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mayankdevnani

  • 2 years ago

An echo repeats two syllables.If the velocity of sound is 330 m/s , then the distance of the reflecting surface is- a) 66.0 m b) 33.0 m c) 99.0 m d) 16.5 m

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  1. mayankdevnani
    • 2 years ago
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    @amistre64 @TuringTest @hartnn @ganeshie8 @ghazi @theyatin

  2. ghazi
    • 2 years ago
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    i am not sure about this , but possibly its 66 m

  3. mayankdevnani
    • 2 years ago
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    yaa its right!!!

  4. mayankdevnani
    • 2 years ago
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    how do you do this? @ghazi

  5. theyatin
    • 2 years ago
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    @mayankdevnani

  6. mayankdevnani
    • 2 years ago
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    no answer is wrong it gets 16.5m @theyatin

  7. mayankdevnani
    • 2 years ago
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    @ganeshie8 @geerky42 @Aperogalics

  8. theyatin
    • 2 years ago
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    oops then i must sleep. . .

  9. mayankdevnani
    • 2 years ago
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    oh!!!! koi baat nahi

  10. ghazi
    • 2 years ago
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    this will help you http://in.answers.yahoo.com/question/index?qid=20120713012737AAWgHSc

  11. theyatin
    • 2 years ago
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    can you explain the concept of syllable??

  12. mayankdevnani
    • 2 years ago
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    i don't know what is syllable??

  13. mayankdevnani
    • 2 years ago
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    @hartnn @ash2326 @jazy @agreene

  14. mayankdevnani
    • 2 years ago
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    @TuringTest

  15. theyatin
    • 2 years ago
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    well its very tough i see it as if minimum frequency gets us time t=0.05s minimum and as we have 2 syllables 0.05*2=0.1 ( as i have studied in link given by gazi) so the distance must be velocity* time=33meter but to strike wall and come back it becomes 66 as total. . . but here question is asking for the distance of surface so i am not quite sure. . .

  16. mayankdevnani
    • 2 years ago
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    @amistre64 @AccessDenied @ash2326 @dumbcow @hartnn @UnkleRhaukus @Callisto

  17. mayankdevnani
    • 2 years ago
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    @Shane_B @surdawi @AccessDenied @ajprincess @Yahoo!

  18. mayankdevnani
    • 2 years ago
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    @henpen

  19. amistre64
    • 2 years ago
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    im assuming the question is saying that you hear your echo 2 times

  20. henpen
    • 2 years ago
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    You need the time between the 2 sounds

  21. amistre64
    • 2 years ago
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    |dw:1352738321407:dw|

  22. theyatin
    • 2 years ago
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    yey!!! that justifies my answer 16.5m which when multiplied by 4 gets an answer of 66m again explaining it. . . we can hear minimum 20hz that leads to T=0.05sec and thus the distance leads to velocity * time = 330 * 0.05=16.5 thus when multilied by 4 gets 66m. . .

  23. mayankdevnani
    • 2 years ago
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    why we multiplied it by 4 @theyatin @amistre64 @henpen

  24. amistre64
    • 2 years ago
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    the distance traveled between the "wall" and the source is quadrupled in order to hear the echo twice, as i pictured it up. assuming that im reading the question correctly to begin with

  25. mayankdevnani
    • 2 years ago
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    oh!!! i see

  26. amistre64
    • 2 years ago
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    imagine you and a friend throwing a ball back and forth; how far does the ball travel relative to your positions if you catch it 2 times and throw it 2 times?

  27. mayankdevnani
    • 2 years ago
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    i think 4 times. right

  28. amistre64
    • 2 years ago
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    correct

  29. mayankdevnani
    • 2 years ago
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    thnx... @amistre64

  30. amistre64
    • 2 years ago
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    hope its the right concept ;) good luck

  31. mayankdevnani
    • 2 years ago
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    what is syllables???

  32. amistre64
    • 2 years ago
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    im thinking that it pertains to the rebound that you hear. and not really referencing words perse

  33. mayankdevnani
    • 2 years ago
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    oh!!!

  34. amistre64
    • 2 years ago
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    if you stand in an echoy place and give a brisk woot!! you can hear how many times it echos back at you

  35. amistre64
    • 2 years ago
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    one of them old dead scientist guys used a long hallway that he knew the length of in order to estimate the speed of sound that way

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