How are earthquakes at continental plate boundaries different from earthquakes at mid-ocean ridges?

Answers

Answer 1

Differences between earthquakes at continental plate boundaries and earthquakes at mid-ocean ridges discussed in the answer.

What is earthquakes at continental plate boundaries and earthquakes at mid-ocean ridges?

When plates travel in the same direction and collide, this happens. The thinner, denser, and more flexible oceanic plate dips beneath the thicker, more rigid continental plate when a continental plate meets an oceanic plate. It's known as subduction.

Deep ocean trenches, like the one that runs down South America's west coast, are created by subduction. The continent's undercut rocks start to melt. On occasion, a chain of volcanoes forms as the molten rock rises to the surface and passes through the continent. Nearly 80% of earthquakes happen along convergent borders, when plates are being forced together.

Shallow earthquakes, often less than 30 km deep, occur in small bands adjacent to plate borders along mid-ocean ridges and transform edges. There are earthquakes in subduction zones at various depths, some of which are more than 700 km deep. Because they occur everywhere throughout the subducting slab that stretches beneath the opposing plate, earthquake bands are wider along subduction zones. Wide areas of distributed earthquakes may be associated with collision zones between continents, such as those between the Eurasian plate and the plates to the south of the African, Arabian, and Indian tectonic arcs.

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Related Questions

If you drag a 50kg block across the floor which has a coefficient of friction of .30, what is
the force needed to accelerated it at 2.0m/s^2 ?

Answers

Answer:

750 Newton

Explanation:

force=mass(acceleration)

In a car race, the average speed of the car was 90 km/hr where the total
covered distance was 450 km.
1. Calculate the average speed of the car in SI unit.

Answers

Explanation:

The speedometer of a motor car showed the following speeds in km per hour at the ends of successive intervals of 3 secs. 38.9, 52.8, 64.4. 73.6, 80.9, 86.1, 90.0, 92.5 Plot the velocity – time graph in convenient units and find from the graph, (a) The distance covered while the speed increased from 80

light of wavelength 520 nm falls on a slit that is 3.20 um wide. Estimate how far the first bright-sh difrrection fringe is

Answers

Answer:

because of the gravity of the earth

HELP PLEASE : )
IT'S SCIENCE

Answers

Answer:

1. A and C are balanced forces as all the forces get cancelled by the opposite and equal force.

2. B and D as in B the Net force is 5N to the right and in D the Net force is 20N to the right.

3. This would be all the unbalance forces which are B and D

4. B would be moving 5N to the right and D would move 20N to the right.

which wave has a higher frequency and why?

Answers

Explanation:

the figure in the left side has higher frequency.

because it has more nos. of wave in 1sec.

Where does the energy released in a nuclear decay reaction come from

A. Electrons

B. Chemical bonds

C. Positrons

D. The binding energy of the nucleus

Answers

The answer is positrons

¿Cuál es la densidad de un metal si una muestra tiene una masa de 63.5 g cuando se mide en el aire y una masa aparente de 55.4 g cuando está sumergida en agua?. Considere la densidad del agua como 1000 kg/m3.

Answers

La densidad del metal como se requiere en la pregunta es 7.8 * 10 ^ 3 Kg / m ^ 3.

Sabemos que ese empuje hacia arriba = peso en aire - peso en líquido

Peso en el aire = 63,5 * 10 ^ -3 Kg * 10 m / s ^ 2 = 0,635 N

Peso en líquido = 55,4 * 10 ^ -3 Kg * 10m / s ^ 2 = 0,554 N

Empuje hacia arriba = 0,635 N - 0,554 N = 0,081 N

Pero ;

Empuje hacia arriba = Volumen * densidad del fluido * aceleración debido a la gratitud

volumen = empuje hacia arriba / densidad del fluido * aceleración debido a la gratitud

volumen = 0.081 N / 1000 * 10

Volumen = 8.1 * 10 ^ -6 m ^ 3

Densidad = masa / volumen

Densidad = 63,5 * 10 ^ -3 Kg / 8,1 * 10 ^ -6 m ^ 3

= 7,8 * 10 ^ 3 kg / m ^ 3

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A guitar string has a fundamental frequency f. The tension in the string is increased by 1.70%. Ignoring the very small stretch of the string. How does the fundamental frequency change?

Answers

The characteristics of the speed of the waves in strings and the resonance allows to find the change in the fundamental frequency when changing the tension is:

 The change in fundamental frequency is: f = 1.08 f₀

The speed of the chord wave is given by the relationship between the tension and the density of the medium.

