All categories

3 years ago

What is likely happening at the boundary between the Nazca plate and the Pacific plate?

2 answers:

8 0

<span>The eastern margin is a convergent boundary subduction zone under the South American Plate and the Andes Mountains, forming the Peru–Chile Trench. The southern side is a divergent boundary with the Antarctic Plate, the Chile Rise, where seafloor spreading permits magma to rise.</span>

Alika [10]3 years ago

8 0

Explanation:

The south-eastern side of the Pacific plate is a divergent boundary (the plates are pulled, and not pushed, apart) with the Nazca Plate forming the East Pacific Rise.

The East Pacific Rise is part of the circumglobal system of active volcanic ridges, all of which define the position of diverging plates where new crust is being created. These ridges represent the locus of global shallow-water earthquakes.

I hope you find this information useful and interesting! Good luck!

You might be interested in

Paola can flex her legs from a bent position through a distance of 20.1 cm. Paola leaves the ground when her legs are straight,

makkiz [27]

The third equation of free fall can be applied to determine the acceleration. So that Paola's acceleration during the flight is 39.80 m/ $s^{2}$ .

Acceleration is a quantity that has a direct relationship with velocity and also inversely proportional to the time taken. It is a vector quantity.

To determine Paola's acceleration, the third equation of free fall is appropriate.

i.e $V^{2}$ = $U^{2}$ ± 2as

where: V is the final velocity, U is the initial velocity, a is the acceleration, and s is the distance covered.

From the given question, s = 20.1 cm (0.201 m), U = 4.0 m/s, V = 0.

So that since Poala flies against gravity, then we have:

$V^{2}$ = $U^{2}$ - 2as

0 = $(4)^{2}$ - 2(a x 0.201)

= 16 - 0.402a

0.402a = 16

a = $\frac{16}{0.402}$

= 39.801

a = 39.80 m/ $s^{2}$

Therefore Paola's acceleration is 39.80 m/ $s^{2}$ .

Visit: brainly.com/question/17493533

7 0

2 years ago

Read 2 more answers

A diver makes 1.0 revolutions on the way from a 9.2-m-high platform to the water. Assuming zero initial vertical velocity, find

____ [38]

Answer:

Average angular velocity ≈ 4.59 rad/s

Explanation:

Using the equation of motion,

H = ut + (1/2)t² ............................ equation 1.

Where H= height, u = initial velocity(m/s), g = acceleration due to gravity(m/s²), t = time(s) u= 0 ∴ ut =0

H =(1/2)gt².................................... equation 2.

making t² the subject of the relation in equation 2,

∴ t² = 2H/g

Where H = 9.2 m, g= 9.8 m/s

∴ t² = ( 2×9.2)/9.8

t = √(2 × 9.2/9.8) = √(18.4/9.8)

t = 1.37 s.

The average angular velocity = θ/t

Where θ = is the number of revolution that the diver makes, t = time

θ = 1 rev.

Since 1 rev = 2π (rad)

t = 1.37 s

Average angular velocity = 2π/t

π = 3.143

Average angular velocity = (2×3.143)/1.37 = 6.286/1.37

Average angular velocity ≈ 4.59 rad/s

8 0

3 years ago

Which is an example of nuclear energy being converted into heat and light energy?

likoan [24]

The answer to the question is A

6 0

2 years ago

Read 2 more answers

A train at rest emits a sound at a frequency of 1057 Hz. An observer in a car travels away from the sound source at a speed of 2

il63 [147K]

Answer:

993.52 Hz

Explanation:

The frequency of sound emitted by the stationery train is 1057 Hz.

The car travels away from the train at 20.6 m/s.

The frequency the observer hears is given by the formula:

$f_o = \frac{v - v_o}{v}f$

where v = velocity of sound = 343 m/s

vo = velocity of observer

f = frequency from source

This phenomenon is known as Doppler's effect.

Therefore:

$f_o = \frac{343 - 20.6}{343} * 1057\\ \\f_o = 322.4 / 343 * 1057\\\\f_o = 993.52 Hz$

The frequency heard by the observer is 993.52 Hz.

4 0

3 years ago

How deep can an object with 6360N hitting on a sponge get ???

borishaifa [10]

Answer:

The sponge must go $\dfrac{636}{m}\ \text{meter}$ deep

Explanation:

If F = 6360 N, then it is required to find how deep can an object with this force hitting on a sponge get.

We know that, F = mgh

m is mass

g is acceleration due to gravity

$h=\dfrac{F}{mg}\\\\h=\dfrac{6360}{10m}\\\\h=\dfrac{636}{m}\ \text{meter}$

So, the sponge must go $\dfrac{636}{m}\ \text{meter}$ deep.

5 0

3 years ago