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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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What is the purpose of a rearview mirror in a car?

Shalnov [3]

The third choice.

The driver wants to see the object that is behind him. The light reflects off the mirror into the eyes of the driver portraying the object behind him

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3 years ago

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I will Give Brainliest To whoever actually answers A 500 kg satellite experiences a gravitational force of 3000 N, while moving

snow_lady [41]

Answer:

$9.7\times 10^{-4}\ rad/s$

Explanation:

Given:

$m=500 kg\\F=3000 N$

$Radius of earth , R=6371 \times 10^3\ m\\Angular speed =\omega\\We\ know\ that\ \\F= m\times \omega^{2} \times R\\\omega^{2}=\frac{F}{m*R} \\\\=\frac{3000}{500*6371 \times 10^3\ m}$

= $\frac{6}{6371 \times 10^3\ m}$

= $9.7\times 10^{-4}\ rad/s$

7 0

3 years ago

What conclusions can be drawn about the existence of nitrogen-14 and nitrogen-15? A) They are isotopes of nitrogen and they cont

AleksandrR [38]

<span>Answer: A) They are isotopes of nitrogen and they contain the same number of protons and electrons but each contains a different number of neutrons - 7 and 8 respectively.

Isotopes are atoms of a chemical element whose nucleus has the same atomic number, Z, but different atomic mass, A. The atomic number corresponds to the number of protons in the atom, therefore the isotopes of an element contain the same number of protons and electrons (atoms have to be neutral particles). The difference in atomic masses arises from the difference in the number of neutrons in the atomic nucleus.
</span>

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3 years ago

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Sarah weighs 500 newtons. She climbs a set of stairs 6 meters high. How much work does she do? 83 J Work equals force times dist

jek_recluse [69]

Answer:

<em>Answer: Work equals force times distance. 3,000 J</em>

Explanation:

Work Done By A Force

When some force $\vec F$ is applied and a displacement $\vec r$ is achieved, the work done by the force is given by

$W=\vec F \cdot \vec r$

Note that the work is a scalar magnitude as the result of the dot-product of two vectors. If the force and the displacement are parallel, then the vectors can be replaced as its magnitudes F,x and the work is

$W=F.x$

The dot product becomes a simple arithmetic product, i.e force times distance.

Sara weighs 500 Nw and she climbs up a 6 meter set of stairs. She needs to lift her weight up, so the force is the weight and the distance is the height of the stairs, thus

$W=500\ Nw.\ 6\ m=3,000\ Joule$

Answer: Work equals force times distance. 3,000 J

7 0

3 years ago

Two cars are heading towards one another. Car A is moving with an acceleration of aA = 4 m/s2. Car B is moving with an accelerat

Paladinen [302]

Answer:

Car B reaches car A in 19.7 s.

Explanation:

Hi there!

The equation of the position of an object moving in a straight line at constant acceleration is as follows:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the object at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration

When both cars meet, their positions are the same. At the meeting point:

position of car A = position of car B

xA = xB

x0A + v0A · t + 1/2 · aA · t² = x0B + v0B · t + 1/2 · aB · t²

Let´s place the origin of the frame of reference at the point where A is located. In that case x0A = 0 and x0B = 2900 m. Since both cars are initially at rest, v0A and v0B = 0. So, the equation gets reduced to this:

1/2 · aA · t² = x0B + 1/2 · aB · t²

If we replace with the data we have and solve for t:

1/2 · 4 m/s² · t² = 2900 m - 1/2 · 11 m/s² · t²

2 m/s² · t² = 2900 m - 5.5 m/s² · t²

5.5 m/s² · t² + 2 m/s² · t² = 2900 m

7.5 m/s² · t² = 2900 m

t² = 2900 m / 7.5 m/s²

t = 19.7 s

Car B reaches car A in 19.7 s.

4 0

3 years ago