**Problem Statement:** A child weighing 200 N is being held back in a swing by a horizontal force of 125 N, as shown in the image. What is the tension \( T \) in the rope that supports the swing in units of Newtons? **Note:** Please enter only the numerical answer. If you include any units in your answer, your answer will be counted as incorrect. **Diagram Explanation:** - The diagram shows a child sitting on a swing. - The swing is supported by a rope with tension labeled as \( T \). - A horizontal force \( F \) of 125 N is applied, pulling the swing backward. - The weight of the child, represented by a downward arrow, is labeled as 200 N. There is a text box for inputting the numerical answer. **Calculation Description (not shown in the image):** To solve the problem, you need to consider the forces acting on the system. Use the Pythagorean theorem to find the tension in the rope: \[ T = \sqrt{(F^2 + \text{Weight}^2)} \] Substitute the known values: \[ T = \sqrt{(125^2 + 200^2)} \] \[ T = \sqrt{15625 + 40000} \] \[ T = \sqrt{55625} \] \[ T = 235.75 \text{ N (approximately)} \] Remember, you must enter only the number 235.75 for the answer without units. The image presents a scenario involving two forces acting on a boat. The question posed is: "What is the x-component of the resultant force in the figure below?" **Diagram Explanation:** - The diagram shows a top-down view of a boat on water with two forces, \( F_A \) and \( F_B \), applied at angles. - **Force \( F_A \):** - Magnitude: \( 40.0 \, \text{N} \) - Angle: \( 45.0^\circ \) with respect to a vertical line, making an effective angle of \( 45.0^\circ \) to the x-axis. - **Force \( F_B \):** - Magnitude: \( 30.0 \, \text{N} \) - Angle: \( 37.0^\circ \) below the negative x-axis. - Coordinate System: - The diagram includes an xy-coordinate system with positive x directed to the right and positive y directed upward. For each force, the x-component can be calculated using trigonometric functions: - **X-component of \( F_A \):** \( F_{Ax} = F_A \cdot \cos(45.0^\circ) \) - **X-component of \( F_B \):** \( F_{Bx} = F_B \cdot \cos(37.0^\circ) \) To find the resultant x-component of the force, the x-components of \( F_A \) and \( F_B \) should be added, keeping in mind the direction associated with the angles.