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Sveta_85 [38]
3 years ago
15

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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If a cup of coffee has temperature 95∘C95∘C in a room where the temperature is 20∘C,20∘C, then, according to Newton's Law of Coo
lina2011 [118]

Answer:

T = 76.39°C

Explanation:

given,

coffee cup temperature = 95°C

Room temperature= 20°C

expression

T( t ) = 20 + 75 e^{\dfrac{-t}{50}}

temperature at t = 0

T( 0 ) = 20 + 75 e^{\dfrac{-0}{50}}

T(0) = 95°C

temperature after half hour of cooling

T( t ) = 20 + 75 e^{\dfrac{-t}{50}}

t = 30 minutes

T( 30 ) = 20 + 75 e^{\dfrac{-30}{50}}

T( 30 ) = 20 + 75 \times 0.5488

T(30) = 61.16° C

average of first half hour will be equal to

T = \dfrac{1}{30-0}\int_0^30(20 + 75 e^{\dfrac{-t}{50}})\ dt

T = \dfrac{1}{30}[(20t - \dfrac{75 e^{\dfrac{-t}{50}}}{\dfrac{1}{50}})]_0^30

T = \dfrac{1}{30}[(20t - 3750e^{\dfrac{-t}{50}}]_0^30

T = \dfrac{1}{30}[(20\times 30 - 3750 e^{\dfrac{-30}{50}} + 3750]

T = \dfrac{1}{30}[600 - 2058.04 + 3750]

T = 76.39°C

4 0
3 years ago
A box is pushed 40 m by a mover. The amount of work done was 2,240 j. How much force was exerted on the box
Georgia [21]

The force exerted on the box is 56 N

Explanation:

The work done by a force on an object is given by

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement of the object

\theta is the angle between the direction of the force and of the displacement

For the box in this problem, we have:

W = 2240 J is the work done

d = 40 m is the displacement of the box

Assuming that the  force is parallel to the displacement, \theta=0

Solving the equation for F, we find the force exerted on the box:

F=\frac{W}{d cos \theta}=\frac{2240}{(40)(cos 0)}=56 N

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

3 0
3 years ago
A fox runs for 12 seconds at a speed of 9.65 m/s. How much distance does it cover?
serg [7]

Answer:

115.8

Explanation:

6 0
3 years ago
Read 2 more answers
The odometer of a car changes from 1048 km to 1096 km in 40
Makovka662 [10]

Answer:

20m/s

Explanation:

it covers 20 metres in a second

3 0
2 years ago
s A horizontal insulating rod of length 11.0-cm and charge 19 nC is in a plane with a long straight vertical uniform line charge
guajiro [1.7K]

Answer:

F = 2.26 ×  10⁻³ N

Explanation:

given,

length of rod = 11 cm

charge  = 19 nC

linear charge density = 3.9 x 10⁻⁷ C/m

electric force at 2 cm away.

E(r) = \dfrac{2K\lambda}{r}

F = E q

F= \dfrac{2K\lambda\ q}{L}\int \dfrac{dr}{r}

integrating from 0.02 to 0.02 + L

F= \dfrac{2K\lambda\ q}{L}[ln(0.02+L)-ln(0.002)]

F= \dfrac{2\times 9 \times 10^9\times 3.9\times 10^{-7}\times 19 \times 10^{-9}}{0.11}[ln(0.02+0.11)-ln(0.002)]

F = 2.26 ×  10⁻³ N

5 0
4 years ago
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