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3 years ago

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

1 answer:

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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What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8 cm from equilibrium and then

Sedbober [7]

Answer:

$x=(0.088m)\cos(\sqrt{\frac{k}{m} } t)$

Explanation:

We first identify the elements of this simple harmonic motion:

The amplitude A is 8.8cm, because it's the maximum distance the mass can go away from the equilibrium point. In meters, it is equivalent to 0.088m.

The angular frequency ω can be calculated with the formula:

$\omega =\sqrt{\frac{k}{m}}$

Where k is the spring constant and m is the mass of the particle.

Now, since the spring starts stretched at its maximum, the appropriate function to use is the positive cosine in the equation of simple harmonic motion:

$x=A\cos(\omega t)$

Finally, the equation of the motion of the system is:

$x=(0.088m)\cos(\omega t)$

$x=(0.088m)\cos(\sqrt{\frac{k}{m} } t)$

7 0

3 years ago

After the slab is charged, and the electrophorus is placed on the slab, the student briefly touches the electrophorus, effective

inna [77]

Answer:

By induction method

Explanation:

Induction method involves charging an electrically neutral body by bringing it in contact with an electrically charged body.

For the electrophus, a charge opposite that on the slab is induced on the side in contact with the slab; driving the opposite charge (this will be the same as that on the slab) to the other end of the elctrophus. Touching the electrophus removes the charge opposite the charge induced on the electrophus by the charged slab either by drawing up charge from the earth or taking the charge to earth (depends on the charge. A negative charge is drawn to earth while a positive charge draws up electrons from the earth)

5 0

3 years ago

(this is from my workbook)

inysia [295]

I’m pretty sure it’s D. variable

7 0

3 years ago

Read 2 more answers

12. A las 10 de la mañana Elena sale a 100 Km/h de una ciudad A con dirección a Madrid. A la misma hora sale Javier desde otra c

Klio2033 [76]

Answer:

a) t = 3.3 [h]

b ) Hora = 13:18 o 1:18 [pm]

c) x = 327.79 [km/h] (Elena)

x = 196.6 [km] (Javier)

Explanation:

Para poder solucionar este problema debemos hacer un planteamiento inicial de ubicacion de las ciudades, este planteamiento nos ayudara a entender el problema de una manera mas facil.

Tenemos las ciudades A & B y la ciudad de Madrid que esta a una distancia x con respecto de B, (ver esquema adjunto).

De manera logica debemos deducir que la Ciudad A debe estar mas lejos de Madrid que la ciudad B de la misma Madrid, ya que en caso contrario Javier nunca alcanzara a Elena, ya que Elena va mas rapido que Javier.

a) Ahora debemos de utilizar la siguiene ecuacion de la cinematica, cuando los cuerpos se mueven a velocidad constante.

$x=x_{o}+v*t$

Donde:

x -xo = Distancia entre el punto inicial y punto final.

v = velocidad [m/s]

t = tiempo [s]

Debemos convertir las velocidades de kilometros por hora a metros por segundo.

$100 [\frac{km}{h} ]*1[\frac{h}{3600s}]*1000[\frac{m}{1km} ] = 27.77[m/s]\\60[\frac{km}{h} ]*1[\frac{h}{3600s}]*1000[\frac{m}{1km} ] = 16.66[m/s]$

Seguidamente formulamos una ecuacion por cada movimiento, luego debemos igualar estas ecuaciones en funcion de la variable x que sera el punto donde se encuentren ambas personas.

Para Elena

$(132000+x) = 27.77*t\\x = 27.77*t - 132000$