Answered

Westonci.ca offers fast, accurate answers to your questions. Join our community and get the insights you need now. Explore thousands of questions and answers from a knowledgeable community of experts on our user-friendly platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.

Determine the magnitude of the vector sum \( V = V_1 + V_2 \) and the angle \( \theta_x \) which \( V \) makes with the positive \( x \)-axis. Complete both graphical and algebraic solutions. Assume \( a = 4 \), \( b = 5 \), \( V_1 = 21 \) units, \( V_2 = 23 \) units, and \( \theta = 59^{\circ} \).

Answers:
[tex]\[
\begin{array}{l}
V = 2.55 \, \text{units}
\end{array}
\][/tex]

Sagot :

GinnyAnswer
To determine the magnitude of the vector sum \( \mathbf{V} = \mathbf{V}_1 + \mathbf{V}_2 \) and the angle \( \theta_x \) which \( \mathbf{V} \) makes with the positive \( x \)-axis, we'll break down the vectors into their components and then sum them.

Step-by-Step Solution:

1. Determine the components of \( \mathbf{V}_1 \) and \( \mathbf{V}_2 \):
- \( \mathbf{V}_1 \) has magnitude 21 units and is along the positive \( x \)-axis (\( \theta = 0^\circ \)).

[tex]\[ V_{1x} = V_1 \cos(0^\circ) = 21 \cos(0^\circ) = 21.0 \text{ units} \][/tex]
[tex]\[ V_{1y} = V_1 \sin(0^\circ) = 21 \sin(0^\circ) = 0.0 \text{ units} \][/tex]

- \( \mathbf{V}_2 \) has magnitude 23 units and makes an angle \( \theta = 59^\circ \) with the positive \( x \)-axis.

[tex]\[ V_{2x} = V_2 \cos(59^\circ) = 23 \cos(59^\circ) \approx 11.85 \text{ units} \][/tex]
[tex]\[ V_{2y} = V_2 \sin(59^\circ) = 23 \sin(59^\circ) \approx 19.71 \text{ units} \][/tex]

2. Sum the components to get the resultant vector \( \mathbf{V} \):
- Sum the \( x \)-components:

[tex]\[ V_x = V_{1x} + V_{2x} = 21.0 + 11.85 = 32.85 \text{ units} \][/tex]

- Sum the \( y \)-components:

[tex]\[ V_y = V_{1y} + V_{2y} = 0.0 + 19.71 = 19.71 \text{ units} \][/tex]

3. Calculate the magnitude of the resultant vector \( \mathbf{V} \):

[tex]\[ V = \sqrt{V_x^2 + V_y^2} = \sqrt{(32.85)^2 + (19.71)^2} \approx 38.31 \text{ units} \][/tex]

4. Determine the angle \( \theta_x \) which \( \mathbf{V} \) makes with the positive \( x \)-axis:

[tex]\[ \theta_x = \tan^{-1} \left( \frac{V_y}{V_x} \right) = \tan^{-1} \left( \frac{19.71}{32.85} \right) \approx 30.97^\circ \][/tex]

Conclusion:

The magnitude of the vector sum [tex]\( \mathbf{V} \)[/tex] is approximately 38.31 units, and the angle [tex]\( \theta_x \)[/tex] which [tex]\( \mathbf{V} \)[/tex] makes with the positive [tex]\( x \)[/tex]-axis is approximately 30.97 degrees.
We hope this was helpful. Please come back whenever you need more information or answers to your queries. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.