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3 years ago

How do the three types of boundary's work together to keep the Earth at equilibrium?

Physics

2 answers:

Maurinko [17]3 years ago

8 0

There are three types: divergent, convergent, and transform boundaries. I hope this helps.

icang [17]3 years ago

4 0

Answer:

Divergent , convergent and transform boundaries.

Explanation:

The three types of the boundaries that are formed on the earth are that of the divergent, convergent and the transform boundary that forms over the earth's surface due to the presence of the tectonic plate. Thus all these boundaries work in a close cyclic manner as the formation of the newer plate is balanced by the depletion of the old ones.

The rocks that are formed in the divergent boundaries of the ocean are carried away from there a source to the shoes and the plates that collide with each other undergo subduction and hence they are again regenerated n the geosphere.

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Water is pumped steadily out of a flooded basement at a speed of 5.4 m/s through a uniform hose of radius 0.83 cm. The hose pass

Gala2k [10]

To solve this problem it is necessary to apply the concepts related to the flow as a function of the volume in a certain time, as well as the potential and kinetic energy that act on the pump and the fluid.

The work done would be defined as

$\Delta W = \Delta PE + \Delta KE$

Where,

PE = Potential Energy

KE = Kinetic Energy

$\Delta W = (\Delta m)gh+\frac{1}{2}(\Delta m)v^2$

Where,

m = Mass

g = Gravitational energy

h = Height

v = Velocity

Considering power as the change of energy as a function of time we will then have to

$P = \frac{\Delta W}{\Delta t}$

$P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)$

The rate of mass flow is,

$\frac{\Delta m}{\Delta t} = \rho_w Av$

Where,

$\rho_w$ = Density of water

A = Area of the hose $\rightarrow A=\pi r^2$

The given radius is 0.83cm or $0.83 * 10^{-2}$ m, so the Area would be

$A = \pi (0.83*10^{-2})^2$

$A = 0.0002164m^2$

We have then that,

$\frac{\Delta m}{\Delta t} = \rho_w Av$

$\frac{\Delta m}{\Delta t} = (1000)(0.0002164)(5.4)$

$\frac{\Delta m}{\Delta t} = 1.16856kg/s$

Final the power of the pump would be,

$P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)$

$P = (1.16856)((9.8)(3.5)+\frac{1}{2}5.4^2)$

$P = 57.1192W$

Therefore the power of the pump is 57.11W

6 0

3 years ago

What are five types of organisms That can reproduce asexually

Sindrei [870]

Answer:

Bacteria, Hydra, Copperheads, Blackworms, and Strawberries

Explanation:

7 0

3 years ago

An airplane starts from rest and accelerates at a constant rate of 3.00 m/s2 for 30.0 s before leaving the ground. a. How far di

rosijanka [135]

The formula we can use in this case is:

d = v0t + 0.5 at^2

v = at + v0

where,

d = distance travelled

v0 = initial velocity = 0 since at rest

t = time travelled

a = acceleration

v = final velocity when it took off

a. d = 0 + 0.5 * 3 * 30^2

d = 1350 m

b. v = 3 * 30 + 0

v = 90 m/s

8 0

3 years ago

What are 3 ways an object can be charged

lorasvet [3.4K]

pile
battery
power sector

8 0

3 years ago

Why would a rocket fall vertically downwards if it was launched vertically upwards

IgorC [24]

Answer:

It would because the shape of the rocket is designed to be able to slice through the air as smooth as possible and now you may be thinking that air is already smooth but when you try to push something as large and heavy like a rocket then the shape of the rocket will be very important. The bottom of the rocket is flatter then the top so it is not designed to fly smoothly through the air. So the rocket would fall vertically downward(If it was still in one piece)because of it's shape. It is easier for the top of the rocket to go smoothly through the air then the bottom.

Explanation:

I am 90% sure this is correct but if I'm not please tell me

3 0

2 years ago