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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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Estimate the magnitude of the electric field due to the proton in a hydrogen atom at a distance of
Lana71 [14]

Electric field due to a point charge is given as

E = \frac{kq}{r^2}

here we know that

q = 1.6 \times 10^{-19} C

also the distance is given as

r = 5.29 \times 10^{-11} m

now we will have

E = \frac{(9\times 10^9)(1.6 \times 10^{-19})}{(5.29 \times 10^{-11})^2}

so we will have

E = 5.14 \times 10^{11} N/C

so above is the electric field due to proton

5 0
3 years ago
| A 1.0 kg stone is dropped from a bridge 100 m above a canyon. What will be the kinetic energy of the stone after it
Mnenie [13.5K]

Answer:

Option D

490 J

Explanation:

When at a height of 100 am above and released, the ball initially posses only potential energy. When it falls, some potential energy is converted to kinetic energy.

Initial potential energy= mgh where m is the mass, g is the acceleration due to gravity and h is height. Substituting 1 Kg for m, 9.81 for g and 100 m for h then

PE initial = 1*9.81*100= 981 J

At 50 m, PE will be 1*9.81*50=490.5 J

Subtracting PE at 50 m from initial PE we get the energy that has been converted to kinetic energy hence

981-490.5= 490.5 J

Approximately, 490 J

8 0
3 years ago
A 0.43 kg hammer is moving horizontally at 5.2 m/s when it strikes a nail and comes to rest after driving the nail 0.014 m into
vagabundo [1.1K]

Answer:

The duration of the impact is 0.005384 seconds

Explanation:

Given

m = 0.43 kg

v = 5.2 m/s

x = 0.014 m

Knowing the formulas

v^{2}_{f} = v^{2}_{i} + 2ax\\0 = 5.2^{2}  + 2a*0.014\\a = - 965.71 m/s^{2} \\\\vf = vi + at\\0 = 5.2 + (965.71)t\\t = 0.005384 s

5 0
3 years ago
Read 2 more answers
The amount of kinetic energy and object has depends on its ?
Lady_Fox [76]
It depends on Mass and velocity
4 0
2 years ago
Can someone pleaseEeeeeeee ASAP answer this❗️❗️❗️❗️ I really need help it is my second time posting this❗️❗️❗️❗️
belka [17]

Answer:

Data:-vi=om/s (b/c as in question penny is dropped from building means before coming to ground its initial state or velocity was considered as zero ) now distance or height h=380m and now we have to find the final velocity vf=? and the time t=?

Explanation:

So applying second eq of motion s=vit+1/2×gt² (here we have taken a gravity b/c when ever body is in vertical position then acceleration due to gravity is applied ) s=0×t+1/2×gt² , s=0+1/2×9.8×t² ,380=4.9t² we have to find t so 4.9t²=380 , t²=380÷4.9 , t²=77.55 now sq root on b/s

\sqrt{t }  =  \sqrt{77.55}

so t=8.806s and now apply 1st eq o²f motion to find out vf so vf=vi+gt , vf=0+9.8×8.806 ,vf=86.298 and if you want to verify that either this is answer is correct or not so put the value of t in second eq of motion and if you got distance same as give in the question so your value of t is considered as correct likewise s=vit+1/2gt² , s=0+1/2×9.8(8.806)²,s=4.9×77.55 ,s=380m (proved) I hope it would be helpfull

3 0
3 years ago
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