- mayankdevnani

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

- jamiebookeater

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- mayankdevnani

- ghazi

i am not sure about this , but possibly its 66 m

- mayankdevnani

yaa its right!!!

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- mayankdevnani

how do you do this? @ghazi

- anonymous

- mayankdevnani

no answer is wrong it gets 16.5m @theyatin

- mayankdevnani

- anonymous

oops then i must sleep. . .

- mayankdevnani

oh!!!! koi baat nahi

- ghazi

this will help you http://in.answers.yahoo.com/question/index?qid=20120713012737AAWgHSc

- anonymous

can you explain the concept of syllable??

- mayankdevnani

i don't know what is syllable??

- mayankdevnani

- anonymous

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. . .

- mayankdevnani

- mayankdevnani

- mayankdevnani

- amistre64

im assuming the question is saying that you hear your echo 2 times

- anonymous

You need the time between the 2 sounds

- amistre64

|dw:1352738321407:dw|

- anonymous

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. . .

- mayankdevnani

- amistre64

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

- mayankdevnani

oh!!! i see

- amistre64

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?

- mayankdevnani

i think 4 times. right

- amistre64

correct

- mayankdevnani

thnx... @amistre64

- amistre64

hope its the right concept ;) good luck

- mayankdevnani

what is syllables???

- amistre64

im thinking that it pertains to the rebound that you hear. and not really referencing words perse

- mayankdevnani

oh!!!

- amistre64

if you stand in an echoy place and give a brisk woot!! you can hear how many times it echos back at you

- amistre64

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.