All categories

2 years ago

Describe the sequence of events in the lithification of a sandstone

2 answers:

6 0

Lithification is a diagenetic process in which loose sediment is converted to hard rocky compaction and cementation. So sediments (sand) are buried the increase in pressure from the weight of the overlaying material pushes the grains closer together. The volume of sediment is reduced and the fluids between the grains are also squeezed out. This leaves the sandstone tightly compressed together this is lithification.

Strike441 [17]2 years ago

5 0

<h3>Answer and explanation;</h3>

Lithification refers to the process that involves compacting, sedimentation and cementing of the mass together under pressure to create a sand stone, a form of sedimentary rock.
Weathering and erosion of rocks creates sediments which collect and are then moved to another location.
At some point the sediment may become heavy for the agent such as water to carry and thus deposited. As this deposit thickens, it becomes heavier and thus lithification process start.
Compaction forces the grains of the deposited sediments closer together, forcing out air and water.
As a result of compaction the sediment's pores becomes smaller. Minerals deposit in these pores as the water squeezes out. As the water moves out the pressure increases, the mineral deposits start hardening thus cementing the entire mass and this results to a sand stone.

You might be interested in

The movie "The Gods Must Be Crazy" begins with a pilot dropping a bottle out of an airplane. A surprised native below, who think

Sergio039 [100]

Answer:

⇔⇔⇔↑∑∑∩∅¬⊕║⊇↔∴∉∵

Explanation:

8 0

2 years ago

Listed following are three possible models for the long-term expansion (and possible contraction) of the universe in the absence

Lynna [10]

Answer:

1) Recollapsing universe

2) critical universe

3) Coasting universe

Explanation:

According to the smallest ration (ratio actual mass density to current density) to largest ration, rank of models for expansion of universe are

1) Recollapsing universe -in this, metric expansion of space is reverse and universe recollapses.

2) critical universe - in this, expansion of universe is very low.

3) Coasting universe - in this, expansion of universe is steady and uniform

6 0

2 years ago

The Bernoulli equation is valid for steady, inviscid, incompressible flows with a constant acceleration of gravity. Consider flo

irina1246 [14]

Answer:

$p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant$

Explanation:

first write the newtons second law:

F $_{s}$ =δma $_{s}$

Applying bernoulli,s equation as follows:

∑ $δp+\frac{1}{2} ρδV^{2} +δγz=0\\$

Where, $δp$ is the pressure change across the streamline and $V$ is the fluid particle velocity

substitute $ρg$ for {tex]γ[/tex] and $g_{0}-cz$ for $g$

$dp+d(\frac{1}{2}V^{2}+ρ(g_{0}-cz)dz=0$

integrating the above equation using limits 1 and 2.

$\int\limits^2_1 \, dp +\int\limits^2_1 {(\frac{1}{2}ρV^{2} )} \, +ρ \int\limits^2_1 {(g_{0}-cz )} \,dz=0\\p_{1}^{2}+\frac{1}{2}ρ(V^{2})_{1}^{2}+ρg_{0}z_{1}^{2}-ρc(\frac{z^{2}}{2})_{1}^{2}=0\\p_{2}-p_{1}+\frac{1}{2}ρ(V^{2}_{2}-V^{2}_{1})+ρg_{0}(z_{2}-z_{1})-\frac{1}{2}ρc(z^{2}_{2}-z^{2}_{1})=0\\p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant$

there the bernoulli equation for this flow is $p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant$

note: $ρ$ =density(ρ) in some parts and change(δ) in other parts of this equation. it just doesn't show up as that in formular

4 0

2 years ago

Because the soles of your shoes have cleats, you can exert a forward force of 100 N even on slippery ice. A 10-kg picnic cooler

Brilliant_brown [7]

Answer:

you must throw 3 snowballs

Explanation:

We can solve this exercise using the concepts of conservation of the moment, let's define the system as formed by the refrigerator and all the snowballs. Let's write the moment

Initial. Before bumping that refrigerator

p₀ = n m v₀

Where n is the snowball number

Final. When the refrigerator moves

pf = (n m + M) v

The moment is preserved because the forces during the crash are internal

n m v₀ = (n m + M) v

n m (v₀ - v) = M v

n = M/m v/(vo-v)

Let's look for the initial velocity of the balls, suppose the person throws them with the maximum force if it slides in the snow (F = 100N), let's use the second law and Newton

F = m a

a = F / m

The distance the ball travels from zero speed to maximum speed is the extension of the arm (x = 1 m), let's look kinematically for the speed of the balls when leaving the arm

v₁² = v₀² + 2 a x

v₁² = 0+ 2 (100/1) 1

v₁ = 14.14 m / s

This is the initial speed for the crash

v₀ = v = 14.14 m / s

Let's calculate

n = M/m v/ (v₀-v)

n = 10/1 3 / (14.14 -3)

n = 2.7 balls

you must throw 3 snowballs

7 0

2 years ago

a projectile is launched straight up at 141 m/s . How fast is it moving at the top of its trajectory? suppose it is launched upw

BARSIC [14]

The velocity of projectile has 2 components, horizontal component vcosθ and vertical component vsinθ, where v is the velocity of projection and θ is the angle between +ve X-axis and projectile motion.

In case 1, θ = 90⁰

So horizontal component is vcos90 = 0

Vertical component at maximum height = 0

So velocity at maximum height = 0 m/s

In case 2, θ = 45⁰

So horizontal component is 141cos45 = 100m/s

Vertical component at maximum height = 0

So velocity at maximum height = 100 m/s

7 0

2 years ago