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dalvyx [7]
3 years ago
11

When reading a seismograph, _____ waves come first, then _____ waves, and, finally, _____ waves. S, P, L L, P, S P, S, L L, S, P

P, L, S
Physics
2 answers:
elena55 [62]3 years ago
8 0
<h3><u>Answer;</u></h3>

P, S, L

When reading a seismograph,<u> </u><u>P waves</u> come first, then <u>S waves,</u> and, finally, <u>L waves</u>.

<h3><u>Explanation;</u></h3>
  • Seismic waves are used to measure the strength of an earthquake using and instrument called  seismograph. The major three seismic waves used are the primary waves (P-waves), secondary waves (S-waves) and the L waves.
  • Secondary waves move through earth by causing particles in rocks to move at right angles to the direction of the waves.  Primary waves on the other hand move through the earth by causing particles of rocks to move back and forth in the same direction as the waves.
  • The <u>primary waves</u> are the first to reach a seismograph after an earthquake . These waves move in a push-and-pull motion.

Naddik [55]3 years ago
6 0
<span>When reading a seismograph, P waves (Fastest) come first, then S waves (Second fastest), and, finally, L </span><span> (Love) R (</span><span>Rayleigh) waves.

Considering answer options: P, S, L waves. Answer
</span>

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Answer:

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We are going to solve this interesting problem

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Let's use the concept of conservation of energy

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         Em₀ = U = m g y₁

final point. At the exit of the ramp

         Em_f = K + U = ½ m v² + ½ I w² + m g y₂

notice that we include the translational and rotational energy, we assume that the height of the exit ramp is y₂

energy is conserved

          Em₀ = Em_f

         m g y₁ = ½ m v² + ½ I w² + m g y₂

angular and linear velocity are related

        v = w r

the moment of inertia of a sphere is

         I = \frac{2}{5} m r²

we substitute

         m g (y₁ - y₂) = ½ m v² + ½ (\frac{2}{5} m r²) (\frac{v}{r})²

         m g h = ½ m v² (1 + \frac{2}{5})

where h is the difference in height between the two sides of the ramp

h = y₂ -y₁

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This is the exit velocity of the vertical movement of the sphere

         v_sphere = 0.845 √2gh

B) is the same case, but for a box without friction

   starting point

          Em₀ = U = mg y₁

   final point

          Em_f = K + U = ½ m v² + m g y₂

          Em₀ = Em_f

          mg y₁ = ½ m v² + m g y₂

          m g (y₁ -y₂) = ½ m v²

          v = √2gh

this is the speed of the box

          v_box = √2gh

to know which body reaches higher in the air we can use the kinematic relations

          v² = v₀² - 2 g y

at the highest point v = 0

           y = vo₀²/ 2g

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           y_sphere = 5/7 2gh / 2g

           y_esfera = 5/7 h

for the block

           y_block = 2gh / 2g

            y_block = h

       

therefore the block reaches higher than the sphere

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