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Answer:
<em>The depth will be equal to</em> <em>6141.96 m</em>
<em></em>
Explanation:
pressure on the submarine = 62 MPa = 62 x 10^6 Pa
we also know that = ρgh
where
ρ is the density of sea water = 1029 kg/m^3
g is acceleration due to gravity = 9.81 m/s^2
h is the depth below the water that this pressure acts
substituting values, we have
= 1029 x 9.81 x h = 10094.49h
The gauge pressure within the submarine = 101 kPa = 101000 Pa
this gauge pressure is balanced by the atmospheric pressure (proportional to 101325 Pa) that acts on the surface of the sea, so it cancels out.
Equating the pressure , we have
62 x 10^6 = 10094.49h
depth h = <em>6141.96 m</em>
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: