Need the answer of Q5 and Q6...

Dear student
Kindly find the answer of your asked query.

time taken by the sound wave to travel the distance of 20 m is t.t=distancesound velocity=20340sfrequency of sound is 280 Hz that means it performs 280 oscillations in 1 seconds.hence oscillations in 20340 seconds are:= 20340×280=23.33 seconds
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Is this ur handwriting Ishika? Its really really really awesome... Well the questions are also nice:-)
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APOLIGIZING NOT CONFIRMED WITH Q.5
QUESTION 6
a.THE MINIMUM TIME LIMIT TO HEAR ANOTHER DISTINCT SOUND IS 0.1SEC
THEREFORE THE MINIMUM DISTANCE = SPEEDOF SOUND * TIME
= 340*0.1=34
THE DISTANCE BETWEEN THE REFLECTOR IS 34/2=17m
​b. In a hotter day speed of the sound will increase because when the temperature is high the vibration of the particles increases . Therefore it will be easier for sound to travel in a hotter day
Q.5 answer may be 16.5 vibrations{not sure}
REGARDS
NAUFAL
 
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