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MaRussiya [10]
3 years ago
10

If the mass of the object below is 28g, what is the density of the object below. Units are in cm below. Please round your answer

to the hundredths place. Зcm 2cm 5cm​
Chemistry
1 answer:
Murljashka [212]3 years ago
3 0

Answer:

d = 0.93 g/cm³

Explanation:

Given data:

Mass of object = 28 g

Volume of object = 3cm×2cm×5cm

density of object = ?

Solution:

Volume of object = 3cm × 2cm ×5cm

Volume of object = 30 cm³

Density of object:

d = m/v

by putting values,

d = 28 g/ 30 cm³

d = 0.93 g/cm³

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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

                       Erosion                                              Erosion

1                             C                                                          a

2                             E                                                         e

3                             H                                                         h

3                         Erosion                                                   m

4                           A                                                         Erosion

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.

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