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velikii [3]
3 years ago
15

Seafloor spreading causes an oceanic plate to collide with a continental plate. If it were to go through the whole tectonic cycl

e, number the steps that would follow.
Physics
2 answers:
Hitman42 [59]3 years ago
7 0
If you're going to ask a question, at least provide the steps or possible answers that you are provided with so that others can help you. Luckily, I know the exact question you are trying to solve. 
The order the steps go in is:
 <span> The oceanic plate would subduct under the continental plate into the asthenosphere. 
From the intense heat and pressure within the mantle, the crust of the oceanic plate would mix with the rock of the mantle. 
Mantle convection would cause magma to rise up through divergent boundaries.</span>
Snowcat [4.5K]3 years ago
4 0

Answer: The entire process is mentioned in  step-wise below-

  1. Firstly, two plates moves opposite to each other, due to this the crust gradually become thin and subsequently rises up.
  2. On continuous thinning of the crust, at one point, the magma will come out to the surface producing a small rift.
  3. Rifting continuously takes place and results in a rift valley,and slowly after millions of years will form an ocean.
  4. Then, the two diverging plate will collide with another oceanic or continental plate, in which the more denser oceanic crust will get subducted beneath the less denser one.
  5. This suducted plate, at much greater depth undergoes partial melting and slowly makes its way up to the crust.
  6. This magma on further pushing towards the surface, initiates sea floor spreading, thus repeating the entire cycle.
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A point charge of -4.28 pC is fixed on the y-axis, 2.79 mm from the origin. What is the electric field produced by this charge a
makkiz [27]

Answer:

E = (-3.61^i+1.02^j) N/C

magnitude E = 3.75N/C

Explanation:

In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:

\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}]              (1)

Where the minus sign means that the electric field point to the charge.

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q = -4.28 pC = -4.28*10^-12C

r: distance to the charge from the point P

The point P is at the point (0,9.83mm)

θ: angle between the electric field vector and the x-axis

The angle is calculated as follow:

\theta=tan^{-1}(\frac{2.79mm}{9.83mm})=74.15\°

The distance r is:

r=\sqrt{(2.79mm)^2+(9.83mm)^2}=10.21mm=10.21*10^{-3}m

You replace the values of all parameters in the equation (1):

\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}

The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C

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