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mina [271]
3 years ago
11

A hovering mosquito is hit by a raindrop that is 50 times as massive and falling at 8.1 m/s , a typical raindrop speed. how fast

is the raindrop, with the attached mosquito, falling immediately afterward if the collision is perfectly inelastic?
Physics
1 answer:
Airida [17]3 years ago
6 0

here we will use the momentum conservation

initial total momentum = final total momentum

P_i = P_f

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

now plug in all data here

m(0) + 50m(8.1) = mv + 50mv

405 = 51 v

v = 7.94 m/s

so the final speed will be 7.94 m/s

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A vertical spring has a spring constant of 2900 N/m. The spring is compressed 80 cm and a 8 kg spider is placed on the spring. T
Serga [27]

Answer:

a)  k_{e} = 928 J , b)U = -62.7 J , c) K = 0 , d) Y = 11.0367 m,  e)  v = 15.23 m / s  

Explanation:

To solve this exercise we will use the concepts of mechanical energy.

a) The elastic potential energy is

      k_{e} = ½ k x²

      k_{e} = ½ 2900 0.80²

      k_{e} = 928 J

b) place the origin at the point of the uncompressed spring, the spider's potential energy

     U = m h and

     U = 8 9.8 (-0.80)

     U = -62.7 J

c) Before releasing the spring the spider is still, so its true speed and therefore the kinetic energy also

      K = ½ m v²

      K = 0

d) write the energy at two points, maximum compression and maximum height

     Em₀ = ke = ½ m x²

     E_{mf} = mg y

     Emo = E_{mf}

     ½ k x² = m g y

     y = ½ k x² / m g

     y = ½ 2900 0.8² / (8 9.8)

     y = 11.8367 m

As zero was placed for the spring without stretching the height from that reference is

     Y = y- 0.80

     Y = 11.8367 -0.80

     Y = 11.0367 m

Bonus

Energy for maximum compression and uncompressed spring

     Emo = ½ k x² = 928 J

     E_{mf}= ½ m v²

     Emo = E_{mf}

     Emo = ½ m v²

      v =√ 2Emo / m

     v = √ (2 928/8)

     v = 15.23 m / s

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anyanavicka [17]

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