Physics, 21.06.2019 22:30 dyamondgeorge1oy4nex

During takeoff, the sound intensity level of a jet engine is 170db at a distance of 34 m. what's the sound intensity level at a distance of 1.0 km?

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Answer from: cocobelle

140.629dB

Explanation:

Sound intensity level $\beta$ is a logarithmic quantity and is defined as:

$\beta=(10dB) log(\frac{I}{I_{o}})$

Where:

is the sound intensity

$I_{o}=10^{-12}W/m^2$ is the reference sound intensity

In this case we are given a sound intensity $\beta_{1}=170dB$ (during takeoff at a distance $r_{1}=34m$ ):

$\beta_{1}=(10dB) log(\frac{I_{1}}{I_{o}})=170dB$ (1)

Now, from (1) we have to find $I_{1}$ .

Dividing both sides of the equation by 10dB :

$\frac{10dB}{10dB} log(\frac{I_{1}}{I_{o}})=\frac{170dB}{10dB}$ (2)

$log(\frac{I_{1}}{I_{o}})=17$ (3)

Applying logarithm properties on both sides:

$\frac{I_{1}}{I_{o}}=10^{17}$ (4)

Isolating $I_{1}$ :

$I_{1}=10^{17}I_{o}$ (5)

$I_{1}=10^{17}(10^{-12}W/m^2)$ (6)

$I_{1}=100000W/m^2$ (7) This is the sound intensity at a distance of 34 m

On the other hand, we know the sound intensity is inversely proportional to the square of the distance :

$I\propto\frac{1}{r^2}$

This means:

$I_{1}r_{1}^2=I_{2}r_{2}^2$ (7)

Where $r_{2}=1km=1000m$

Finding $I_{2}$ from (7):

$I_{2}=\frac{I_{1}r_{1}^2}{r_{2}^2}$ (8)

$I_{2}=\frac{(100000W/m^2)(34m)^2}{(1000m)^2}$ (9)

$I_{2}=115.6W/m^2$ (9) This is the sound intensity at a distance of 1km

Now, knowing the value of $I_{2}$ we will be able to find the sound intensity level $\beta_{2}$ at a distance of 1km:

$\beta_{2}=(10dB) log(\frac{I_{2}}{I_{o}})$ (10)

$\beta_{2}=(10dB) log(\frac{115.6W/m^2}{100000W/m^2})$ (11)

Finally:

$\beta_{2}=140.629dB$ (12) This is the sound intensity level at a distance of 1 km.

Answer from: Quest

Caredbajshjn i belgajsbajbabasbbaba

Answer from: Quest

force is based upon mass and acceleration.the statement is not always true.

it can be explained from newton's second law of motion.

as per newton's second law of motion the rate of change of momentum is directly proportional to the applied force and takes place along the direction of force.

mathematically it can be written as -

$f\alpha \frac{dp}{dt}$

⇒ $f= k \frac{dp}{dt}$

here k is the proportionality constant whose value is 1.

$f=\frac{dp}{dt}$

$f= \frac{d[mv]}{dt}$

$f= m\frac{dv}{dt} +v\frac{dm}{dt}$

$f=ma+v\frac{dm}{dt}$ [a is the acceleration=dv/dt]

hence we see that force not only depends on mass and acceleration but also on velocity and rate of change of mass. force f will depend on mass and acceleration only when mass is constant.

hence the statement that force is based upon mass and acceleration is false.

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