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joja [24]
3 years ago
6

Answer this please nowww

Physics
1 answer:
Viefleur [7K]3 years ago
4 0
(3) The frictional force exerted by the floor on the box
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single goose sounds a loud warning when an intruder enters the farmyard. Some distance from the goose, you measure the sound lev
Goryan [66]

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

     The number of geese is N = 26

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

            I is the intensity for one goose in W/m^2

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

               

               

           

4 0
3 years ago
who brought astronomy out of the dark ages by declaring: "finally we shall place the sun himself at the center of the universe.
solong [7]

Nicolaus Copernicus was the one who brought astronomy out of the dark ages.

In one of his books "On the Revolutions of the Heavenly Bodies", he published the renowned line which declares his theory.

Nicolaus Copernicus, A polish astronomer put forward the theory that the is th Sun one that rests in the middle or center of the Universe and the planet Earth revolves around it on its axis every day.which was considered to be called the Heliocentric system.

To know more about the Heliocentric system refer to the link brainly.com/question/3491738?referrer=searchResults.

To know about Sir Nicolaus Copernicus' work refer to the link brainly.com/question/6699117?referrer=searchResults.

#SPJ4

6 0
2 years ago
Consider a horizontal layer of the dam wall of thickness dx located a distance x above the reservoir floor. What is the magnitud
adoni [48]

Answer:

Explanation:

Attached is the solution

6 0
3 years ago
When the starter motor on a car is engaged, there is a 320 A current in the wires between the battery and the motor. Suppose the
Salsk061 [2.6K]

Answer:

0.5

Explanation:

because it is V uwbsusvegwjosnfvehdbuxbdusndgdghqbwbwbbeuehdbdhdhdhdudhdhdudhdnskowoqllqlqlqnebe

8 0
3 years ago
A sound source is moving at 80 m/s toward a stationary listener that is standing in still air (a) Find the wavelength of the sou
Setler [38]

Answer:

a. wavelength of the sound, \vartheta = 1.315\vartheta_{o}

b. observed frequecy, \lambda = 0.7604\lambda_{o}

Given:

speed of sound source, v_{s} = 80 m/s

speed of sound in air or vacuum, v_{a} = 343 m/s

speed of sound observed, v_{o} = 0 m/s

Solution:

From the relation:

v = \vartheta \lambda        (1)

where

v = velocity of sound

\vartheta = observed frequency of sound

\lambda = wavelength

(a) The wavelength of the sound between source and the listener is given by:

\lambda = \frac{v_{a}}{\vartheta }         (2)

(b) The observed frequency is given by:

\vartheta = \frac{v_{a}}{v_{a} - v_{s}}\vartheta_{o}

\vartheta = \frac{334}{334 - 80}\vartheta_{o}

\vartheta = 1.315\vartheta_{o}                (3)

Using eqn (2) and (3):

\lambda = \frac{334}{1.315} = \frac{1}{1.315}\frac{v_{a}}{\vartheta_{o}}

\lambda = 0.7604\lambda_{o}

4 0
3 years ago
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