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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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The horizontal beam in Fig. E11.14 weighs 190 N, and its center of gravity is at its center. Find (a) the tension in the cable a
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Answer:

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Explanation:

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According to figure,

The angle is

\sin\theta=\dfrac{3}{5}

The horizontal component is

T_{x}=T\cos\theta

The vertical component is

T_{y}=T\sin\theta

(a). We need to calculate the tension in the cable

Using formula of net torque acting on the pivot

\sum\tau=F_{b}\times r+F_{w}\times r'-T\sin \theta\times r'

Put the value into the formula

0=190\times2+300\times 4-T\sin\theta\times 4

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(b). We need to calculate the horizontal components of the force exerted on the beam at the wall

Using formula of horizontal component

F_{x}=T\cos\theta

Put the value into the formula

F_{x}=658.33\times\dfrac{4}{5}

F_{x}=526.66\ N

(c). We need to calculate the vertical components of the force exerted on the beam at the wall

Using formula of vertical component

F_{y}=F_{b}+F_{w}-T\sin\theta

Put the value into the formula

F_{y}=190+300-658.33\times\dfrac{3}{5}

F_{y}=95.002\ N

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(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

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