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klasskru [66]
3 years ago
14

A 300 gg bird flying along at 5.7 m/sm/s sees a 10 gg insect heading straight toward it with a speed of 29 m/sm/s (as measured b

y an observer on the ground, not by the bird). The bird opens its mouth wide and enjoys a nice lunch.What is the bird's speed immediately after swallowing? _____m/s
Physics
1 answer:
liberstina [14]3 years ago
8 0

Answer:

bird's speed  = 4.58 m/s

Explanation:

given data

mass of bird mb = 300 g

velocity of bird vb = 5.7 m/s

mass of insect mi = 10 g

velocity of insect vi = - 29 m/s

solution

here momentum of system is constant

we apply here conservation of momentum so

mb × v + mi × v = mb × vb + mi × vi   ....................1

v ( mb + mi ) = mb × vb + mi × vi  

put here value and we get

v ( 300 + 10 ) = { 300 × 5.7 }  +  { 10 × (-29) }

v = \frac{1710-290}{310}  

v = 4.58 m/s

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Answer:

The answer to your question is    vo = 5.43 m/s

Explanation:

Data

distance = d= 5.8 m

height = 3 m

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angle = 60°

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g = 9.81 m/s²

Formula

              hmax = vo²sinФ/ 2g

Solve for vo²

              vo² = 2ghmax / sinФ

Substitution

              vo² = 2(9.81)(3 - 1.7) / 0.866

Simplification

              vo² = 19.62(1.3) / 0.866

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              vo² = 29.45

Result

              vo = 5.43 m/s

               

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