A moth of length 1.0 cm is flying around 1.0 m from a bat as soon as the bat emits a sound tide at 80.0 kHz . The temperature of waiting is around 10.0 C. To feeling the visibility of the moth utilizing echolocation, the bat must emit a sound through a wavelength same to or less than the length of the insect.

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v=sqrt (ratio of warmth capacities*RT/M)

where ratio of warm capacities ( 1.4 because that air), T is the pure temperature in kelvins (which is same to the Celsius temperature plus 273.15 C), is the molar mass of the gas (for air, the typical molar mass is M=28.810^-3 kg/mol), and also R is the universal gas continuous (R=8.314Jmol^-1*K^-1). The wavelength of the 80.0-kHz wave emitted by the bat is 4.23 mm.

How lengthy after the bat emits the wave will it hear the echo from the moth?

3 Answers

Frankly, ns don"t recognize of propagation rate calculation based upon temperature and R and also all the other offered data. I think that the a concern of replacing values, yet if the frequency and wavelength are recognized (and there"s no error in this information) why not apply directly:

v = freq x wavelenght = 80000 x 0.00423 m/s = 338.4 m/s

(in truth it is usually considered that the speed of sound at sea level is 336 to 340 m/s approximately)

Then d = v t => t = d / v = 2 * 1m / <338.4 (m/s)>

t = 0.0059 s = 5.9 ms


Ps. While ns was typing Paul C posted his reply. I read about the hold-up due come the wave momentum and I don"t think it might be of much influence, considering the wave duration is

T = 1/f = 12.5 µs, giving less than 1% error.

actually you"ve done every the tough work currently in finding the end the wavelength, either the or the question offered you too lot info. So we have the right to ignore many of the nonense and also hope that the guy who wrote the inquiry knew enough around what he to be on around and say:

velocity=wavelength x frequency


transit time = 2metres/velocity


Now from a practical suggest of view this won"t in reality be the case; it"ll take a tiny longer. Factor being is the the wave doesn"t bounce off the moth choose it would certainly off an immovable wall. The wave has a specific amount that momentum and the itty bitty moth i m sorry doesn"t have actually much mass gets several of this moved to it. This collision in between the moth and also the sound wave will offer a slight delay time; true it"s a very tiny amount of time but going on that our answer is exact to a portion of a millisecond it will probably transform the result noticibly.

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Sorry I lost a aspect of ten top top the frequency, no wonder the velocity appeared so slow. So splitting the time by ten offers 5.91ms. The guy below got it right however left turn off a decimal location for part reason. When providing answers come a concern thats data is accurate to 3sig figs the answer should additionally be precise to 3 sig figs. Think girlfriend should offer him the point out anyway. Together you can see the 1/f phase delay is that the very same order together the accuracy that the answer therefore it will certainly be important. As sound is compression waves the hold-up is mass dependent so i can"t placed a worth to it v the amount of information given.