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Otrada [13]
3 years ago
7

A person standing a certain distance from an airplane with four equally noisy jet engines is experiencing a sound level of 145 d

B . What sound level would this person experience if the captain shut down all but one engine?
Physics
1 answer:
Katyanochek1 [597]3 years ago
5 0

Answer:

The sound level is 138.97 dB.

Explanation:

Given that,

Sound level L= 145 dB

We need to calculate the intensity

Using formula of sound intensity

L=10\log(\dfrac{I}{I_{0}})

Put the value into the formula

145=10\log(\dfrac{I}{I_{0}})

\log(\dfrac{I}{I_{0}})=\dfrac{145}{10}

\log(\dfrac{I}{I_{0}})=14.5

\dfrac{I}{I_{0}}=10^{14.5}

I=10^{14.5}\times I_{0}

I=10^{14.5}\times10^{-12}

I=316.2

We need to calculate the noise of one engine

Using formula of intensity

I'=\dfrac{I}{4}

Put the value into the formula

I'=\dfrac{316.2}{4}

I'=79.05

We need to calculate the intensity level due to one engine

Using formula of sound intensity

L'=10\log(\dfrac{I'}{I_{0}})

L'=10\log(\dfrac{79.05}{10^{-12}})

L'=138.97\ dB

Hence, The sound level is 138.97 dB

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local AM radio station broadcasts at a frequency of 685.9 kHz. Calculate the wavelength at which they are broadcasting. Waveleng
natta225 [31]

Answer:

λ = 437 m.

Explanation:

  • Since a radio wave is a electromagnetic type of wave, it propagates approximately at the same speed of light in vacuum, 3*10⁸ m/s.
  • As in any wave, there exist a fixed relationship between the propagation speed, the wavelength and the frequency of the radio wave, as follows:

       c = \lambda* f (1)

  • Replacing by the values of the speed and the frequency, and solving for the wavelength λ, we get:

       \lambda = \frac{c}{f} =\frac{3e8 m/s}{685.9e31/s} =437 m  (2)

8 0
2 years ago
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oksian1 [2.3K]

Answer:

Explanation:

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6 0
3 years ago
You need to focus a 10 mW, 632.8 nm Gaussian laser beam that is 5.0 mm in diameter into a sample. You have access to a lens with
Anna [14]

Answer:

ee that the lens with the shortest focal length has a smaller object

           

Explanation:

For this exercise we use the constructor equation or Gaussian equation

        \frac{1}{f}  = \frac{1}{p} + \frac{1}{q}

where f is the focal length, p and q are the distance to the object and the image respectively.

Magnification a lens system is

          m = \frac{h'}{h} = - \frac{q}{p}

             h ’= -\frac{h q}{p}

In the exercise give the value of the height of the object h = 0.50cm and the position of the object p =∞

Let's calculate the distance to the image for each lens

f = 6.0 cm

           \frac{1}{q} = \frac{1}{f }  - \frac{1}{p}

as they indicate that the light fills the entire lens, this indicates that the object is at infinity, remember that the light of the laser rays is almost parallel, therefore p = inf

          q = f = 6.0 cm

for the lens of f = 12.0 cm q = 12.0 cn

to find the size of the image we use

           h ’= h q / p

where p has a high value and is the same for all systems

           h ’= h / p q

Thus

f = 6 cm h ’= fo 6 cm

 

f = 12 cm h ’= fo 12  cm

therefore we see that the lens with the shortest focal length has a smaller object

8 0
3 years ago
Quartz gold and calcite are examples of?coal is what
givi [52]
Quartz, gold and calcite are examples of minerals, coal is a fossil fuel
5 0
3 years ago
A biker can ride at 12 m/s on a level road when there is no wind and at 7.5 m/s on a level road when there is a head wind of 5 m
alexandr402 [8]

Answer:

The answer is "1.94 \ m^2".

Explanation:

Formula:

F= \frac{1}{2} \times C_D \times density \times area \times  velocity^2\\\\Power (P) = F \times velocity \\\\P = \frac{C_D}{2} \times density \times area \times  velocity^3\\

Given value:

\ P = 10.25 \ W \\\\\ density = 1.2 \ kg/m^3 \\\\\ velocity= 2.5 \\\\

10.25 = \frac{C_D A}{2} \times 1.2 \times 2.5^3\\\\C_D A= \frac{10.25 \times 2 }{1.2 \times 2.5^3}\\\\C_D A = 1.94 m^2\\

8 0
3 years ago
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