Answer:
Part A
The sound intensity at the location 150 from the firework = 4.3[tex]\overline 5[/tex] W/m²
Part B
The power the firework is emitting is approximately 1,231.5 kW
Explanation:
Part A
The data of the firework sound heard by the two friends are;
Let 'A' represent the location of the friend at a point 150 m from the fireworks and let 'B' represent the location of the friend watching from a distance of 700 m from the firework
The distance of 'A' from the sound, r₁ = 150 m
The distance of 'B' from the sound, r₂ = 700 m
The intensity at which 'B' hears the sound, I₂ = 0.2 W/m²
The relationship between sound intensity and distance is given as follows;
[tex]\dfrac{I_2}{I_1} = \left (\dfrac{r_1}{r_2} \right )^2[/tex]
[tex]\therefore {I_1} = \dfrac{I_2}{ \left (\dfrac{r_1}{r_2} \right )^2} = I_2 \times \left (\dfrac{r_2}{r_1} \right )^2[/tex]
Plugging in the values gives;
[tex]\therefore {I_1} = 0.2 \times \left (\dfrac{700}{150} \right )^2 = \dfrac{196}{45} = 4.3\overline 5[/tex]
The sound intensity at location 'A', I₁ = 4.3[tex]\overline 5[/tex] W/m²
Part B
The relationship between power, 'P', and intensity, 'I', is presented as follows;
[tex]I = \dfrac{P}{4 \cdot \pi \cdot r^2}[/tex]
P = I·4·π·r² = 4·I·π·r²
Therefore, at point 'A', where the distance, r₁ = 150 m, and the intensity, I₁ = 4.3[tex]\overline 5[/tex] W/m², we have;
P = 4 × 4.3[tex]\overline 5[/tex] W/m² × π × (150 m)² = 1231504.32021 W
The power the firework is emitting, P ≈ 1,231.5 kW.