          [tex]v= \sqrt{\frac{T}{\mu } }[/tex]  

Where v is the velocity of the wave, T the tension of the string and μ the density

In a rope held at the ends, a process of standing waves occurs, two at the point where it is attached we have a node and a anti-node in the center.

             2L = n λ

Where L is the length of the chord and call the wavelength

Wave speeds are related to wavelength and frequency.

        v = λ f

We substitute.

            [tex]\sqrt{\frac{T}{\mu } } = \frac{2L}{n} \ \ f[/tex]  

For the fundamental frequency n = 1

            f₀ = [tex]f_o = \sqrt{\frac{T}{\mu } } \ \ \frac{1}{2L}[/tex]  

They indicate that the tension increases 1.70%

           T = T₀ + 0.17 T₀

           T = 1.17 T₀

We substitute.

         [tex]f = \sqrt{1.17 } \ \sqrt{\frac{T_o}{\mu } } \ \ \frac{1}{2L}[/tex]

         f = ra1.17 f₀

         f = 1.08 f₀

In conclusion, using the characteristics of the velocity of the waves in strings and the resonance we can find the change in the fundamental frequency when changing the tension is:

 The change in fundamental frequency is: f = 1.08 f₀

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Why are saturn's rings so noticeable?.

Answers

Answer:

Why are Saturn's rings so noticeable? They are made of lots of fresh, bright icy particles from a recent breakup.

Explanation:

what is the direction of the rotational velocity vector of the jar lid when you open a jar of pickles?

Answers

Answer:

Explanation:

The rotational velocity vector is directed away from the jar

PLEASE HELP!!!!!
How does the law of conservation of energy apply to an object rolling down different surfaces?

Answers

Answer:

it states that energy can either be gained or lost but it only changes its form.

Explanation:

for example:as a ball is still on the table it posses a potential energy of 100j and a k.e of 0j,as it falls it gains k.e so the midpoint the p.e is equal to the k.e (50j equally) as it approches the ground it completely gains k.e (100j) and the p.e is 0j.

total energy is 100j so it has been converted from p.e to k.e.

hope u have understood.

Virtual image formed by concave mirror

Answers

Answer:

yes

Explanation:

because it is a diverging mirror

Rubies are used in electronics.
Please select the best answer from the choices provided
) Т
) F

Answers

False, rubies are not used in electronics because they are bad conductors.

Every atom contains a
.............. which is positively charged

Answers

Answer:

Proton

Explanation:

The nucleus has an overall positive charge as it contains the protons. Every atom has no overall charge (neutral). This is because they contain equal numbers of positive protons and negative electrons.

Dark matter is inferred to exist because it explains the abundance of helium in the universe today. we can observe its gravitational influence on visible matter. it explains how the expansion of the universe can be accelerating. we see lots of dark patches in the sky

Answers

we can observe its gravitational influence on visible matter.

Dark matter is a term used in astrophysics to refer to matter that corresponds to approximately 80% of the matter in the universe.

Dark matter is related to the motion of galaxies because visible matter makes up only a small part of the cluster and the galaxies show signs of being composed mainly of a halo of dark matter concentrated in their center.

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find the impulse of a 20.0 KG object accelerates to8.1 M/S from 2.3 M/S

Answers

Answer:

116 N•s

Explanation:

change in momentum = mvf-mvi

= m(Vf - Vi)

= 20( 8.1-2.3)

change in momentum= 116 kg•m/s

: .impulse is 116 N•s

Suppose that a cart is moving along a road at a constant velocity. Give two examples of ways to cause the cart to decelerate, explaining how the net force is affected. (2 points)

Answers

The cart moving at a constant velocity will decelerate when a braking force is applied or the road is made to become rougher.

Deceleration is also known as negative acceleration. It occurs when velocity decreases with time. There are two ways by which the cart moving at constant velocity can be made to decelerate. These two ways are;

By applying a braking force: A braking force causes the velocity to decrease with time. When a braking force is applied the frictional force exceeds the forward force hence the net force decreases.By increasing the roughness of the road: When the roughness of the road is increased, the frictional force also increases causing the net force on the cart to decrease hence it decelerates.

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28
If it takes 1,697 m to stop a car initially moving at 46 m/s, what distance is required to stop a car moving at a quarter
of the initial speed under the same conditions?

