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Paul [167]
3 years ago
9

Why is faulting most likely to occur near earth's surface and not deep within earth ?

Physics
1 answer:
melisa1 [442]3 years ago
7 0
Because of tectonics plates and the top aka surface of the earth is where the pressure is released and the reactions occur for example the roads splitting... this is just how i would answer it.
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Gravity pulls downward on a rock with a force of 800 N. If you pull upward on the rock with a force of 400 N, what is the total
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The answer is
a. 400 N downward
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This graph indicates that the kinetic energy of an object is increasing. What is the most reasonable explanation for this observ
soldi70 [24.7K]

B) The object's velocity doubled.

Explanation:

The graph is missing: find it in attachment.

The kinetic energy of an object is the energy possessed by the object due to its motion. It is calculated as

K=\frac{1}{2}mv^2

where

m is the mass of the object

v is the velocity of the object

We notice that:

  • The  kinetic energy is directly proportional to the mass
  • The kinetic energy is proportional to the square of the velocity

In the graph, one of the two quantities (either mass or speed) is represented on the x-axis, while the quantity on the y-axis is the kinetic energy.

First of all, we notice that the relationship is not linear: this means that the quantity on the x-axis cannot be the mass, so it must be the velocity.

Moreover, we notice that when the quantity on the x-axis increases from 1 to 2 (so, it doubles), the kinetic energy increases by a factor of 4. This means that the object's velocity has doubled, therefore

B) The object's velocity doubled.

Learn more about kinetic energy:

brainly.com/question/6536722

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3 years ago
When did agriculture first began?​
timofeeve [1]

Answer:

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Through what potential difference would you need to accelerate an alpha particle, starting from rest, so that it will just reach
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Answer:

\Delta V    = 1.8 \times 10^7 V

Explanation:

GIVEN

diameter = 15 fm  =15 \times 10^{-15}m

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K_{i}+U_{i} =K_{f}+U_{f}

there will be some initial kinetic  energy but after collision kinetic energy will zero

K_{i} + 0 = 0 + \frac{1}{4 \pi \epsilon _{0}} \frac{(2e)(92e)}{7.5 \times 10^{-15}}

on solving these equations we get kinetic energy initial

KE_{i} = 5.65\times 10 ^{-12} \times \frac  {1 eV}{1.6 \times 10^{-19}}

KE_{i} = 35.33 J ..............(i)

That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2  e

and gains kinetic energy K  =e∆V  ..........(ii)

 by accelerating through a potential difference ∆V

Thus the alpha particle will

just reach the {238}_U nucleus after being accelerated through a potential difference  ∆V

equating (i) and second equation we get

e∆V  = 35.33 Me V

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