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Fiesta28 [93]
2 years ago
7

HELP!!!!! PLEASE! :( GIVING 65 POINTS AND WILL GIVE BRAINLIEST.

Physics
1 answer:
Alja [10]2 years ago
5 0

Explanation:

The given problem should be solved using basic sedimentary laws proposed by Nicolas Steno and other group of geologists.

Some of the laws that would find application in this problem are:

  • Law of superposition of strata which states that "in an undeformed sequence of sedimentary rocks, the oldest strata is at the bottom and the youngest on top"
  • Law of lateral continuity states that "sedimentary rocks are laid down down horizontally and laterally in space until they taper at edges of basins"
  • Law of cross-cutting "events such as folding, fracturing, intrusion are younger than than the rock layers that they affect".

Based on this premise, we can derive the geologic column for the area:

                       Chart A                                               Chart B

                       <em> Erosion                                              Erosion</em>

1                             C                                                          a

2                             E                                                         e

3                             H                                                         h

3                      <em>   Erosion                                                   </em>m

4                           A                                                      <em>   Erosion</em>

5                            B                                                              c

6                            F                                                               j

7                            D                                                              l

8                            L                                                              g

9                            G                                                              k

10                           J                                                              f

11                            i                                                                i

12                                                                                          d

13                                                                                           b

The above was derived using the sedimentary laws for relative dating.

Observations 1

In Record A unconformity will occur between layers H and A

    Record B unconformity will occur between layers c and m

An unconformity is a huge break in the stratigraphic record. It is usually a time gap as a result of non-deposition of sediments, erosion of a rock layers amidst other factors.

The type of unconformity here is called an angular unconformity.

In this kind of unconformity, parallel beds are deposited on the tilted or folded layer below resulting in an angular discordance of some sort.

The surface between H and A is an erosional surface of the tilted beds in record A. This was followed by the deposition of new beds that are parallel to the tilted ones below.

In record be, a folded and eroded surface separate c and m.

Analysis 2

In record A, the rock layer K is older than F but younger than D. This rock layer has been tilted. This interpretation is based on the law of superposition of strata. If the sequences are reconstructed without the tilt, layer K will be on top of D and F will be on top of K.

Analysis 3

Layer c is younger than j but older than the parallel lying m on top of it. This geologic structure is called a synform. In synform, the rock strata has a concave shape. Usually, we can assume that this is a synformal syncline in which the youngest bed is at the core of the fold. This pits layer c as the youngest in the folded beds. The age increases outward in a synform.

Analysis 4

Layers C and E in record A are both eroded beds. Layer C has been eroded to the level of layer E.

We can establish that when layer C was exposed and eroded, the overburden on the earth was relieved and part of bed E re-adjusted up to the level of E.

Analysis 5

Reconstructing the geologic history of A

To form a sedimentary rock sequence, a basin must be available. Sedimentary basins are natural depressions on the earth crust in which sediments collects and are deposited.

Sediments i - A were deposited layer by layer into the basin. They became compacted and lithified to form sedimentary rocks.

With the passage of time, these layers became tilted by natural forces. This could be an earthquake, stress e.t.c

After the tilting, a period of non-deposition occurred in which the forces of denudation prevailed to cause the erosion of the bed when it was uplifted and exposed.

After this period, layers H-C was then deposited, compacted and lithified. The whole rock sequence was then uplifted exposing some rock layers. Erosion washed parts of layers C and E to what we can see in the present day.

Analysis 6

In record B, sequences b-c were laid in the basin of deposition where they were transformed into sedimentary rocks.

The rocks were then subjected to a compressive stress regime. The layers buckled and a synformal sycline fold formed.

After this period, the rock was uplifted and exposed. Erosion caused the washing away of parts of the top layer.

A period of deposition shortly followed and parallel layers m-a were deposited. This layer was then transformed into sedimentary rocks.

After this the layer became exposed and part of a was eroded to its present day level.

Learn more:

Sedimentary rocks brainly.com/question/9131992

Alignment of sedimentary rocks brainly.com/question/8655172

#learnwithBrainly

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(4A) The mass of Earth is 5.972 * 10^24 kg, and the radius of Earth is 6,371 km.
faltersainse [42]

Answer:

x₁ = 345100 km

Explanation:

The direction of the attraction forces between the earth and the object, and between the moon and the object, are in opposite direction and  (along the straight line between the centers of earth and moon) and as gravity is always attractive, the net force will become zero when both forces are equal. According to this:

Let  call "x₁"  distance between center of the earth and the object, and

"x₂" the distance between center of the moon and the object, Mt mass of the earth, Ml mass of the moon, m₀ mass of the object

we can express:

F₁  ( force between earth and the object )

F₁ = K *  Mt * m₀/ ( x₁)²        K is a gravitational constant

F₂  (force between mn and the object)

