Question

Question is in the picture

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

(a)  D = 72.05 dB  (2 d.p.)

(b)  I = 2.51 W/m²  (2 d.p.)

Step-by-step explanation:

Given function:

$\boxed{D(I)=10 \log_{10}(I)+120, \quad I\geq 0}$

where:

• D is the sound intensity level (in dB).
• I is the sound intensity (in W/m²).

Part (a)

Given:

• $I=1.6 \times 10^{-5}\; \sf W/m^2$

Substitute the given value of I into the given function and solve for D:

$\begin{aligned}\implies D(1.6 \times 10^{-5}) & =10 \log_{10}(1.6 \times 10^{-5})+120\\& = 10 \log_{10}(0.000016)+120\\& = 10 (-4.795880017)+120\\& = -47.95880017+120\\&=72.04119983\\&=72.04\; \sf dB\;(2\:d.p.)\end{aligned}$

Part (b)

Given:

• D = 124 dB

Substitute the given value of D into the given function and solve for I:

$\begin{aligned}\implies 124 & =10 \log_{10}(I)+120\\4 & =10 \log_{10}(I)\\10 \log_{10}(I)&=4\\\log_{10}(I)&=0.4\\10^{\log_{10}(I)}&=10^{0.4\\I&=2.511886432\\I&=2.51\; \sf W/m^2\; (2\:d.p.)\end{aligned}$