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Pavel [41]
3 years ago
13

As the plates continues to grind against each other,what geologic events could take place

Physics
2 answers:
motikmotik3 years ago
8 0
<span>As the plates continue to grind against each other the geologic events could take place is the Earthquake

Explanation
:
</span>
An earthquake (also called<span> a quake, tremor or temblor) </span>is that the<span> shaking of the surface of </span>the planet<span>, </span>ensuing<span> from the </span><span>sudden </span>release<span> of energy </span>within the<span> Earth's </span>layer that makes unstable<span> waves.
Earthquakes </span>will point<span> size from </span>those who are therefore weak<span> that </span>they can not<span> be felt to those violent enough to toss </span>people<span> around and destroy whole cities.
The seismicity, or </span>unstable activity<span>, of </span>an area, is that the<span> frequency, </span>sort<span> and size of earthquakes </span>practised<span> over an </span>amount of your time<span>. </span>

<span>The phenomena of the tectonic drifts </span>it isn't<span> forever and at one </span>purpose can<span> stop.
The </span>results of<span> what you </span>call<span> grind are earthquakes, erosion, land formation, Tectonic movement into the ocean or out of the ocean.

This </span>development can<span> continue </span>until purpose<span> that, a gap between the crust and </span>mantle are going to be massive<span> enough to clap the crust into the </span><span>mantel.</span>

-BARSIC- [3]3 years ago
3 0
As tectonic plates continue to grind against each other, the most common event that can take place would be an Earthquake, although volcanic eruptions can also occur. 
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Explain why some people see objects nearby clearly, but objects far away appear blurry. Also, explain how this condition can be
Maru [420]

Myopia

Explanation:

myopia is a common vision condition in which you can see objects near to you clearly, but objects farther away are blurry. It occurs when the shape of your eye causes light rays to refract incorrectly, focusing images in front of your retina instead of on your retina. It can be corrected corrected with eyeglasses, contact lenses or refractive surgery.

7 0
3 years ago
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A ball is thrown straight up. What are the velocity and acceleration of the ball at the highest point in its path?
zubka84 [21]

Answer:

b. v = 0, a = 9.8 m/s² down.

Explanation:

Hi there!

The acceleration of gravity is always directed to the ground (down) and, near the surface of the earth, has a constant value of 9.8 m/s². Since the answer "b" is the only option with an acceleration of 9.8 m/s² directed downwards, that would solve the exercise. But why is the velocity zero at the highest point?

Let´s take a look at the height function:

h(t) = h0 + v0 · t + 1/2 g · t²

Where

h0 = initial height

v0 = initial velocity

t = time

g = acceleration due to gravity

Notice that the function is a negative parabola if we consider downward as negative (in that case "g" would be negative). Then, the function has a maximum (the highest point) at the vertex of the parabola. At the maximum point, the slope of the tangent line to the function is zero, because the tangent line is horizontal at a maximum point. The slope of the tangent line to the function is the rate of change of height with respect to time, i.e, the velocity. Then, the velocity is zero at the maximum height.

Another way to see it (without calculus):

When the ball is going up, the velocity vector points up and the velocity is positive. After reaching the maximum height, the velocity vector points down and is negative (the ball starts to fall). At the maximum height, the velocity vector changed its direction from positive to negative, then at that point, the velocity vector has to be zero.

8 0
3 years ago
What is unusual about the results of mass determinations of clusters of galaxies?
Art [367]

Answer:

I think it's bigger than most galaxies

3 0
2 years ago
Read 2 more answers
A ball initially at rest rolls down a hill with an acceleration of 3.3 m/s2. If it accelerates for 7.5 s, how far will it move?
goldfiish [28.3K]

<u>Answer:</u>

  Ball will move 92.8125 meter along the cliff in 7.5 seconds.

<u>Explanation:</u>

We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this case initial velocity = 0 m/s, acceleration = 3.3 m/s^2, we need to calculate displacement when time = 7.5 seconds.

Substituting

  s=0*7.5+\frac{1}{2} *3.3*7.5^2\\ \\ =92.8125 meter

  So ball will move 92.8125 meter along the cliff in 7.5 seconds.

5 0
3 years ago
The electric company charges $0.08/kilowatt-hour. If a 2 kilowatt appliance is run for 3hours, how much will you have to pay to
asambeis [7]

Answer:

$ 0.48

Explanation:

We can calculate this quantity easily using successive products and taking into account the units.

\frac{0.08}{kw*h}*2[kw]*3[hr]\\ \\=0.48

The amount is $ 0.48

3 0
3 years ago
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