Question

Sound pulses emitted by dolphin travel through
20 °C ocean water at a rate of 1450 m/s. In 20°C
air, these pulses would travel 342.9 m/s. How can
you account for this differences in speed ?

Answer 1

Explanation:

we know that molecules of water are much closer to each other when compared to molecules of air... since they are closer to each other , they can vibrate faster and transmit waves at faster speeds when compared to air in which the molecules are placed far apart from each other

here is numerical proof

for gas

$v_{solid} = \sqrt{\frac{GRT}{M} }$ ( where G = atomicity , R = gas constant, T = temperature and M = mass )

after substituting all constants for air we get :

                 = $20.05 \sqrt{T}$ = 343m/s at 20 degree celcius

for liquid

$v_{liquid}  = \sqrt{\frac{B}{d} }$ (where B is bulk modulus and d = density)

after substituting constants for water at 20°C we get :

$v_{liquid}$ = 1450 m/s

Answer 2

Answer:

we know that molecules of water are much closer to each other when compared to molecules of air... since they are closer to each other , they can vibrate faster and transmit waves at faster speeds when compared to air in which the molecules are placed far apart from each other

here is numerical proof

for gas

v_{solid} = \sqrt{\frac{GRT}{M} }vsolid=MGRT ( where G = atomicity , R = gas constant, T = temperature and M = mass )

after substituting all constants for air we get :

                 = 20.05 \sqrt{T}20.05T = 343m/s at 20 degree celcius

for liquid

v_{liquid}  = \sqrt{\frac{B}{d} }vliquid =dB (where B is bulk modulus and d = density)

after substituting constants for water at 20°C we get :

v_{liquid}vliquid = 1450 m/s

Explanation:

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