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Art [367]
3 years ago
6

Birds which are famous for performing a ritualized mating dance

Physics
2 answers:
dlinn [17]3 years ago
6 0
<span>Birds which famous for performing a ritualized mating dance are: The Grebes

Grebes most commonly live near a fresh water source.

They conducted this ritual because When two Grebes with opposite sex dances near each other, they will release some sort of pheromones that make them attracted to one another. </span>
AVprozaik [17]3 years ago
6 0

Birds which famous for performing a ritualized mating dance are:

The Grebes

Grebes most commonly live near a fresh water source.

They conducted this ritual because When two Grebes with opposite sex dances near each other, they will release some sort of pheromones that make them attracted to one another. 

Explanation:

Hedge Sparrow Albatross Peacock Blue Manakin Grebe are also included.

Albatrosses are primitive birds that start their ritualized mating dance with shared grooming and then drive into a mating dance where they pat their bills and open their mouths. The gooney birds are typically observed as falling into four genera, yet there is disagreement over the quantity of specie.

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The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and
nikitadnepr [17]

Complete question:

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.

Answer:

The exit velocity is 629.41 m/s

Explanation:

Given;

initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}

k = 1.4

T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}}\\\\T_2 = 1200(\frac{80}{150})^{\frac{1.4-1 }{1.4}}\\\\T_2 = 1002.714K

Work done is given as;

W = \frac{1}{2} *m*(v_i^2 - v_e^2)

inlet velocity is negligible;

v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41  \ m/s

Therefore, the exit velocity is 629.41 m/s

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Answer:

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Explanation:

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