Answers

Answer:

Approximately [tex]106\; \rm m[/tex], [tex](1/16)[/tex] of the original distance, assuming that the acceleration is constant.

Explanation:

Let [tex]v[/tex] and [tex]u[/tex] denote the final and initial velocity of the vehicle ([tex]v = 0\; \rm m \cdot s^{-1}[/tex] since this vehicle has come to a stop.)Let [tex]a[/tex] denote the acceleration of the vehicle (a constant under the assumption.) Let [tex]x[/tex] denote the displacement of the vehicle.

If the acceleration on this vehicle is constant, the SUVAT equations would apply:

[tex]\displaystyle v^{2} - u^{2} = 2\, a\, x[/tex].

Rearrange this equation to find an expression for [tex]x[/tex], the displacement required for the vehicle to come to a stop:

[tex]\begin{aligned} x &= \frac{v^{2} - u^{2}}{2\, a} \\ &= \frac{-u^{2}}{2\, a} && (\text{$v = 0\; \rm m \cdot s^{-1}$})\end{aligned}[/tex].

Thus, under these assumptions, [tex]x[/tex] would be proportional to [tex]u^{2}[/tex]. In other words, the distance required for this vehicle to come to a stop would be proportional to the square of the initial velocity of this vehicle.

If the initial velocity [tex]u[/tex] is reduced to [tex](1/4)[/tex] (a quarter) of the initial value, the distance required for stopping this vehicle would be [tex](1/4)^{2} = (1/16)[/tex] of the initial value:

[tex]\begin{aligned}1,\!697\; {\rm m} \times \frac{1}{16} \approx 106\; \rm m\end{aligned}[/tex].

two objects are sitting 6m apart, one object has a mass of 100kg and the other has a mass of 200kg. what is the gravitational attraction between them

Answers

F=G(m1m2/r^2)

F= 6.67x10^-11(100 x 200/ 36)

F= 3.7 x 10^-8

How does the diameter of a black hole (size of the event horizon) depend on the mass inside the black hole

Answers

Answer:

The greater the mass, the greater is the diameter

Explanation:

The greater the mass, the greater is the diameter. almost all galaxies contain supermassive black holes.

How does water temperature affect density?

Answers

it expands, increasing in volume. The warmer the water, the more space it takes up, and the lower its density.

if a car travels around a circular track with a radius of 100 m once every 10s what will its average angular velocity around the track be?

Answers

Answer:

0.628 radians per second

Explanation:

Angular velocity

= 2pi/T

= 2(3.14159)/10

= 0.628 radians per second

The average angular velocity around the track is 62.8 m/s, using the circumference of the track as the total distance.

Average angular velocity:

What information do we have?

Radius of track = 100 m

Time taken = 10s

Circumference of track = 2πr

Circumference of track = 2(3.14)(100)

Circumference of track = (2)(314)

Circumference of track = 628 m

Average angular velocity around the track = Circumference of track / Time taken

Average angular velocity around the track = 628 / 10

Average angular velocity around the track = 62.8 m/s

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Jade has a refractive index of 1.61. If light approaches the gem at an angle of 80.0°, what is the angle of refraction?

Calculate the index of refraction of a material if the angle of incidence is 60° and the angle of refraction is 50°

Answer using calculations & full work and thoughts shown.

Answers

From Snell's law,  the angle of refraction is 38 degree approximately and the index of refraction is 1.13

The given refractive index is 1.61

If light approaches the gem at an angle of 80.0°, that means the angle of incidence is 80.0°

To calculate the angle of refraction, we will use Snell's law which state that:

n = sin i / sin r

where n = refractive index

Substitute all the parameters into the formula

1.61 = sin 80 / sin r

make sin r the subject of formula

sin r = sin 80 / 1.61

sin r = 0.611

r = [tex]Sin^{-1}[/tex](0.611)

r = 37.7

r = 38 degree (Approximately)

We will use the same formula to calculate the index of refraction of a material if the angle of incidence is 60° and the angle of refraction is 50°. That is,

n = sin i / sin R

Where n = index of refraction

i = Incidence

R = refraction

Substitute all the parameters to the formula

n = sin 60 / sin 50

n = 0.866 / 0.766

n = 1.13

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Please help :[ Hold a flat piece of paper horizontally. Lift it high in front of you, and drop it. Then crumple a piece of paper into a tight ball, hold it high in front of you, and drop it from the same height. Compare the motion of the two pieces of paper.

