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weqwewe [10]
3 years ago
5

A student drew the following model:

Physics
2 answers:
Gelneren [198K]3 years ago
8 0

Answer: C Tectonic Plates

Explanation:

Morgarella [4.7K]3 years ago
5 0

Answer:

C. Tectonic plates.

Explanation:

The series given in the question is mantle...convection...volcano...cooling...crust..motion and then tectonic plate is next.

here the process started in mantle where hot lava was moving and due to convection, this molten lava came on the the surface in the form of volcano and cooled down and rocky crust was formed and the moving crust is known as tectonic plates.

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prohojiy [21]

Answer:

bending of light

Explanation:

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3 years ago
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A subducting oceanic plate
beks73 [17]

Maybe this would help understand it better.

<span>Tectonic plates can transport both continental crust and oceanic crust, or they may be made of only one kind of crust. Oceanic crust is denser than continental crust. At a subduction zone, the oceanic crust usually sinks into the mantle beneath lighter continental crust</span>

4 0
3 years ago
If a quasar is moving away from us at v/c = 0.8, what is the measured redshift?
Delicious77 [7]

Answer:

The measured redshift is z =2

Explanation:

Since the object is traveling near light speed, since v/c = 0.8, then we have to use a redshift formula for relativistic speeds.

z= \sqrt{\cfrac{c+v}{c-v}}-1

Finding the redshift.

We can prepare the formula by dividing by lightspeed inside the square root to both numerator and denominator to get

z= \sqrt{\cfrac{1+\cfrac vc}{1-\cfrac vc}}-1

Replacing the given information

z= \sqrt{\cfrac{1+0.8}{1-0.8}}-1

z= \sqrt{\cfrac{1.8}{0.2}}-1\\z= \sqrt{9}}-1\\z=3-1\\z=2

Thus the measured redshift is z = 2.

4 0
3 years ago
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Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a
34kurt

Answer:

E=\frac{KQ}{2\sqrt 2a^2}

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}

Substitute x=a and R=a

Then, we get

E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}

E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}

E=\frac{KQa}{2\sqrt 2a^3}

E=\frac{KQ}{2\sqrt 2a^2}

Where K=9\times 10^9 Nm^2/C^2

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=\frac{KQ}{2\sqrt 2a^2}

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