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kodGreya [7K]
2 years ago
12

Change in inherited characteristics over time is called what?

Physics
1 answer:
adell [148]2 years ago
6 0

Answer:

characteristics

Explanation:

Change in inherited characteristics over time is called what?Change in inherited characteristics over time is called what?Change iChange in inherited characteristics over time is called what?n inherited characteristics over time is called what?Change in inherited characteristics over time is called what?

Change in inherited characteristics over time is called what?Change in inherited characteristics over time is called what?Change in inherited characteristics over time is called what?

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olga55 [171]

Answer:

Peer Review

Explanation:

When scientist check other scientists the process is dubbed "peer review".

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3 years ago
Which statement accurately describes a sample of water during parts a and c of the heating curve
vivado [14]

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A and C is about 12 cm away from each other.

Explanation:

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With the simplified model of the eye, what corrective lens (specified by focal length as measured in air) would be needed to ena
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Answer:

Please see the attached picture for the complete answer.

Explanation:

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3 years ago
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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