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The speed of sound in air is 340 m/s.A loudspeaker placed between two walls X and Y, but nearer to wall x than wall y,...

The speed of sound in air is 340 m/s.A loudspeaker placed between two walls X and Y, but nearer to wall x than wall y, is sending out constant sound pulses.How far is the speaker from wall Y if it is 200 m from wall X and the time between the two echoes received is 0.176 s.

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Jim
t(2)-t(1)=0.176
t(2)=2x/340
t(1)=400/340
2x/340-400/340=0.176
2x-400=0.176(340)
2x=59.84+400

X= 229.9 m

jim items answered the question on September 17, 2018 at 09:07

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