Let's break down the problem: **Given:** - Wind velocity (\( V \)) = 100 km/h = 27.78 m/s (since \( 1 \text{ km/h} = 0.2778 \text{ m/s} \)) - Diameter of circular sign (\( D \)) = 2.5 m - Air temperature = 27°C - Air pressure = 101 kPa - Drag coefficient (\( C_d \)) = 1.3 **To Find:** - Drag force (\( F_d \)) on the sign **Formula for Drag Force:** \[ F_d = \frac{1}{2} C_d \rho A V^2 \] Where: - \( C_d \) = drag coefficient - \( \rho \) = density of air - \( A \) = frontal area of the sign - \( V \) = velocity of wind --- ### Step 1: Find Air Density (\( \rho \)) at 27°C and 101 kPa The density of air at standard conditions (0°C, 101.325 kPa) is about 1.293 kg/m³. At 27°C, we use the ideal gas law: \[ \rho = \frac{P}{R T} \] Where: - \( P \) = pressure = 101,000 Pa - \( R \) = specific gas constant for dry air = 287 J/(kg·K) - \( T \) = temperature in K = 27 + 273.15 = 300.15 K \[ \rho = \frac{101000}{287 \times 300.15} \approx 1.174 \text{ kg/m}^3 \] --- ### Step 2: Find Frontal Area (\( A \)) \[ A = \frac{\pi}{4} D^2 = \frac{\pi}{4} (2.5)^2 \approx 4.91 \text{ m}^2 \] --- ### Step 3: Calculate Drag Force (\( F_d \)) \[ F_d = \frac{1}{2} \times 1.3 \times 1.174 \times 4.91 \times (27.78)^2 \] First, calculate \( (27.78)^2 \): \[ (27.78)^2 = 772.97 \] Now plug in the values: \[ F_d = 0.5 \times 1.3 \times 1.174 \times 4.91 \times 772.97 \] \[ = 0.65 \times 1.174 \times 4.91 \times 772.97 \] \[ = 0.7631 \times 4.91 \times 772.97 \] \[ = 3.7479 \times 772.97 \] \[ = 2895.6 \text{ N} \] --- ### **Answer:** The nearest value in the options is **2896 N**. So, the correct answer is: **2896 N**