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2 years ago

7

Why do the layers of sedimentary rocks parallel to earth's surface?

1 answer:

Vedmedyk [2.9K]2 years ago

6 0

Answer:

why do the layers of sedimentary rocks parallel to earth's surface?

Why are sedimentary rocks found on or close to Earth's surface?

Explanation: Erosion breaks existing rocks and turns them into sediments. ... When subjected to heat and pressure deeper in the earth sedimentary rocks are changed into igneous and metamorphic rocks. Because erosion only happens on the earth's surface sedimentary rocks are only formed on the earth's surface.

Explanation:

Why do sedimentary rocks have layers and how these layers are formed?

The different groups of sediments could have been deposited through wind, water, ice, and/or gravity at different intervals of time and compacted on top of each other, until they create a sedimentary rock that has several different types of sediments (possibly from different rock types) in the form of layers.

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You will get 50pts and the check mark.

Tamiku [17]

B) A ladybug crawling forward at constant rate of 2.5 m/s

6 0

3 years ago

Read 2 more answers

Please Help!!!! I WILL GIVE BRAINLIEST!!!!!!!!!!!!!

Bas_tet [7]

Given info

d = 0.000250 meters = distance between slits

L = 302 cm = 0.302 meters = distance from slits to screen

$\theta_8 = 1.12^{\circ}$ = angle to 8th max (note how m = 8 since we're comparing this to the form $\theta_m$ )

$x_n = x_5 = 3.33 \text{ cm} = 0.0333 \text{ meters}$ (n = 5 as we're dealing with the 5th minimum )

---------------

Method 1

$d\sin(\theta_m) = m\lambda\\\\0.000250\sin(\theta_8) = 8\lambda\\\\8\lambda = 0.000250\sin(1.12^{\circ})\\\\\lambda = \frac{0.000250\sin(1.12^{\circ})}{8}\\\\\lambda \approx 0.000 000 61082633\\\\\lambda \approx 6.1082633 \times 10^{-7} \text{meters}\\\\ \lambda \approx 6.11 \times 10^{-7} \text{ meters}\\\\ \lambda \approx 611 \text{ nm}$

Make sure your calculator is in degree mode.

-----------------

Method 2

$\Delta x = \frac{\lambda*L*m}{d}\\\\L*\tan(\theta_m) = \frac{\lambda*L*m}{d}\\\\\tan(\theta_m) = \frac{\lambda*m}{d}\\\\\tan(\theta_8) = \frac{\lambda*8}{0.000250}\\\\\tan(1.12^{\circ}) = \frac{\lambda*8}{0.000250}\\\\\lambda = \frac{1}{8}*0.000250*\tan(1.12^{\circ})\\\\\lambda \approx 0.00000061094306 \text{ meters}\\\\\lambda \approx 6.1094306 \times 10^{-7} \text{ meters}\\\\\lambda \approx 611 \text{ nm}\\\\$

-----------------

Method 3

$\frac{d*x_n}{L} = \left(n-\frac{1}{2}\right)\lambda\\\\\frac{0.000250*3.33}{302.0} = \left(5-\frac{1}{2}\right)\lambda\\\\0.00000275662251 \approx \frac{9}{2}\lambda\\\\\frac{9}{2}\lambda \approx 0.00000275662251\\\\\lambda \approx \frac{2}{9}*0.00000275662251\\\\\lambda \approx 0.00000061258279 \text{ meters}\\\\\lambda \approx 6.1258279 \times 10^{-7} \text{ meters}\\\\\lambda \approx 6.13 \times 10^{-7} \text{ meters}\\\\\lambda \approx 613 \text{ nm}\\\\$

There is a slight discrepancy (the first two results were 611 nm while this is roughly 613 nm) which could be a result of rounding error, but I'm not entirely sure.

7 0

3 years ago

Matthew is looking at two different atoms. One has 8 protons and 10 neutrons. The other substance has 8 protons and 12 neutrons.

Lunna [17]

Answer:

They are the same element.

Explanation:

Atom >>>>> Proton >>>> Neutron

A >>>>>>>> 8 >>>>>>>>> 10

B >>>>>>>> 8 >>>>>>>>> 12

From the table above we can see that both atoms have the same proton number.

Therefore, they are the same element because they have the same proton number which means that they have the same atomic number. The element in this case is existing as an isotope in that the atoms have the same proton number but different neutron number.

6 0

3 years ago

An FM radio station broadcasts electromagnetic radiation at a frequency of 93.5 MHz. The wavelength of this radiation is _______

Alborosie

Answer:

3.21

Explanation:

The relation between frequency and wavelength is shown below as:

$c=frequency\times Wavelength
$

c is the speed of light having value $3\times 10^8\ m/s$

Given, Frequency = 93.5 MHz = $93.5\times 10^{6}\ Hz$

Thus, Wavelength is:

$Wavelength=\frac{c}{Frequency}$

$Wavelength=\frac{3\times 10^8}{93.5\times 10^{6}}\ m$

$Wavelength=3.21 \ m$

<u>Answer - A.</u>

7 0

3 years ago

You are working as an assistant to an air-traffic controller at the local airport, from which small airplanes take off and land.

Alika [10]

Answer:

d = 2021.6 km

Explanation:

We can solve this distance exercise with vectors, the easiest method s to find the components of the position of each plane and then use the Pythagorean theorem to find distance between them

Airplane 1

Height y₁ = 800m

Angle θ = 25°

cos 25 = x / r

sin 25 = z / r

x₁ = r cos 20

z₁ = r sin 25

x₁ = 18 103 cos 25 = 16,314 103 m = 16314 m

z₁ = 18 103 sin 25 = 7,607 103 m= 7607 m

2 plane

Height y₂ = 1100 m

Angle θ = 20°

x₂ = 20 103 cos 25 = 18.126 103 m = 18126 m

z₂ = 20 103 without 25 = 8.452 103 m = 8452 m

The distance between the planes using the Pythagorean Theorem is

d² = (x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²2

Let's calculate

d² = (18126-16314)² + (1100-800)² + (8452-7607)²

d² = 3,283 106 +9 104 + 7,140 105

d² = (328.3 + 9 + 71.40) 10⁴

d = √(408.7 10⁴)

d = 20,216 10² m

d = 2021.6 km

7 0

3 years ago