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Lemur [1.5K]
3 years ago
15

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

Physics
1 answer:
Ira Lisetskai [31]3 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   
You might be interested in
Desperate for help! please!!!!!!!!
lys-0071 [83]

0.4029 \mathrm{m} / \mathrm{s}^{2} is the acceleration of the box.

<u>Explanation:</u>

Given data:

Mass of the box = 3.74 kg

Flat friction-less ground is pulled forward by a 4.20 N force at a 50.0 degree angle and pulled back by a 2.25 N force at a 122 degree angle.

First, we need to find the net horizontal force acting on the box. With the given data, the equation can be formed as below.  Net horizontal force acting on the box (F) is given by

F=\left(4.20 \times \cos 50^{\circ}\right)+\left(2.25 \times \cos 122^{\circ}\right)

F=(4.20 \times 0.64278)+(2.25 \times-0.5299)

F = 2.699676 – 1.192275 = 1.507 N

Next, find acceleration of the box using Newton's second law of motion. This states that the link between mass (m) of an objects and the force (F) required to accelerate it. The equation can be given as

F=m \times acceleration\ (a)

\text {acceleration }(a)=\frac{F}{m}=\frac{1.507}{3.74}=0.4029 \mathrm{m} / \mathrm{s}^{2}

8 0
3 years ago
You test a moon buggy on Earth. When the buggy hits a bump, it oscillates up and down on its springs with a period of 4 seconds.
Blizzard [7]

Answer:

Remains same

Explanation:

T = Time period of oscillation

m = mass

k = spring constant

Time period of oscillation is given as

T = 2\pi \sqrt{\frac{m}{k} }

we know that as we move from earth to moon, the value of spring constant "k"  and mass "m" remains unchanged because they do not depend on the acceleration due to gravity.

Time period depends on spring constant inversely and directly on the mass.

hence the time period remains the same.

3 0
3 years ago
100 + 50 + BRAINLIST PLEASE HELP!!!
irina1246 [14]

Answer:

Explanation:

1. What are the forces acting on the block when it is hanging freely from the spring scale? What is the net force on the block? What are the magnitudes of each of the forces acting on the block? Explain.

When a block is hanging freely, two forces are acting on it = tension force from the spring scale and gravity force on the block itself. The net force is zero as the block is not accelerating.  The magnitudes of tension and gravity force are the same but in opposite directions.

2. What are the forces that act on the block when it is placed on the ramp and is held in place by the spring scale? What is the net force acting on the block? Explain. (Assume that the ramps are frictionless surfaces.)

There are three forces acting on the block when it is placed on the ramp and is held in place by the spring scale: as in 1, there are tension and gravity but there is a third force - reaction force from the ramp surface on the block that is perpendicular to the surface. Again the block is not moving so the net force is zero.

3. What is the magnitude of normal force acting on the block when it is resting on the flat surface? How does the normal force change as the angle of the ramp increases? Explain. (Assume that the ramps are frictionless surfaces.)

On flat surface, the normal force is equal to the gravity force of the block i.e. its weight. On a vertical surface, the normal force is equal to zero. For the angle of ramp, θ, the normal force = weight * cos θ.

6 0
3 years ago
Read 2 more answers
An object with mass 100 kg moved in outer space. When it was at location &lt;8, -30, -4&gt; its speed was 5.5 m/s. A single cons
Alenkasestr [34]

Answer:

v = ( 6.41 i^ + 8.43 j^ + 2.63 k^ ) m/s

Explanation:

We can solve this problem using the kinematic relations, we have a three-dimensional movement, but we can work as three one-dimensional movements where the only parameter in common is time (a scalar).

X axis.

They indicate the initial position x = 8 m, its initial velocity v₀ = 5.5 m / s, the force Fx₁ = 220 N x₁ = 14 m, now the force changes to Fx₂ = 100 N up to the point xf = 17 m. The final speed is wondered.

As this movement is in three dimensions we must find the projection of the initial velocity in each axis, for this we can use trigonometry

the angle fi is with respect to the in z and the angle theta with respect to the x axis.

               sin φ = z / r

                Cos φ = r_p / r

               z = r sin φ

               r_p = r cos φ

the modulus of the vector r can be found with the Pythagorean theorem

               r² = (x-x₀) ² + (y-y₀) ² + (z-z₀) ²

               r² = (14-8) 2 + (-21 + 30) 2+ (-7 +4) 2

               r = √126

               r = 11.23 m

Let's find the angle with respect to the z axis (φfi)

                φ = sin⁻¹ z / r

                φ = sin⁻¹ ( \frac{-7+4}{11.23} )

                φ = 15.5º

Let's find the projection of the position vector (r_p)

                r_p = r cos φ

                r_p = 11.23 cos 15.5

                r_p = 10.82 m

This vector is in the xy plane, so we can use trigonometry to find the angle with respect to the x axis.