Answers

The one you lifted in front of you ( the flat piece of paper being holded horizontally ) goes down so slow. While the one being crumpled into a ball goes down fast. Subscribe to DreamfanBG for support.

Answer:

Answer below

Explanation:

The flat piece of paper just gently glided to the ground, as if it were a feather. It didn’t go down in a straight line, it swayed from side to side. The crumpled piece of paper seemed like it had more weight to it (although I know that it didn’t), and it fell way quicker than the flat piece of paper.

_____________________________________________________

(Hope this helps can I pls have brainlist (crown)☺️)

pease HELP Me out with this science problem no point of answering a fake question cause it will be deleted and brainly will give you a BANN WARNING!!

Answers

Answer: Increasing the amount of sugar.

Explanation: Increasing the amount of reactants increases the rate of the reaction. This is given by the collision theory, which states that a chemical reaction must involve the collisions of particles together, as well as with the walls of the container it is in. By increasing the amount of reactant in the container you are increasing the collision frequency, and the rate of the reaction is dependant on the collision frequency multiplied by the fraction of collisions that are effective. Increasing the temperature increases the collision frequency as well as the fractions of collisions that are effective, so it’s definitely not that one. Hope this helped:)

If the acceleration of an object increases, what effect does this have on the Force and Mass?
A Force Increases, Mass Increases
B Force Decreases, Mass Decreases
C Force increases, Mass remains the same
D Force Decreases, Mass remains the same​

Answers

Answer:

Force Increases and Mass remains the same.

Explanation:

what are the slight fluctuations seen in maps of the cosmic background radiation?

Answers

If you look very closely, there are slight fluctuations from place to place. farther into space-further back in time-then when we look at the farthest galaxies. What are the slight fluctuations seen in maps of the cosmic background radiation? A reduction in the curvature of space.

The beginning of the formation of galaxies and clusters of galaxies are the slight fluctuations seen in maps of the cosmic background radiation. So, the correct option is C.

What is Cosmic background radiation?

The cosmic background radiation is defined as the microwave radiation that fills all of space which is a remnant that provides an important source of data on the early universe. With a standard optical telescope, the background space between stars and galaxies is almost completely dark.

It is also called the Cosmic Microwave Background (CMB) or electromagnetic radiation filling the universe which is a residual effect of the Big Bang that happened 13.8 billion years ago. The beginning of the formation of galaxies and clusters of galaxies are the slight fluctuations seen in maps of the cosmic background radiation.

Therefore, the correct option is C.

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Your question is incomplete, most probably the complete question is:

What are the slight fluctuations seen in maps of the cosmic background radiation?

a. uncertainties in the map

b. variations in the instrument's sensitivity

c. the beginning of the formation of galaxies and clusters of galaxies

d. dark matter

e. none of the above

Match the following terms to the correct definition:

1.law of conservation of energy:
2.thermal energy:
3.temperature:
4.kinetic energy:
5.potential energy:

A.the energy associated with the motion and positions of particles
B.the stored energy an object possess, often due to its position relative to other objects around it
C.a scientific principle that states that energy can neither be created nor destroyed by ordinary chemical or physical means.
D.the energy that an object possesses due to its motion
E.a measure of the average kinetic energy of the particles in an object due to their random motions

Answers

The terms with their correct definition is matched as follows:

The law of conservation of energy is a scientific principle that states that energy can neither be created nor destroyed by ordinary chemical or physical means.

Thermal energy is the measure of the average kinetic energy of the particles in an object due to their random motions.

Temperature is the energy associated with the motion and positions of particles.

Kinetic energy is the energy that an object possesses due to its motion or movement. It is measured in Joules.

Potential energy is the stored energy an object possess, often due to its position relative to other objects around it.

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8) If a dog is running at 25 feet per second, how far will he travel after I minute of running? Efe runs for 30 minutes at a speed of 40 miles per hour, how far will run? 10) Turbo the Snail moves across the ground at a pace of 12 feet per day If the garden is 48 feet away, how many days will it take for the snail to get there?​

Answers

the dog will run 1,500 feet
Efe will run 20 miles in 30 minutes
Turbo will take 4 days to get there

What is the acceleration as an object goes from +20 m/s to +26 m/s over 1.4 s?

Answers

Answer:

4.3 m/s^2

Explanation:

a = vf - vi / t

26 - 20 / 1.4

= 4.28 ( round your numbers)

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