Step 1: Resolve the first displacement into components. The first displacement is $20.0 \text{ m}$ east. $$d_{1x} = 20.0 \text{ m}$$ $$d_{1y} = 0 \text{ m}$$ Step 2: Resolve the second displacement into components. The second displacement is $20.0 \text{ m}$ at $30^\circ$ north of east. $$d_{2x} = 20.0 \cos(30^\circ) \text{ m}$$ $$d_{2y} = 20.0 \sin(30^\circ) \text{ m}$$ Using $\cos(30^\circ) = \frac{\sqrt{3}}{2}$ and $\sin(30^\circ) = \frac{1}{2}$: $$d_{2x} = 20.0 \left(\frac{\sqrt{3}}{2}\right) = 10\sqrt{3} \text{ m} \approx 17.32 \text{ m}$$ $$d_{2y} = 20.0 \left(\frac{1}{2}\right) = 10.0 \text{ m}$$ Step 3: Sum the x-components and y-components to find the resultant components. $$R_x = d_{1x} + d_{2x} = 20.0 \text{ m} + 10\sqrt{3} \text{ m} = (20.0 + 10\sqrt{3}) \text{ m}$$ $$R_x \approx 20.0 + 17.32 = 37.32 \text{ m}$$ $$R_y = d_{1y} + d_{2y} = 0 \text{ m} + 10.0 \text{ m} = 10.0 \text{ m}$$ Step 4: Calculate the magnitude of the resultant displacement. The magnitude $R$ is given by the Pythagorean theorem: $$R = \sqrt{R_x^2 + R_y^2}$$ $$R = \sqrt{(20.0 + 10\sqrt{3})^2 + (10.0)^2}$$ $$R = \sqrt{(400 + 400\sqrt{3} + 300) + 100}$$ $$R = \sqrt{800 + 400\sqrt{3}}$$ Using $\sqrt{3} \approx 1.732$: $$R = \sqrt{800 + 400(1.732)}$$ $$R = \sqrt{800 + 692.8}$$ $$R = \sqrt{1492.8}$$ $$R \approx 38.637 \text{ m}$$ Rounding to one decimal place, the magnitude is $38.6 \text{ m}$. Comparing with the given options: A. 28.3 m B. 40.0 m C. 38.6 m D. 10.4 m The closest option is C. The final answer is $\boxed{\text{C. } 38.6 \text{ m}}$. Send me the next one 📸