A jet engine emits sound uniformly in all directions, radiating an acoustic power of 2.35 × 10 5 2.35×105 W. Find the intensity of the sound at a distance of 51.3 51.3 m from the engine and calculate the corresponding sound intensity level.

Intensity of sound = 7.11W/m²

Corresponding sound intensity level = 128.5dB

Explanation:

(a) Intensity (I) of a sound is related to the power (P) of the sound and coverage area (A) of the sound as follows;

I  =            -----------------(i)

From the question;

P = acoustic power = 2.35 x 10⁵W

distance = radius of coverage = r = 51.3m

Let's first calculate the area of coverage (A) of the sound as follows;

A = 4 x π x r²                  [substitute the value of r = 51.3 and take π = 3.142]

A = 4 x 3.142 x 51.3²

A = 33075.08m²

Now, substitute the values of P and A into equation (i) as follows;

I =

I = 7.11W/m²

Therefore, the intensity of the sound at that distance is 7.11W/m².

(b) To calculate the sound intensity level, β, (basically in decibels), the following relation is used;

β = log₁₀[]               ----------------------(ii)

Where;

I₀ = reference sound intensity level (threshold of hearing) = 10⁻¹²W/m²

I = Intensity of sound = 7.11W/m² [as calculated above]

Substitute these values into equation (ii) as follows;

β = 10 x log₁₀[]

β = 10 x log₁₀[7.11 x 10¹²]

β = 10 x 12.85dB

β = 128.5dB

Therefore, the corresponding sound intensity level is 128.5dB

1= 26.17

2= 1.17

3=30.86

4=5.86

just took the assigment on edge

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west