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anonymous
 one year ago
help me in this problem about sound please
anonymous
 one year ago
help me in this problem about sound please

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Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Ok, lets do a first, did you attempt it?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1\[T_{15} = 288.15 K~~~\text{and}~~~T_0 = 273.15K\] and you need to find \[\frac{ T_{15} }{ T_0 }\] at this point it's plug and chug

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Hey, are you there?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hey! , I'm sorry for the delay.. I was studying :( . I think that nobody was going to answer me u_u . Are you there?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Hey, did you try the problem?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hey, I don't know what equation i have to use . :(

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2I think that we have to use this equation for the sound speed v_S: \[\Large {v_S} = \sqrt {\frac{{\gamma RT}}{M}} \] where R is the gas constant, \gamm is the ratio C_p/C_v between specific heats at constant pressure and volume, and M is the mass of a mole of our gas

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2so we can write: \[\Large \frac{{{v_S}\left( {{T_{15}}} \right)}}{{{v_S}\left( {{T_0}} \right)}} = \sqrt {\frac{{{T_{15}}}}{{{T_0}}}} = \sqrt {\frac{{288.15}}{{273.15}}} = ...\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1We need to use \[v_{sound} = \sqrt{\frac{ \Gamma RT }{ M }}\] where gamma is a constant, R is gas constant, M is molecular mass of gas, and T is absolute temperature is what I can find as far as the second problem goes http://hyperphysics.phyastr.gsu.edu/hbase/sound/souspe3.html for the third problem it seems we will need to know the doppler effect.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Yes exactly @Michele_Laino haha!

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2:) @Astrophysics we gave the answer at the same time

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1I was looking for the relationship as I could not recall it, but I knew it has to do with a gas constant from chemistry and it was the same one you seem to have used!

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.2yes! you are right!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you guys @Michele_Laino @Astrophysics .. i really appreciate your help.
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