# Step-by-Step Solution Let's break down the problem and solve for the Mach number. --- ## **Step 1: Write the Given Data** - **Temperature, \( T \) = 42°C** - **Velocity of airplane, \( V \) = 295 km/h** --- ## **Step 2: Convert Units Where Necessary** ### **Convert Temperature to Kelvin** \[ T(K) = T(^\circ C) + 273.15 = 42 + 273.15 = 315.15~K \] ### **Convert Speed to m/s** \[ V = 295~\text{km/h} \times \frac{100~\text{m}}{1~\text{km}} \times \frac{1~\text{h}}{360~\text{s}} = 81.94~\text{m/s} \] --- ## **Step 3: Calculate the Speed of Sound at 42°C** The speed of sound in air is given by: \[ a = \sqrt{\gamma R T} \] Where: - \( \gamma \) (ratio of specific heats for air) = **1.4** - \( R \) (specific gas constant for air) = **287 J/kg·K** - \( T \) = **315.15 K** \[ a = \sqrt{1.4 \times 287 \times 315.15} \] \[ a = \sqrt{1.4 \times 287 \times 315.15} \] \[ = \sqrt{1.4 \times 287 = 401.8} \] \[ = \sqrt{401.8 \times 315.15} = \sqrt{126585.57} \] \[ a \approx 356.1~\text{m/s} \] --- ## **Step 4: Calculate the Mach Number** The Mach number is defined as: \[ M = \frac{V}{a} \] Substitute the values: \[ M = \frac{81.94}{356.1} \approx .23 \] --- ## **Step 5: Final Answer** ### **Final Answer** \[ \boxed{M \approx .23} \] --- ### **Assumptions Made** - Air behaves as an ideal gas. - The specific heat ratio (\( \gamma \)) for air is 1.4. - The specific gas constant (\( R \)) for air is 287 J/kg·K. - The altitude effect on temperature and pressure is negligible (close to ground before landing). --- ## **Summary Table** | Parameter | Value | |-----------|-------| | Temperature (\( T \)) | 42°C (315.15 K) | | Airplane Speed (\( V \)) | 295 km/h (81.94 m/s) | | Speed of Sound (\( a \)) | 356.1 m/s | | **Mach Number (\( M \))** | **.23** | --- **Thus, the Mach number of the airplane is approximately .23.**