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Alekssandra [29.7K]
4 years ago
6

How do rocks within Earth change as P waves pass? How do rocks within Earth change as P waves pass? Rocks within Earth both expa

nd and contract as P waves pass. Rocks within Earth neither expand nor contract as P waves pass. Rocks within Earth expand as P waves pass. Rocks within Earth contract as P waves pass.
Physics
1 answer:
ohaa [14]4 years ago
6 0

Rocks within Earth both expand and contract as P waves pass

Explanation:

Rocks within the earth both expands and contracts as P-waves passes through them. P-waves are elastic waves.

  • Elastic waves behaves in such a way that they do not cause permanent deformation of rocks.
  • They can be said to cause elastic deformation when they travel through rocks.
  • They simply temporarily expand and contract the rock within a short period by causing the vibration of particles of the medium.
  • After a short while, the rock returns back to its original position as if nothing has happened to it.
  • These elastic waves are better called seismic waves.
  • P-waves are primary waves that can travel through any medium.

Learn more:

Earthquakes brainly.com/question/11292835

#learnwithBrainly

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Answer:

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b) 24.3 m/s

Explanation:

a)

  • In order to get the total distance traveled since you see the deer till the car comes to an stop, we need to take into account that this distance will be composed of two parts.
  • The first one, is the distance traveled at a constant speed, before stepping on the brakes, which lasted the same time that the reaction time, i.e., 0.5 sec.
  • We can find this distance simply applying the definition of average velocity, as follows:

       \Delta x_{1} = v_{1o} * t_{react} = 20 m/s * 0.5 s = 10 m (1)

  • The second part, is the distance traveled while decelerating at -11 m/s2, from 20 m/s to 0.
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  • where v₁f = 0, v₁₀ = 20 m/s, a = -11 m/s².
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       \Delta x_{2} = \frac{-(20m/s)^{2}}{2*(-11m/s)} = 18.2 m (3)

  • So, the total distance traveled was the sum of (1) and (3):
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  • Since the initial distance between the car and the deer was 39 m, after travelling 28.2 m, the car was at 10.8 m from the deer when it came to a complete stop.

b)

  • We need to find the maximum speed, taking into account, that in the same way that in a) we will have some distance traveled at a constant speed, and another distance traveled while decelerating.
  • The difference, in this case, is that the total distance must be the same initial distance between the car and the deer, 39 m.
  • ⇒Δx = Δx₁ + Δx₂ = 39 m. (5)
  • Δx₁, is the distance traveled at a constant speed during the reaction time, so we can express it as follows:

       \Delta x_{1} = v_{omax} * t_{react} = 0.5* v_{omax} (6)

  • Δx₂, is the distance traveled while decelerating, and can be obtained  using (2):

        v_{omax} ^{2} = 2* a* \Delta x_{2} (7)

  • Solving for Δx₂, we get:

       \Delta x_{2} = \frac{-v_{omax} ^{2} x}{2*a}  = \frac{-v_{omax} ^{2}}{(-22m/s2)} (8)

  • Replacing (6) and (8) in (5), we get a quadratic equation with v₀max as the unknown.
  • Taking the positive root in the quadratic formula, we get the following value for vomax:
  • v₀max = 24.3 m/s.
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