A tunnel under a river is 2.00 km long. (a) At what frequencies can the air in the tunnel resonate? (b) Explain whether it would
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Answer:
447 K
Explanation:
25 C = 25 + 273 = 298 K
Assuming ideal gas, we can apply the ideal gas law
Since pressure is tripled, then . Volume is halved, then
First we need to find the speed of the dolphin sound wave in the water. We can use the following relationship between frequency and wavelength of a wave:
where
v is the wave speed
its wavelength
f its frequency
Using
and 
, we get
We know that the dolphin sound wave takes t=0.42 s to travel to the tuna and back to the dolphin. If we call L the distance between the tuna and the dolphin, the sound wave covers a distance of S=2 L in a time t=0.42 s, so we can write the basic relationship between space, time and velocity for a uniform motion as:
and since we know both v and t, we can find the distance L between the dolphin and the tuna:
where
v is the wave speed
f its frequency
Using
We know that the dolphin sound wave takes t=0.42 s to travel to the tuna and back to the dolphin. If we call L the distance between the tuna and the dolphin, the sound wave covers a distance of S=2 L in a time t=0.42 s, so we can write the basic relationship between space, time and velocity for a uniform motion as:
and since we know both v and t, we can find the distance L between the dolphin and the tuna: