |\vec{v}| \cdot \sin(\theta_v) = 5 \cdot \sin(140^\circ)$

Using a calculator: $v_x = 5 \cdot \cos(140^\circ) \approx -3.83$ $v_y = 5 \cdot \sin(140^\circ) \approx 3.83$

Thus, the components of $\vec{v}$ are: $\vec{v} = (-3.83, 3.83)$

2. Vector $\vec{w}$:
   • Magnitude = 3
   • Direction angle = 70°

The components of $\vec{w}$ are: $w_x = |\vec{w}| \cdot \cos(\theta_w) = 3 \cdot \cos(70^\circ)$ $w_y = |\vec{w}| \cdot \sin(\theta_w) = 3 \cdot \sin(70^\circ)$

Using a calculator: $w_x = 3 \cdot \cos(70^\circ) \approx 1.03$ $w_y = 3 \cdot \sin(70^\circ) \approx 2.82$

Thus, the components of $\vec{w}$ are: $\vec{w} = (1.03, 2.82)$

Step 2: Find the components of the resultant vector

Now, add the components of $\vec{v}$ and $\vec{w}$ to get the components of the resultant vector $\vec{R}$.

$R_x = v_x + w_x = -3.83 + 1.03 = -2.80$ $R_y = v_y + w_y = 3.83 + 2.82 = 6.65$

So the resultant vector $\vec{R}$ has components: $\vec{R} = (-2.80, 6.65)$

Step 3: Calculate the magnitude of the resultant vector

The magnitude of the resultant vector $\vec{R}$ is given by: $|\vec{R}| = \sqrt{R_x^2 + R_y^2}$

Substitute the values: $|\vec{R}| = \sqrt{(-2.80)^2 + (6.65)^2} = \sqrt{7.84 + 44.22} = \sqrt{52.06} \approx 7.2$

Thus, the magnitude of the resultant vector is approximately: $|\vec{R}| \approx 7.2$

Step 4: Calculate the angle $\theta$ of the resultant vector

The angle $\theta$ of the resultant vector with respect to the positive x-axis is given by: $\theta = \tan^{-1}\left(\frac{R_y}{R_x}\right)$

Substitute the values: $\theta = \tan^{-1}\left(\frac{6.65}{-2.80}\right) \approx \tan^{-1}(-2.38) \approx -1.21 \, \text{radians}$

Convert this to degrees: $\theta = -1.21 \times \frac{180}{\pi} \approx -69.3^\circ$

Since the angle is negative, the resultant vector lies in the second quadrant. To get the angle with respect to the positive x-axis, we add 180°: $\theta = 180^\circ - 69.3^\circ = 110.7^\circ$

Thus, the angle of the resultant vector is approximately: $\theta \approx 110.7^\circ$

Final Answer:

• Magnitude of the resultant vector: $\boxed{7.2}$
• Angle with respect to the positive x-axis: $\boxed{110.7^\circ}$

Would you like further clarification or details on any part of the process?

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Here are some related questions that may interest you:

1. How would the magnitude and direction change if the vectors were at 90° and 45°, respectively?
2. What happens to the angle of the resultant if the vectors are orthogonal (at 90° to each other)?
3. How can we use vector addition to solve real-world physics problems, like force addition?
4. How do you find the resultant of more than two vectors?
5. What would the angle of the resultant be if both vectors were pointing in the same direction?

Tip: When calculating angles in trigonometry, be mindful of the quadrant in which the angle lies, especially when using inverse trigonometric functions.