Jawaban:

A. 43.98 dB

B. 0.00314 W

Penjelasan:

A.

r1 = 5 m

TI1 = 50 dB

jika 1 mesin pada jarak 100 m

TI2 = TI1 - 20 \: log \: ( \frac{r2}{r1} )TI2=TI1−20log(

r1

r2

)

TI2= 50 - 20 \: log \: ( \frac{100}{5} ) = 23.98TI2=50−20log(

5

100

)=23.98

jika n = 100 mesin pada jarak 100 m

TI2 = TI1 + 10 \: log \: nTI2=TI1+10logn

TI2 = 23.98 + 10 \: log \: 100 = 43.98TI2=23.98+10log100=43.98

B.

TI = 10 \: log \: ( \frac{I}{Io} )TI=10log(

Io

I

)

43.98 = 10 \: log \:( \frac{I}{ {10}^{ - 12} } )43.98=10log(

10

−12

I

)

I = {10}^{4.398} \times {10}^{ - 12} = 2.5 \times {10}^{ - 8}I=10

4.398

×10

−12

=2.5×10

−8

\begin{gathered}I = \frac{P}{A} \\ 2.5 \times {10}^{ - 8} = \frac{P}{4 \: \pi \times {100}^{2} } \\P = 10 \: \pi \times {10}^{ - 4} = 3.14 \times {10}^{ - 3} \end{gathered}

I=

A

P

2.5×10

−8

=

4π×100

2

P

P=10π×10

−4

=3.14×10

−3