A physicist at a fireworks display times the lag between seeing an explosion and hearing its sound, and finds it to be 0.600 s. (Enter your answers to at least four decimal places. Use the following relationship between Kelvin and Celsius: T(K)= T(°C) + 273.15.) (a) How far away (in m) is the explosion if air temperature is 24.0°C and if you neglect the time taken for light to reach the physicist? 207.19 x m (b) Calculate the distance to the explosion (in m) taking the speed of light into account. Note that this distance is negligibly greater. 207.198 x m

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Chapter16: Waves
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A physicist at a fireworks display times the lag between seeing an explosion and hearing its sound, and finds it to be 0.600 s. (Enter your answers to at least four decimal places. Use the following relationship between Kelvin and Celsius:
T(K) = T(°C) + 273.15.)
(a) How far away (in m) is the explosion if air temperature is 24.0°C and if you neglect the time taken for light to reach the physicist?
207.19
Xm
(b) Calculate the distance to the explosion (in m) taking the speed of light into account. Note that this distance is negligibly greater.
207.198
Xm
Transcribed Image Text:A physicist at a fireworks display times the lag between seeing an explosion and hearing its sound, and finds it to be 0.600 s. (Enter your answers to at least four decimal places. Use the following relationship between Kelvin and Celsius: T(K) = T(°C) + 273.15.) (a) How far away (in m) is the explosion if air temperature is 24.0°C and if you neglect the time taken for light to reach the physicist? 207.19 Xm (b) Calculate the distance to the explosion (in m) taking the speed of light into account. Note that this distance is negligibly greater. 207.198 Xm
Expert Solution
Step 1

(a)

The speed of sound in air is expressed as a function of the temperature of air is 

         vsound=331.31+T(°C)273.15

Where, 

T is the temperature of the air .

Given,

T=24°C

So,

Speed of sound vsound=331.31+24273.15

                                        =345.5482523 m/s

Given,

The time interval between seeing the explosion and hearing the its sound is t=0.600 s

So, 

  The distance to the explosion(ignoring time time taken for to reach the physicist) is

          d=vsound×t

             =345.5482523×0.600

             =207.3289 m

 

 

                                                   

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