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Annette [7]
3 years ago
15

What happens to Earth’s surface when horizontal pressure is applied to rocks?

Biology
2 answers:
irakobra [83]3 years ago
8 0

Answer:

Option (C)

Explanation:

In a sedimentary sequence of rocks, when the rocks are subjected to stress (pressure) horizontally, then it gives rise to the formation of a series of folds, which are characterized by distinct crests and troughs, which eventually forms long ridges and valleys. These ridges and valleys are oriented in such a way that they are directly perpendicular to the direction of the horizontal stress applied on these rocks.

These folds are very prominent in nature sometimes but most of the time, in an outcrop, they are covered by other rocks and sediments. These folds can be microscopic to macroscopic in scale.

These are some of the most important structures found in the sedimentary rock types.

Thus, the correct answer is option (C).

Hoochie [10]3 years ago
5 0
<span>What happens to them would basically happen to everything only the earth's surface is more robust. The answer is c. they fold to create ridges and valleys perpendicular to the direction to which pressure is applied. This means that it takes the form based on the force that is applied to it which is in this case horizontal, meaning from sides instead of vertical which would be from up or down.</span>
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In the following diagram, the voltage is 15 volts and the resistance is 3.0 ohms. Use Ohm's law to determine the curre
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Answer:

Explanation:

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The rate at which the concentration of a drug in the bloodstream decreases is proportional to the concentration at any time t. I
Westkost [7]

Answer:

C(t) = C_oe^{-kt}

Explanation:

From the given information:

At any given time (t), let c(t) represent the concentration of the drug present in bloodstream.

Deriving the equation:

\dfrac{dC}{dt} decrease proportionally to Concentration C

i.e

\dfrac{dc}{dt} \alpha - C

\dfrac{dc}{dt} = -k C

\dfrac{dc}{c} = -k dt

\int \dfrac{dc}{c} = -k \int dt

㏑(C) = -kt + λ

where,

λ is the integration constant.

Integrating at t = 0, concentration of blood = Co g/mL

C(0) = Co

㏑(C₀) = 0 + λ

λ = ㏑(C₀)

From ㏑(C) = -kt + λ

㏑(C) = -kt + ㏑C₀

㏑(C)  - ㏑C₀ =  -kt

\dfrac{C}{C_o} = e^{-kt}

C(t) = C_oe^{-kt}

∴

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C(t) = C_oe^{-kt}

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