To solve this problem, we need to use the Doppler effect for sound waves.
1. **Given:**
- Frequency of SONAR \( f_0 = 40.0 \, \text{kHz} \)
- Speed of sound in water \( v = 1450 \, \text{m/s} \)
- Speed of the enemy submarine \( v_s = 360 \, \text{km/h} = \frac{360 \times 1000}{3600} \, \text{m/s} = 100 \, \text{m/s} \)
2. **Step 1: Find the frequency detected by the enemy submarine:**
Since the enemy submarine is moving towards the source, the detected frequency \( f_1 \) is given by:
\[
f_1 = f_0 \left(\frac{v + v_s}{v}\right)
\]
Substituting the values:
\[
f_1 = 40 \times 10^3 \left(\frac{1450 + 100}{1450}\right) = 40 \times 10^3 \left(\frac{1550}{1450}\right)
\]
\[
f_1 \approx 42.76 \, \text{kHz}
\]
3. **Step 2: Find the frequency of the reflected wave as observed by the SONAR:**
The SONAR now receives this reflected sound, which acts as if it's coming from a moving source (the enemy submarine). So, the frequency observed by the SONAR \( f_r \) is:
\[
f_r = f_1 \left(\frac{v + v_s}{v}\right)
\]
\[
f_r = 42.76 \times 10^3 \left(\frac{1550}{1450}\right) \approx 45.75 \, \text{kHz}
\]
4. **Final Answer:**
The closest option to the calculated value is \( 46 \, \text{kHz} \).
**So, the correct answer is A. 46kHz.**