                 cos θ = x / r_p

                 θ = cos⁻¹ x / r_p

                 θ = cos⁻¹ ( \frac{14-8}{10.82})  

                 θ = 56.3º

taking the angles we can decompose the initial velocity.

               sin φ = v_z / v₀

               cos φ = v_p / v₀

               v_z = v₀ sin φ

               v_z = 5.5 sin 15.5 = 1.47 m / z

               v_p = vo cos φ

               v_p = 5.5 cos 15.5 = 5.30 m / s

                 

               cos θ = vₓ / v_p

                sin θ = v_y / v_p

                vₓ = v_p cos θ

                v_y = v_p sin θ

                vₓ = 5.30 cos 56.3 = 2.94 m / s

                v_y = 5.30 sin 56.3 = 4.41 m / s

 

                 

we already have the components of the initial velocity

                v₀ = (2.94 i ^ + 4.41 j ^ + 1.47 k ^) m / s

let's find the acceleration on this axis (ax1) using Newton's second law

                Fₓx = m aₓ₁

                aₓ₁ = Fₓ / m

                aₓ₁ = 220/100

                aₓ₁ = 2.20 m / s²

Let's look for the velocity at the end of this interval (vx1)

Let's be careful if the initial velocity and they relate it has the same sense it must be added, but if the velocity and acceleration have the opposite direction it must be subtracted.

                 vₓ₁² = v₀ₓ² + 2 aₓ₁ (x₁-x₀)

                 

let's calculate

                 vₓ₁² = 2.94² + 2 2.20 (14-8)

                 vₓ₁ = √35.04

                 vₓ₁ = 5.92 m / s

to the second interval

they relate it to xf

                   aₓ₂ = Fₓ₂ / m

                   aₓ₂ = 100/100

                   aₓ₂ = 1 m / s²

final speed

                    v_{xf}²  = vₓ₁² + 2 aₓ₂ (x_f- x₁)

                    v_{xf}² = 5.92² + 2 1 (17-14)

                    v_{xf} =√41.05

                    v_{xf} = 6.41 m / s

We carry out the same calculation for each of the other axes.

Axis y

acceleration (a_{y1})

                      a_{y1} = F_y / m

                      a_{y1} = 460/100

                      a_[y1} = 4.60 m / s²

the velocity at the end of the interval (v_{y1})

                      v_{y1}² = v_{oy}² + 2 a_{y1{ (y₁ -y₀)

                      v_{y1}2 = 4.41² + 2 4.60 (-21 + 30)

                      v_{y1} = √102.25

                       v_{y1} = 10.11 m / s

second interval

acceleration (a_{y2})

                      a_{y2} = F_{y2} / m

                      a_{y2} = 260/100

                      a_{y2} = 2.60 m / s2

final speed

                     v_{yf}² = v_{y1}² + 2 a_{y2} (y₂ -y₁)

                     v_{yf}² = 10.11² + 2 2.60 (-27 + 21)

                      v_{yf} = √ 71.01

                      v_{yf} = 8.43 m / s

here there is an inconsistency in the problem if the body is at y₁ = -27m and passes the position y_f = -21 m with the relationship it must be contrary to the velocity

z axis

 

first interval, relate (a_{z1})

                      a_{z1} = F_{z1} / m

                      a_{z1} = -200/100

                      a_{z1} = -2 m / s

the negative sign indicates that the acceleration is the negative direction of the z axis

the speed at the end of the interval

                    v_{z1}² = v_{zo)² + 2 a_{z1} (z₁-z₀)

                    v_{z1}² = 1.47² + 2 (-2) (-7 + 4)

                    v_{z1} = √14.16

                    v_{z1} = -3.76 m / s

second interval, acceleration (a_{z2})

                    a_{z2} = F_{z2} / m

                    a_{z2} = 210/100

                    a_{z2} = 2.10 m / s2

final speed

                    v_{fz}² = v_{z1}² - 2 a_{z2} | z_f-z₁)

                    v_{fz}² = 3.14² - 2 2.10 (-3 + 7)

                     v_{fz} = √6.94

                     v_{fz} = 2.63 m / s

speed is     v = ( 6.41 i^ + 8.43 j^ + 2.63 k^ ) m/s

5 0
2 years ago
When landing after a spectacular somersault, a 45.0 kg gymnast decelerates by pushing straight down on the mat. Calculate the fo
STALIN [3.7K]
So the force that the gymnast with a mass m=45 kg, has to exert against the ground to stop if her acceleration is a=8*g, where g=9.81 m/s², can be obtained from the Newtons second law: F=m*a, where F is the force, m is the mass and a is acceleration. 

F=m*a=m*8*g=45*8*9.81=3531.6 N. 

So the force that a gymnast has to exert on the mat in order to stop is F=3531.6 N.  
6 0
3 years ago
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