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kow [346]
3 years ago
11

A hiker determines the length of a lake by listening for the echo of her shout reflected by a cliff at the far end of the lake.

She hears the echo 3.7s\; s after shouting. The speed of sound in air is 343 m/s\rm m/s. Determine the length of the lake. Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
Nataliya [291]3 years ago
8 0

To solve the exercise it is necessary to take into account the definition of speed as a function of distance and time, and the speed of air in the sound, as well

v=\frac{d}{t}

Where,

V= Velocity

d= distance

t = time

Re-arrange the equation to find the distance we have,

d=vt

Replacing with our values

d= (343)(3.7)

d= 1269.1m

It is understood that the sound comes and goes across the entire lake therefore, the length of the lake is half the distance found, that is

L_{lake} = \frac{d}{2}

L_{lake} = \frac{1269.1}{2}

L_{lake} = 634.55m

Therefore the length of the lake is 634,55m

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Answer:

The sound level of the 26 geese is  Z_{26}= 96.15 dB

Explanation:

From the question we are told that

    The  sound level is Z_1 =  81.0 \ dB

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Generally the intensity level of sound is mathematically represented as

        The intensity of sound level in dB  for one  goose is mathematically represented as

                       Z_1 = 10 log [\frac{I}{I_O} ]

Where I_o is the  threshold level of intensity with value  I_o = 1*10^{-12} \  W/m^2

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For 26 geese the intensity would be  

          I_{26} = 26 * I

   Then  the intensity of 26 geese in dB is  

              Z_{26} = 10 log[\frac{26 I }{I_o} ]

               Z_{26} = 10 log (\ \ 26 *  [\frac{ I }{I_o} ]\ \ )

               Z_{26} = 10 log (\ \ 26  \ \ ) *   (\ \  10 log [\frac{ I }{I_o} ]\ \ )

 From the law of logarithm we have that

              Z_{26} = 10 log 26 +  10 log [\frac{I}{I_0} ]

                    = 14.15 + 82

                    Z_{26}= 96.15 dB

               

               

           

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