1. **State the problem:** We need to find the sound level in decibels at a distance $d_2 = 4000$ m from an explosion, given the maximum acoustic pressure $\Delta P_{max} = 10.0$ Pa at $d_1 = 500$ m, speed of sound $v = 343$ m/s, and sound absorption rate of 7.00 dB/km. 2. **Relevant formula:** Sound level in decibels is given by $$L = 20 \log_{10}\left(\frac{\Delta P}{P_0}\right)$$ where $P_0 = 20 \times 10^{-6}$ Pa is the reference sound pressure. 3. **Calculate initial sound level at $d_1$:** $$L_1 = 20 \log_{10}\left(\frac{10.0}{20 \times 10^{-6}}\right) = 20 \log_{10}(5 \times 10^{5})$$ $$= 20 \times (\log_{10}5 + 5) = 20 \times (0.69897 + 5) = 20 \times 5.69897 = 113.9794 \text{ dB}$$ 4. **Calculate geometric spreading loss:** Sound intensity decreases with distance squared, so pressure decreases inversely with distance. The change in level due to distance from $d_1$ to $d_2$ is $$\Delta L_{distance} = 20 \log_{10}\left(\frac{d_1}{d_2}\right) = 20 \log_{10}\left(\frac{500}{4000}\right)$$ $$= 20 \log_{10}(0.125) = 20 \times (-0.90309) = -18.0618 \text{ dB}$$ 5. **Calculate absorption loss:** Absorption rate is 7.00 dB/km, distance difference is $$d_2 - d_1 = 4000 - 500 = 3500 \text{ m} = 3.5 \text{ km}$$ Absorption loss: $$\Delta L_{absorption} = 7.00 \times 3.5 = 24.5 \text{ dB}$$ 6. **Calculate total sound level at $d_2$:** $$L_2 = L_1 + \Delta L_{distance} - \Delta L_{absorption} = 113.9794 - 18.0618 - 24.5 = 71.4176 \text{ dB}$$ **Final answer:** The sound level at $d_2 = 4000$ m is approximately **71.4 dB**.