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3 years ago

What happens to Earth’s plates during an earthquake?

Physics

1 answer:

vazorg [7]3 years ago

4 0

A) Claim 1: Plates move, which can cause earthquakes.

Explanation:

The Plate Tectonic Theory proves the claim of plate move, causing earthquakes.

This theory states that the earth’s crust along with the uppermost mantle is formed of several thin but large surfaced rigid patch work of plate-like structures called tectonic plates.

There are about 15 large slabs on the earth’s outer surface and constitutes the lithosphere. Lithosphere of the earth is represented by the oceanic and continental crust layer and the uppermost mantle layer.

These plates move or slide relative with each other. These plates form divergent, convergent, or transform boundaries. Slips or faults along these boundaries forms subduction zones leading to great stress. This prevents normal gliding motion resulting in earthquakes.

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Two people are standing on a 3.12-m-long platform, one at each end. The platform floats parallel to the ground on a cushion of a

Ann [662]

Answer:

The magnitude of displacement is 0.082m

Explanation:

While the ball is in motion,we have MV + mv= 0 ...eq1

Where M = combined mass of the platform and the two people.

V = velocity of the platform

m = mass of the ball

v = velocity of the ball

The distance that the platform moves is given by:

X = Vt ...eq2

Where t is the time that the ball is in the air.

The time the ball is in the air is given by:

L/(v-V) ...eq3

Where L is the length of the platform

The quantity(v-V) = velocity of the ball relative to the platform.

Combining eq2 and eq3

X = (V/(v - V))L

From eq1 , the ratios of the velocities is V/v = -m/M

X = (V/v)L / (1 - (V/v) = (-m/M) L /(1+ (m/M))

X = -mL/(M + m)

X = - (3.36kg × 3.12m) /( 119kg + 3.36kg)

X = - 10.48/ 122.36

X = -0.082m

The minus sign implies that the displacement of the platform is in the opposite direction to the displacement of the ball.

Therefore, the distance moved by the platform is the magnitude of this displacement 0.082m.

7 0

3 years ago

Read 2 more answers

a stone is dropped from rest at an initial height h above the surface of the earth. Show that the speed with which it strikes th

Ksju [112]

Answer:

$v=\sqrt{2gh}$

Explanation:

We could use conversation of energy. Total distance the stone will cover will be

$h=\frac{1}{2} g t^{2}$

the Final velocity will be

$v=gt\\v=g\sqrt{\frac{2h}{g} }\\ v=\sqrt{2gh}$

5 0

3 years ago

A 6 kg cart starting from rest rolls down a hill and at the bottom has a speed of 10 m/s. What is the height of the hill?

Arisa [49]

Answer:

h = 5.09 m

Explanation:

Applying the Law of conservation of energy to this situation, we can write:

$Kinetic\ Energy\ Gained\ by\ the\ Cart = Potential\ Energy\ Lost\ by\ the\ Cart\\\frac{1}{2}mv^2 = mgh\\\\h = \frac{v^2}{2g}$

where,

h = height of the hill = ?

v = speed of cart at the end = 10 m/s

g = acceleration due to gravity = 9.81 m/s²

Therefore,

$h = \frac{(10\ m/s)^2}{(2)(9.81\ m/s^2)}\\\\$

h = 5.09 m

4 0

2 years ago

A pirate fires his cannon parallel to the water but 3.5 m above the water. The cannonball leaves the cannon with a velocity of 1

Bumek [7]

Answer:

202.8m

Explanation:

Given that A pirate fires his cannon parallel to the water but 3.5 m above the water. The cannonball leaves the cannon with a velocity of 120 m/s. He misses his target and the cannonball splashes into the briny deep.

First calculate the total time travelled by using the second equation of motion

h = Ut + 1/2gt^2

Let assume that u = 0

And h = 3.5

Substitute all the parameters into the formula

3.5 = 1/2 × 9.8 × t^2

3.5 = 4.9t^2

t^2 = 3.5/4.9

t^2 = 0.7

t = 0.845s

To know how far the cannonball travel, let's use the equation

S = UT + 1/2at^2

But acceleration a = 0

T = 2t

T = 1.69s

S = 120 × 1.69

S = 202.834 m

Therefore, the distance travelled by the cannon ball is approximately 202.8m.

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2 years ago

Which radiation has a higher frequency than visible light

Bogdan [553]

Infrared radiation lies between the visible and microwave portions of the electromagnetic spectrum. Infrared waves have wavelengths longer than visible and shorter than microwaves, and have frequencies which are lower than visible and higher than microwaves.

6 0

2 years ago

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