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2 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]2 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]2 years ago

3 0

I believe that its waves

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Which statement is true?

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2 years ago

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Bond [772]

Answer:

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Explanation:

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2 years ago

The ball rolling along a straight and level path. The ball is rolling at a constant

Aleksandr-060686 [28]

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6 0

2 years ago

If an atomic nucleus were the size of a dime how far away might one of its electrons be?

mihalych1998 [28]

The radius of a nucleus of hydrogen is approximately $r_{n1}=1\cdot 10^{-15}m$ , while we can use the Borh radius as the distance of an electron from the nucleus in a hydrogen atom: $r_{e1}=5.3 \cdot 10^{-11}m$

The radius of a dime is approximately $r_{n2} = 9\cdot 10^{-3}m$ : if we assume that the radius of the nucleus is exactly this value, then we can find how far is the electron by using the proportion
$r_{n1}:r_{e1}=r_{n2}:r_{e2}$
from which we find
$r_{e2}= \frac{r_{e1} r_{n2}}{r_{n1}}= \frac{(5.3 \cdot 10^{-11}m)(9\cdot 10^{-3}m)}{1 \cdot 10^{-15}m}=477 m$

So, if the nucleus had the size of a dime, we would find the electron approximately 500 meters away.

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2 years ago

An isolated conducting sphere has a 10 cm radius. One wire carries a current of 1.000 002 0 A into it. Another wire carries a cu

romanna [79]

Answer:

t = 5.56 ms

Explanation:

Given:-

- The current carried in, Iin = 1.000002 C

- The current carried out, Iout = 1.00000 C

- The radius of sphere, r = 10 cm

Find:-

How long would it take for the sphere to increase in potential by 1000 V?

Solution:-

- The net charge held by the isolated conducting sphere after (t) seconds would be:

qnet = (Iin - Iout)*t

qnet = t*(1.000002 - 1.00000) = 0.000002*t

- The Volt potential on the surface of the conducting sphere according to Coulomb's Law derived result is given by:

V = k*qnet / r

Where, k = 8.99*10^9 ..... Coulomb's constant

qnet = V*r / k

t = 1000*0.1 / (8.99*10^9 * 0.000002)

t = 5.56 ms

7 0

2 years ago