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nydimaria [60]
3 years ago
5

Describe the sequence of events in the lithification of a sandstone

Physics
2 answers:
guajiro [1.7K]3 years ago
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 <span>together this is lithification.</span>
Strike441 [17]3 years ago
5 0
<h3><u>Answer and explanation;</u></h3>
  • <em><u>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.</u></em>
  • <em><u>Weathering and erosion of rocks creates sediments </u></em>which collect and are then moved to another location.
  • <em><u>At some point the sediment may become heavy for the agent such as water to carry and thus deposited. </u></em>As this deposit thickens, it becomes heavier and thus lithification process start.
  • <em><u>Compaction forces</u></em> the grains of the deposited sediments closer together, forcing out air and water.
  • <em><u>As a result of compaction the sediment's pores becomes smaller. Minerals deposit in these pores as the water squeezes out. </u></em>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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~~~NEED HELP ASAP~~~<br>Please solve each section and show all work for each section.
anastassius [24]

Explanation:

<u>Forces</u><u> </u><u>on</u><u> </u><u>Block</u><u> </u><u>A</u><u>:</u>

Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass m_A as

x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)

y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)

Substituting (2) into (1), we get

\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)

where f_N= \mu_kN, the frictional force on m_A. Set this aside for now and let's look at the forces on m_B

<u>Forces</u><u> </u><u>on</u><u> </u><u>Block</u><u> </u><u>B</u><u>:</u>

Let the x-axis be (+) up along the inclined plane. We can write the forces on m_B as

x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)

y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)

From (5), we can solve for <em>N</em> as

N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension<em> </em><em>T</em><em> </em> is given by

T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)

Substituting (7) into (4) we get

m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba

Collecting similar terms together, we get

(m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag

or

a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)

Putting in the numbers, we find that a = 1.4\:\text{m/s}. To find the tension <em>T</em>, put the value for the acceleration into (7) and we'll get T = 21.3\:\text{N}. To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get N = 50.9\:\text{N}

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