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3 years ago

What is the main evidence for the structure of the Earth that occur from the generation of an earthquake?

Physics

2 answers:

Alik [6]3 years ago

4 0

Answer;

-Waves

Explanation;

-The main evidence for the structure of the Earth comes from studying waves generated during an earthquake. When stress builds up and rocks at depth break, then elastic waves, just like sound waves are sent from the point of rupture. These waves speed out and move through the earth.

-When seismometers are put on the surface of the Earth, these waves can be measured. In particular, if we have enough seismic stations and the quake is big enough, then we can take all the travel time information and invert it. That means we can obtain a graph that shows the density of rock as a function of depth.

-Once we know the density we can estimate what the actual materials are. We find that the earth has a small inner core made out of iron surrounded by a larger outer liquid iron core and that is surrounded by a rocky mantle and rocky crust.

xeze [42]3 years ago

3 0

I believe that its waves

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3 years ago

Two ice skaters, each with a mass of 72.0 kg, are skating at 5.45 m/s when they collide and stick together. If the angle between

Rudiy27

Answer:

The skater 1 and skater 2 have a final speed of 2.02m/s and 2.63m/s respectively.

Explanation:

To solve the problem it is necessary to go back to the theory of conservation of momentum, specifically in relation to the collision of bodies. In this case both have different addresses, consideration that will be understood later.

By definition it is known that the conservation of the moment is given by:

$m_1v_1+m_2v_2=(m_1+m_2)v_f$

Our values are given by,

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As the skater 1 run in x direction, there is not component in Y direction. Then,

Skate 1:

$v_{x1}=5.45m/s$

$v_{y1}=0$

Skate 2:

$v_{x2} = 5.45*cos105= -1.41m/s$

$v_{y2} = 5.45*sin105 = 5.26m/s$

Then, if we applying the formula in X direction:

m_1v_{x1}+m_2v_{x2}=(m_1+m_2)v_{fx}

75*5.45-75*1.41=(75+75)v_{fx}

Re-arrange and solving for v_{fx}

v_{fx}=\frac{4.04}{2}

v_{fx}=2.02m/s

Now applying the formula in Y direction:

$m_1v_{y1}+m_2v_{y2}=(m_1+m_2)v_{fy}$

$0+75*5.25=(75+75)v_{fy}$

$v_{fy}=\frac{5.25}{2}$

$v_{fy}=2.63m/s$

Therefore the skater 1 and skater 2 have a final speed of 2.02m/s and 2.63m/s respectively.

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