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Lemur [1.5K]
4 years ago
15

What is the rock cycle and how does it change the lithosphere?

Physics
1 answer:
Ira Lisetskai [31]4 years ago
8 0
The rock cycle is a basic concept in geology that describes the time-consuming transitions through geologic time among the three main rock types: sedimentary, metamorphic, and igneous. As the adjacent diagram illustrates, each of the types of rocks is altered or destroyed when it is forced out of its equilibrium conditions. An igneous rock such as basalt may break down and dissolve when exposed to the atmosphere, or melt as it is subducted under a continent. Due to the driving forces of the rock cycle, plate tectonics and the water cycle, rocks do not remain in equilibrium and are forced to change as they encounter new environments. The rock cycle is an illustration that explains how the three rock types are related to each other, and how processes change from one type to another over time. This cyclical aspect makes rock change a geologic cycle and, on planets containing life, a biogeochemical cycle.

Plate movements drive the rock cycle by pushing rocks back into the mantle, where they melt and become magna again. Plate movements also cause the folding, faulting and uplift of the crust that move rocks through the rock cycle.

sources: wikapedia, Harmonybaddie on brainly   
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A lone neutron spontaneously decays into a proton plus an _______
frosja888 [35]

Answer:

A lone neutron spontaneously decays into a proton plus an electron.

Explanation:

In an atom, nuclei contain protons and neutrons, which are the fundamental particles of an atom. Neutrons are stable and uncharged particles inside a nucleus.

For 15 times during its lifetime, a free neutron decays and breaks down into more smaller particles.This breakdown causes problems in nuclear reactors, as they start decaying and emit radiations of different wavelengths.

A neutron undergoes the decaying process to produce an electron, a proton, and energy.

The reaction of neutron decay:

  n0 → p+ + e− + νe

5 0
3 years ago
When sitting in the tree the cat has a total of 1375J in its Gravitational potential store. What is the maximum amount of energy
Murrr4er [49]

Answer:

1375J

Explanation:

The gravitational potential/potential energy of the at the top of the tree which is the energy by virtue of its position.

P.E = mgh

mass = m

Acceleration due to gravity = g

height = h

At the top of the tree, the value of h (height) is high resulting in the gravitational potential. When the cat lands on the ground, the value of h is zero, the the gravitational potential would be zero and all the potential energy have been converted to other forms of energy.

Therefore, the total gravitational potential store is equal to the maximum amount of energy that can be transferred which is equal to 1375J.

4 0
3 years ago
Does it take 18 months for twins to be born, and 27 months for triplets, and so on?
Nezavi [6.7K]
No it all takes 9 or 10 months
3 0
3 years ago
A 12.0-cm long cylindrical rod has a uniform cross-sectional area A = 5.00 cm2. However, its density increases linearly from 2.6
andriy [413]

Answer:

(a) The constants required describing the rod's density are B=2.6 and C=1.325.

(b) The mass of the road can be found using A\int_0^{12}\left(B+Cx)dx

Explanation:

(a) Since the density variation is linear and the coordinate x begins at the low-density end of the rod, we have a density given by

2.6\frac{g}{cm^3}+\frac{18.5\frac{g}{cm^3}-2.6\frac{g}{cm^3}}{12 cm}x = 2.6\frac{g}{cm^3}+1.325x\frac{g}{cm^2}

recalling that the coordinate x is measured in centimeters.

(b) The mass of the rod can be found by having into account the density, which is x-dependent, and the volume differential for the rod:

m=\int\rho dv=\int\left(B+Cx\right)Adx=5\int_0^{12}\left(2.6+1.325x\right)dx=126.6,

hence, the mass of the rod is 126.6 g.

7 0
3 years ago
A planet of mass m 6.75 x 1024 kg is orbiting in a circular path a star of mass M 2.75 x 1029 kg. The radius of the orbit is R 8
Umnica [9.8K]

Answer:

The orbital period of the planet is 387.62 days.

Explanation:

Given that,

Mass of planetm =6.75\times10^{24}\ kg

Mass of star m'=2.75\times10^{29}\ kg

Radius of the orbitr =8.05\times10^{7}\ km

Using centripetal and gravitational force

The centripetal force is given by

F = \dfrac{mv^2}{r}

F=m\omega^2r

We know that,

\omega=\dfrac{2\pi}{T}

F=m(\dfrac{2\pi}{T})^2r....(I)

The gravitational force is given by

F = \dfrac{mm'G}{r^2}....(II)

From equation (I) and (II)

m(\dfrac{2\pi}{T})^2r=\dfrac{mm'G}{r^2}

Where, m = mass of planet

m' = mass of star

G = gravitational constant

r = radius of the orbit

T = time period

Put the value into the formula

T^2=\dfrac{4\pi^2R^3}{m'G}

T^2=\dfrac{4\times(3.14)^2\times(8.05\times10^{7})^3}{2.75\times10^{29}\times6.67\times10^{-11}}

T=2\times3.14\times\sqrt{\dfrac{(8.05\times10^{10})^3}{2.75\times10^{29}\times6.67\times10^{-11}}}

T =3.34\times10^{7}\ s

T= 387.62\days

Hence, The orbital period of the planet is 387.62 days.

4 0
3 years ago
Read 2 more answers
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