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3 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]3 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.

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antoniya [11.8K]

A) Moment of inertia about an axis passing through the point where the two segments meet : $I_A=\frac{1}{12} M L^2$

B) Moment of inertia passing through the point where the midpoint of the line connects to its two ends: $I x=\frac{1}{3} M L^2$

What is Moment of inertia?

The term "moment of inertia" refers to a physical quantity that quantifies a body's resistance to having its speed of rotation along an axis changed by the application of a torque (turning force). The axis might be internal or exterior, fixed or not.

A) The moment of inertia about an axis passing through the point where the two segments meet is $I_A=\frac{1}{12} M L^2$ given that the rod is bent at the center and distance from all the points to the axis remains the same, the moment of inertia about the center will remain the same.

B) Determine the moment of inertia about an axis passing through the point midpoint of the line which connects the two ends

First step: determine the distance between the ends ( d )

After applying Pythagoras theorem $\mathrm{d}=\frac{\sqrt{2}}{2} L$

Next step : determine distance between the two axis $(\mathrm{x})$

After applying Pythagoras theorem

$\mathrm{x}=\frac{\sqrt{2}}{4} L$$$

Final step : Calculate the value of $\mathrm{I}_{\mathrm{x}}$

applying Parallel Axis Theorem

$I_x=I_8+M x^2$

$\begin{aligned}& =\frac{1}{12} M L^2+\frac{1}{4} M L^2 \\& \therefore \quad I x=\frac{1}{3} M L^2 \\&\end{aligned}$

Hence we can conclude that Moment of inertia about an axis passing through the point where the two segments meet: $I_A=\frac{1}{12} M L^2$ , Moment of inertia passing through the point where the midpoint of the line connects its two ends: $I x=\frac{1}{3} M L^2$

To learn more about moment of inertia visit:brainly.com/question/15246709

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