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3 years ago

WILL GIVE BRAINLIEST!!!

Physics

2 answers:

sp2606 [1]3 years ago

5 0

Answer:

When two plates collides of varying densities, one tends to subduct introducing volcanic material to the earth surface.

Explanation:

The ring of fire is a geographical area where seismic and volcanic activities are prevalent. This region extend around the edge of the pacific plates and including smaller plates like the Juan de Faca plate, coccus plate and the Philippine plate . The ring of fire is a place known for earthquake and volcanism due to the active plate boundary.

The plate boundaries around the rings of fire are known for subduction activities.

Usually, in a convergence boundary where continental crust collides with oceanic crust the more dense oceanic crust tends to subduct. The continental crust is less dense and possess more buoyancy. The less dense and buoyant nature of the continental crust will cause it to override the oceanic crust.

Generally, when oceanic crust and continental crust collides the oceanic crust subduct reaching the aesthenosphere area, Aesthenosphere material like magma are introduced to the earth surface through the subduction zone. The magma may creep through cracks and fissures or erupt to the earth surface. The subduction zone form trenches which is a prerequisite for creating mountains, volcanisms and earthquakes.

marshall27 [118]3 years ago

4 0

Hello!

Subduction volcanism occurs where two plates are converging on one another. One plate containing oceanic lithospehe descends beneath the adjacent plate, consuming the oceanic lithosphere into the earth's mantle.

I really hope this helped you! :)

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The following currents are measured in the same direction in a three-branch parallel circuit: 250 mA, 300 mA, and 800 mA. What i

Ksju [112]

The current in the junction is 1350 mA or 1.350 A.

Explanation:

As per Kirchoff's first law, the algebraic sum of current meeting at any junction should be equal to the algebraic sum of current leaving the junction. As in the present case, three parallel branch circuit is given the current in 250 mA, 300 mA and 800 mA, respectively, the sum of these three current will be equal to the current in the junction.

So,

I₁+I₂+I₃ = I₄

So I₁,I₂ and I₃ are the current passed in the three parallel branches and I₄ is the current in the junction.

250 + 300 + 800 = 1350 mA

So the current in the junction is 1350 mA or 1.350 A.

5 0

3 years ago

An object with charge q = −6.00×10−9 C is placed in a region of uniform electric field and is released from rest at point A. Aft

sergeinik [125]

Answer:

a) 80 V

b) The magnitude of the electric field is 100 N/C and the direction of the electric field is from point B to point A where the electric field is toward the negative charge

Explanation:

Given :

We are given an object with charge q = -6.00 x I0^-9 C starts moving from the rest at point A, which means its kinetic energy at point A is zero ( $K_{A}$ = 0) to the point B at distance l = 0.500m where its kinetic energy is ( $K_{B}$ = 5.00 x 10^-7J) . Also, the electric potential of q at point A is VA = + 30.0 v.

Required :

(a) We are asked to find the electric potential VB

(b) We want to determine the magnitude and the direction of the electric field E.

Solution

(a) We are given the values for VA, $K_{B}$ and q so we want to find a relationship between these three parameters and VB to get the value of VB.

As we have two states, at points A and B , where the charge moved from A to B due to the applied electric field. The mechanical energy of the object is conservative during this travel, and we can apply eq(1) in this situation:

$K_{A} +U_{A} =K_{B} +U_{B}$ .........................................(1)

Where $K_{A}$ = 0 and the potential energy U of the charge is given by U = q V

where V is the electric potential. So, equation (1) will be in the form :

0+qVA= $K_{B}$ +qVB (Divide by q)

VA= $K_{B}$ /q + VB (solve for VB)

VB=VA- $K_{B}$ /q .......................................(2)

We get the relation between VB, VA and $K_{B}$ , now we can plug our values for VA, $K_{B}$ and q into equation (2) to get VB

VB=VA- $K_{B}$ /q

=30V-(5.00 x 10^-7J)/(-6.00 x I0^-9)

=80 V

(b) After we calculated VB we can use equation a to get the electric field E that applied to the charge q, where the potential difference between the two points equals the integration of the electric field multiplied by the distance l between the two points

VA-VB = $\int\limits^1_0 {E} \, dl$ ...................................(a)

$=E\int\limits^1_0 {} \, dl$

VA-VB=E $l$ (solve for E)

E= VA-VB/ $l$ ..................................(3)

Now let us plug our values for VA, Vs and l into equation (3) to get the electric field E

E= VA-VB/ $l$

=-100 N/C

The magnitude of the electric field is 100 N/C and the direction of the electric field is from point B to point A where the electric field is toward the negative charge

5 0

3 years ago

How does the height of a ramp affect the distance that a toy car will travel?

lyudmila [28]

Answer:

the higher the ramp the less distance it will travel

8 0

3 years ago

A perfect bouncy ball would continue bouncing the same height forever. But in reality, the ball stops bouncing because energy is

galina1969 [7]

Answer:

D. all answers are true

6 0

3 years ago

2 Which is true of a parallel circuit?

mars1129 [50]

Answer:

fxb

Explanation:bfffffffff

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3 years ago