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3 years ago

Which term describes a gap in the geologic record that occurs when sedimentary rocks cover an eroded surface?

Physics

2 answers:

svet-max [94.6K]3 years ago

4 0

Answer: Option (D)

Explanation:

Unconformity refers to the surface that clearly demarcates the period of non-deposition. The rocks found below this are older and the rocks that are above this are the younger rocks. This unconformity is exposed to the surface due to the rapid erosion and slow accumulation of sediments.

The different types of unconformity are as follows-

Angular unconformity- here, sedimentary rocks initially gets deposited, then it gets tilted due to tectonic activities and are eroded. Again new sedimentary rocks deposits forming an angle which is known as the angular unconformity.
Non-conformity- here igneous rocks are intruded and undergoes erosion takes place. Then new sedimentary rocks are deposited and this of contact is known as the non-conformity.
Disconformity- when some sequence of rocks are deposited, undergoes erosion and again new sequence of rocks gets deposited, which is known as disconformity.

Thus, the correct answer is option (D).

Lemur [1.5K]3 years ago

3 0

The term that describes a gap in the geologic record that occurs when sedimentary rocks cover an eroded surface is called unconformity.

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3 years ago

6)the speed of light is approximately 186,000 mi/sec. It takes light from a particular star approximately 9 yrs to reach Earth.

NeTakaya

Answer:

5.2791264*10¹³

Explanation:

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3 years ago

A very long insulating cylinder has radius R and carries positive charge distributed throughout its volume. The charge distribut

blsea [12.9K]

Answer:

1. $E(r) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R})$

2. $E(r) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r}$

3.The results from part 1 and 2 agree when r = R.

Explanation:

The volume charge density is given as

$\rho (r) = \alpha (1-\frac{r}{R})$

We will investigate this question in two parts. First r < R, then r > R. We will show that at r = R, the solutions to both parts are equal to each other.

1. Since the cylinder is very long, Gauss’ Law can be applied.

$\int {\vec{E}} \, d\vec{a} = \frac{Q_{enc}}{\epsilon_0}$

The enclosed charge can be found by integrating the volume charge density over the inner cylinder enclosed by the imaginary Gaussian surface with radius ‘r’. The integration of E-field in the left-hand side of the Gauss’ Law is not needed, since E is constant at the chosen imaginary Gaussian surface, and the area integral is

$\int\, da = 2\pi r h$

where ‘h’ is the length of the imaginary Gaussian surface.

$Q_{enc} = \int\limits^r_0 {\rho(r)h} \, dr = \alpha h \int\limits^r_0 {(1-r/R)} \, dr = \alpha h (r - \frac{r^2}{2R})\left \{ {{r=r} \atop {r=0}} \right. = \alpha h (\frac{2Rr - r^2}{2R})\\E2\pi rh = \alpha h \frac{2Rr - r^2}{2R\epsilon_0}\\E(r) = \alpha \frac{2R - r}{4\pi \epsilon_0 R}\\E(r) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R})$

2. For r> R, the total charge of the enclosed cylinder is equal to the total charge of the cylinder. So,

$Q_{enc} = \int\limits^R_0 {\rho(r)h} \, dr = \alpha \int\limits^R_0 {(1-r/R)h} \, dr = \alpha h(r - \frac{r^2}{2R})\left \{ {{r=R} \atop {r=0}} \right. = \alpha h(R - \frac{R^2}{2R}) = \alpha h\frac{R}{2} \\E2\pi rh = \frac{\alpha Rh}{2\epsilon_0}\\E(r) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r}$

3. At the boundary where r = R:

$E(r=R) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R}) = \frac{\alpha}{4\pi \epsilon_0}\\E(r=R) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r} = \frac{\alpha}{4\pi \epsilon_0}$

As can be seen from above, two E-field values are equal as predicted.

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Answer:

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Explanation:

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Introducing the constant of proportionality

dT(t)/dt = k[T5 - T(t)]

which is the desired differential equation

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