F₂ = K * Ml * m₀ / (x₂)²

Then:

F₁ = F₂               K*Mt*m₀ / x₁²   =  K*Ml*m₀ /x₂²

Or  simplifying the expression

Mt/ x₁²  =  Ml/ x₂²

We know that   x₁   +  x₂  = 384000 Km then

x₁ =  384000 - x₂

Mt/( 384000 - x₂)²  =  Ml / x₂²

Mt *  x₂²  =  Ml *( 384000 - x₂)²

We need to solve for x₂

Mt *  x₂²  =  Ml *[ ( 384000)² + x₂² - 768000*x₂]

By substitution:

5.972*10∧24*x₂² = 7.348*10∧22 * [ 1.47*10∧11 ] + 7.348*10∧22*x₂² -

                                7.348*10∧22*768000*x₂

Simplifying by 10∧22

5.972*10²*x₂²  = 7.348* [ 1.47*10∧11 ] + 7.348*x₂²- 7.348*768000*x₂

Sorting out

5.972*10²*x₂²- 7.348*x₂² = 10.80*10∧11 - 56,43* 10∧5*x₂

(597,2 - 7,348 )* x₂²  = 10.80*10∧11 - 56.43*10∧5*x₂

590x₂²  + 56.43*10∧5*x₂ - 10.80*10∧11 = 0

Is a second degree equation

x₂  =  -56.43*10∧5 ± √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  = -56.43*10∧5 + √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  =  -56.43*10∧5 + √3184*10∧10 + 254880*10∧10  / 1160

x₂ ₁  = -56.43*10∧5 + 10∧5 [ √3184 + 254880 ] /1160

x₂ ₁  =  -56.43*10∧5 + 508* 10∧5  / 1160

x₂ ₁  =  451.27*10∧5/1160

x₂ ₁  =  4512.7*10∧4 /1160

x₂ ₁  = 3.89*10∧4  km (distance between the moon  and the object)

x₂ ₁  = 38900 km

x₂ = 38900 km

We dismiss the other solution because is negative and there is not a negative distance

Then the distance between the earth and the object is:

x₁  = 384000 - x₂

x₁ = 384000 - 38900

x₁ = 345100 km

5 0
3 years ago
Help with these please someone?
DIA [1.3K]

Answer:

8. 2.75·10^-4 s^-1

9. No, too much of the carbon-14 would have decayed for radiation to be detected.

Explanation:

8. The half-life of 42 minutes is 2520 seconds, so you have ...

1/2 = e^(-λt) = e^(-(2520 s)λ)

ln(1/2) = -(2520 s)λ

-ln(1/2)/(2520 s) = λ ≈ 2.75×10^-4 s^-1

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9. Reference material on carbon-14 dating suggests the method is not useful for time periods greater than about 50,000 years. The half-life of C-14 is about 5730 years, so at 65 million years, about ...

6.5·10^7/5.73·10^3 ≈ 11344

half-lives will have passed. Whatever carbon 14 may have existed at the time will have decayed completely to nothing after that many half-lives.

7 0
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Light with wavelength in air ( lambdaair ) is incident on a oil slick ( noil = 1. 25) floating on the ocean ( nwater = 1. 33). w
crimeas [40]

<u>26mm</u> is the thinnest thickness of oil that will brightly reflect the light.

What is wavelength ?

The distance over which a periodic wave's shape repeats is known as the wavelength in physics.  It is a property of both traveling waves and standing waves as well as other spatial wave patterns. It is the distance between two successive corresponding locations of the same phase on the wave, such as two nearby crests, troughs, or zero crossings.  The spatial frequency is the reciprocal of wavelength. The Greek letter lambda () is frequently used to represent wavelength. The term wavelength is also occasionally used to refer to modulated waves, their sinusoidal envelopes, or waves created by the interference of several sinusoids.

To learn more about wavelength visit:

brainly.com/question/16051869

#SPJ4

4 0
1 year ago
A tetrahedron has an equilateral triangle base with 25.0-cm-long edges and three equilateral triangle sides. The base is paralle
djverab [1.8K]

Answer:

a. 7.046 Nm²/C

b. 2.348 Nm²/C

Explanation:

Data given:

Base of equilateral triangle = 25.0 cm = 0.25 m

Strength of electric field = 260 N/C

In order to find the electric flux we first have to find out the area of triangle.

Area of triangle = \frac{\sqrt{3} }{4} a^{2}

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Lets find electric flux,

      Electric Flux = E. A

                          = 260×0.0271

                          = 7.046 Nm²/C

Now we can find the electric flux through each of the three sides.

Electric flux through three sides = \frac{7.046}{3}

                                                = 2.348 N m²/C

       

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(Physical Science) What is the wavelength of light waves if their frequency is 5.0 x 10^14 Hz and the speed of light is 3 x 10^8
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