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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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goblinko [34]

The plane's velocity of 35.11 m/s is actually due in a north-eastward direction. The 12 m/s velocity is the vertical component of the plane's velocity, hence it is pointing northwards. We will use the formula:

Vy = Vsin∅

To determine the angle ∅ at which the plane is flying. This is:

12 = 35.11 * sin∅
∅ = 20.0 degrees

The eastward velocity is:

Vx = Vcos∅
Vx = 35.11 * cos(20)
Vx = 33.0 m/s

The plane's eastward velocity is 33.0 m/s

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3 years ago

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A 50.0 g toy car is released from rest on a frictionless track with a vertical loop of radius R (loop-the-loop). The initial hei

Mariana [72]

Answer:

the speed of the car at the top of the vertical loop $v_{top} = 2.0 \sqrt{gR \ \ }$

the magnitude of the normal force acting on the car at the top of the vertical loop $F_{N} = 1.47 \ \ N$

Explanation:

Using the law of conservation of energy ;

$mgh = mg (2R) + \frac{1}{2}mv^2_{top}\\\\mg ( 4.00 \ R) = mg (2R) + \frac{1}{2}mv^2_{top}\\\\g(4.00 \ R) = g (2R) + \frac{1}{2}v^2 _{top}\\\\v_{top} = \sqrt{2g(4.00R - 2R)}\\\\v_{top} = \sqrt{2g(4.00-2)R$

$v_{top} = 2.0 \sqrt{gR \ \ }$

The magnitude of the normal force acting on the car at the top of the vertical loop can be calculated as:

$F_{N} = \frac{mv^2_{top}}{R} \ - mg\\\\F_{N} = \frac{m(2.0 \sqrt{gR})^2}{R} \ - mg\\\\F_{N} = [(2.0^2-1]mg\\\\F_{N} = [(2.0)^2 -1) (50*10^{-3} \ kg)(9.8 \ m/s^2]\\\\$

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3 years ago

how do you Compute the approximate acceleration of gravity for an object above the earth's surface, assigning accel gravity with

solong [7]

The approximate acceleration of gravity for an object above the earth's surface is 9.8 m/s².

<h3>Acceleration due to gravity </h3>

The acceleration due to gravity of an object above the Earth's surface is calculated from Newton's second law of motion and Newton's law of universal gravitation.

<h3>Force between the object and Earth</h3>

$F_g = \frac{GmM}{R^2}$

<h3>Weight of the object</h3>

$F_g = mg\\$

Solve (1) and (2)

mg = GmM/R²

g = GM/R²

where;

M is mass of Earth
G is gravitational constant
R is radius of the Earth = 6,371 km

g = (6.673 x 10⁻¹¹ x 5.98 x 10²⁴)/(6,371 x 10³)²

g = 9.8 m/s²

Learn more about acceleration due to gravity here: brainly.com/question/88039

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2 years ago

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dem82 [27]

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3 years ago

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likoan [24]

Explanation:

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V = 600 V

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