Let's answer **only the last question** with explanation: ### What reaction force does the ground exert on Dave's skateboard? #### **Step 1: Understand the scenario** - Water exits the nozzle at high velocity, creating a **reaction force** (thrust) that pushes Dave (and his skateboard) in the opposite direction. - The ground must exert a force **equal and opposite** to this reaction to accelerate Dave. - We’re asked for the **reaction force applied by the ground** (i.e., the horizontal force). #### **Step 2: Find the momentum flux (thrust) from the water jet** The **thrust** (reaction force) generated is given by the **momentum flux** of the water leaving the nozzle: \[ F = \dot{m} V_2 \] where: - \(\dot{m}\) = mass flow rate of water - \(V_2\) = velocity of water at the nozzle exit #### **Step 3: Calculate mass flow rate (\(\dot{m}\))** \[ \dot{m} = \rho A_2 V_2 \] - \(\rho = 1000~\text{kg/m}^3\) - \(A_2 = \text{area of nozzle exit} = \frac{\pi d_2^2}{4}\) - \(d_2 = 2.5~\text{cm} = 0.025~\text{m}\) - \(V_2 = 30~\text{m/s}\) Plug in the values: \[ A_2 = \frac{\pi}{4}(0.025)^2 = \frac{\pi}{4} \times 0.000625 = 0.0004909~\text{m}^2 \] \[ \dot{m} = 1000 \times 0.0004909 \times 30 = 14.727~\text{kg/s} \] #### **Step 4: Calculate thrust** \[ F = \dot{m} V_2 = 14.727 \times 30 = 441.8~\text{N} \] #### **Step 5: Account for angle** The nozzle is directed at a \(5^\circ\) angle above the horizontal. The horizontal component (the only one that accelerates Dave) is: \[ F_{x} = F \cos{\theta} \] where \(\theta = 5^\circ\). \[ F_x = 441.8 \cos{5^\circ} = 441.8 \times 0.9962 = 440.1~\text{N} \] #### **Final Answer (with explanation):** \[ \boxed{440~\text{N}} \] **The reaction force that the ground exerts on Dave’s skateboard is approximately 440 N in the horizontal direction, opposite to the jet’s direction.** This is the force required to balance the momentum flux of the water exiting the nozzle at a 5° angle, so Dave and his skateboard accelerate forward.