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andreyandreev [35.5K]
3 years ago
15

The sound source of a ship’s sonar system operates at a frequency of 22.0 kHz . The speed of sound in water (assumed to be at a

uniform 20∘C) is 1482 m/s . What is the difference in frequency between the directly radiated waves and the waves reflected from a whale traveling straight toward the ship at 4.95 m/s ? Assume that the ship is at rest in the water.
Physics
1 answer:
Degger [83]3 years ago
6 0

Answer:

147.456077993 Hz

Explanation:

f_0 = Frequency of the sonar = 22 kHz

v_w = Velocity of the whale = 4.95 m/s

v = Speed of sound in water = 1482 m/s

The difference in frequency is given by

\Delta f=f_0\times\dfrac{2v_{w}}{v-v_w}\\\Rightarrow \Delta f=22000\times\dfrac{2\times 4.95}{1482-4.95}\\\Rightarrow \Delta f=147.456077993\ Hz

The difference in frequency is 147.456077993 Hz

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A 1 036-kg satellite orbits the Earth at a constant altitude of 98-km. (a) How much energy must be added to the system to move t
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Answer:

a) The Energy added should be 484.438 MJ

b) The  Kinetic Energy change is -484.438 MJ

c) The Potential Energy change is 968.907 MJ

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Let 'm' be the mass of the satellite , 'M'(6×10^{24} be the mass of earth , 'R'(6400 Km) be the radius of the earth , 'h' be the altitude of the satellite and 'G' (6.67×10^{-11} N/m) be the universal constant of gravitation.

We know that the orbital velocity(v) for a satellite -

v=\sqrt{\frac{Gmm}{R+h} }         [(R+h) is the distance of the satellite   from the center of the earth ]

Total Energy(E) = Kinetic Energy(KE) + Potential Energy(PE)

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h = h_{i} = 98 km = 98000 m

∴Initial Energy (E_{i})  = \frac{1}{2}mv^{2} + \frac{-GMm}{(R+h_{i} )}

Substituting v=\sqrt{\frac{GMm}{R+h_{i} } } in the above equation and simplifying we get,

E_{i} = \frac{-GMm}{2(R+h_{i}) }

Similarly for final condition,

h=h_{f} = 198km = 198000 m

∴Final Energy(E_{f}) = \frac{-GMm}{2(R+h_{f}) }

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∴ ΔE = E_{f} - E_{i}

        = \frac{GMm}{2}(\frac{1}{R+h_{i} } - \frac{1}{R+h_{f} })

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ΔE = 484.438 MJ

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