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Please solve this problem

$Howdoastronautscommunicateonthesurfaceofmoon\phantom{\rule{0ex}{0ex}}intheabsenseofanyatmosphere?\phantom{\rule{0ex}{0ex}}Yousetyourwatchbythesoundofadis\mathrm{tan}tsiren.Will\phantom{\rule{0ex}{0ex}}itgofastorslow?Explain.\phantom{\rule{0ex}{0ex}}Apersons\mathrm{tan}dingneararailwaylinehearstwodistinct\phantom{\rule{0ex}{0ex}}soundsofanapproachingtrain.Thetimeintervalbetween\phantom{\rule{0ex}{0ex}}thesesoundsis2s.Howmuchisthedis\mathrm{tan}ceofthetrain?\phantom{\rule{0ex}{0ex}}Velocityofsoundinair=340rn{s}^{-l}.Velocityofsoundin\phantom{\rule{0ex}{0ex}}Steel=16timesvelocityofsoundinair.$

If the person standing near a railway line hears two distinct sounds of a train, then the sounds are - one travelling through steel and the another travelling through air.

Let the distance of train be d from the person.

speed of sound in air = 340 m/s

speed of sound in steel = $340\times 16=5440m/s$

time taken by sound travelling through steel to reach the person =

time taken by sound travelling through air to reach the person =

$time=\frac{dis\mathrm{tan}ce}{speed}\phantom{\rule{0ex}{0ex}}{t}_{s}=\frac{d}{5440}\phantom{\rule{0ex}{0ex}}{t}_{a}=\frac{d}{340}$

given that

$\frac{d}{340}-\frac{d}{5440}=2\phantom{\rule{0ex}{0ex}}ord\left(\frac{16-1}{5440}\right)=2\phantom{\rule{0ex}{0ex}}d=\frac{2\times 5440}{15}\phantom{\rule{0ex}{0ex}}d=725m$

Distance of train is 725m

Kindly post other questions in separate thread.

Regards

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