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NemiM [27]
3 years ago
7

A seafloor plate slides under a continental plate during subduction. which type of plate boundary best describes this situation

Physics
1 answer:
Leto [7]3 years ago
5 0

Answer:

The situation where a seafloor plate slides under a continental plate during subduction is best described by convergent plate boundary.

Explanation:

The earth’s crust is broken down into tectonic plates that can move independently. They can interact in three different ways: converge (move toward one another), diverge (move away from one another) or transform (slide past one another). The three kinds of plate margins (boundaries where plates meet) are oceanic-oceanic, continental-continental, and continental-oceanic.

The regions where the plates are moving towards one another are known as convergent plate boundaries. During the convergence of continental and oceanic plates, the more-dense oceanic plate sinks below the less-dense continental plate and the oceanic plate is forced down further into the mantle. This is known as subduction. When the plate enters the mantle, the inside pressure breaks the rock. The broken rocks begins to melt from the heat due to the friction and as a result magma is formed. This magma rises toward the surface by breaking through the crust and forms a chain of volcanoes known as a volcanic arc such as the Cascade Mountains of North America and the Andes Mountains of South America.

During the convergence of two oceanic plates, one of the plates sinks underneath the other and forms an ocean trench (deep depression). The plate that sinks further down into the mantle starts to melt and as a result magma rises toward the surface and forms a chain of volcanic islands behind the ocean trench.

During the convergence of two continental plates, they buckle and compress to form complex mountains ranges of great height such as the Himalayas.

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A 62-kg person jumps from a window to a fire net 20.0 m directly below, which stretches the net 1.4 m. Assume that the net behav
gayaneshka [121]

Answer:

a) x = 0.098

b) x = 2.72 m

Explanation:

(a) To find the stretch of the fire net when the same person is lying in it, you can assume that the net is like a spring with constant spring k. It is necessary to find k.

When the person is falling down he acquires a kinetic energy K, this energy is equal to the elastic potential energy of the net when it is max stretched.

Then, you have:

K=U\\\\\frac{1}{2}mv^2=\frac{1}{2}kx^2        (1)

m: mass of the person = 62kg

k: spring constant = ?

v: velocity of the person just when he touches the fire net = ?

x: elongation of the fire net = 1.4 m

Before the calculation of the spring constant, you calculate the final velocity of the person by using the following formula:

v^2=v_o^2+2gy

vo: initial velocity = 0 m/s

g: gravitational acceleration = 9.8 m/s^2

y: height from the person jumps = 20.0m

v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(20.0m)}=14\frac{m}{s}

With this value you can find the spring constant k from the equation (1):

mv^2=kx^2\\\\k=\frac{mv^2}{x^2}=\frac{(62kg)(14m/s)^2}{(1.4m)^2}=6200\frac{N}{m}

When the person is lying on the fire net the weight of the person is equal to the elastic force of the fire net:

W=F_e\\\\mg=kx

you solve the last expression for x:

x=\frac{mg}{k}=\frac{(62kg)(9.8m/s^2)}{6200N/m}=0.098m

When the person is lying on the fire net the elongation of the fire net is 0.098m

b) To find how much would the net stretch, If the person jumps from 38 m, you first calculate the final velocity of the person again:

v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(38m)}=27.29\frac{m}{s}

Next, you calculate x from the equation (1):

x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(62kg)(27.29m/s)^2}{6200N/m}}\\\\x=2.72m

The net fire is stretched 2.72 m

5 0
3 years ago
A string exerts a force of 20 N on a box at an angle of 38° from the horizontal. What is the horizontal
Gnoma [55]

Answer:

15.76N

Explanation:

horizontal component =Fcos¢ = 20cos38 = 15.76N

4 0
2 years ago
Read 2 more answers
reviews the approach taken in problems such as this one. A bird watcher meanders through the woods, walking 0.916 km due east, 0
Verizon [17]

Answer:

Displacement: 2.230 km    Average velocity: 1.274\frac{km}{h}

Explanation:

Let's represent displacement by the letter S and the displacement in direction 49.7° as A. Displaement is a vector, so we need to decompose all the bird's displacement into their X-Y compoments. Let's go one by one:

  • 0.916 km due east is an horizontal direction and cane be seen as  direction towards the negative side of X-axis.
  • 0.928 km due south is a vertical direction and can be seen as a direction towards the negative side of Y-axis.
  • 3.52 km in a direction of 49.7° has components on X and Y  axes. It is necessary to break it down using trigonometry,

First of all. We need to sum all the X components and all the Y componets.

∑Sx = Ax -0.916 ⇒  ∑Sx = [tex]3.52cos(49.7) - 0.916

∑Sx = 1.361 km

∑Sy = Ay - 0.918 ⇒ ∑Sy = 3.52sin(49.7) - 0.918

∑Sy = 1.767

The total displacement is calculated using Pythagoeran therorem:

S_{total} =\sqrt{Sx^{2}+ Sy^{2} } ⇒

S_{total} = 2.230 km

With displacement calculated, we can find the average speed as follows:

V = S/t  ⇒  V = \frac{2.230}{1.750}

V = 1.274\frac{km}{h}

7 0
3 years ago
Part A.)Six boxes held at rest against identical walls.
krok68 [10]

Answer:

Explanation:

When a body is held against a vertical wall , to keep them in balanced position , normal force is applied on their surface . this force creates normal reaction which acts against the normal force and it is equal to the normal force as per newton's third law . Ultimately friction force is created which is proportional to normal force and it acts in vertically upward direction . It prevents the body from falling down .

Hence normal force = reaction force .

From second law also net force is zero , so if  normal force is N and reaction force is R

R - N = mass x acceleration = mass x 0 = 0

R = N .

Ranking normal force from  highest  to smallest

150 N , 130 N , 120 N

B )

Frictional force is equal to the weight of the body because the body is held at rest .

Ranking of frictional force form largest to smallest

7 kg , 5 kg , 3 kg , 1 kg .

Here frictional force is irrespective of the normal force acting on the body  because frictional force adjusts itself so that it becomes equal to weight in all cases here because it always balances the weight of the body .

6 0
3 years ago
HELP PLEASE 
-BARSIC- [3]
Angel !  You have a formula, and you have an example that's
completely worked out.  The ONLY POSSIBLE reason that you
could still need help is that you're letting math scare you.

I'll do 'A' for you, 'B' most of the way, and get 'C' set up.
If THAT's not enough for you to run with and finish them all,
then you and I should both be embarrassed.

Write the formula on the wall:

                                   °F  =  (9/5) °C  +  32°

A).  Convert 35° C        °F  =  (9/5)(35°) + 32°

(9/5)(35) = 63              °F  =      63°      + 32°

                                     °F  =  95°
____________________________________

B).  Convert 80°F to °C 
                         The formula:    °F  =  (9/5) °C  +  32°

°F = 80                                     80  =  (9/5)°C  +  32

Subtract 32 from each side:      48 = (9/5)°C

Multiply each side by  5 :        240  =  (9) C

Now you take over:
_________________________________________

C).  Convert 15°C to °F.
                         The formula:    °F  =  (9/5) °C  +  32°

°C = 15                                      °F = (9/5) 15° + 32

(9/5) (15) = 27     
                             Go !              °F  = 

7 0
4 years ago
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