Explain the origin of the rock cleavage seen in slate and themineral cleavage seen in amphibole or calcite.

Answers

Answer 1
Answer:

Answer:

Cleavage of a mineral is defined as the direction along which the mineral breaks over the flat mineral surface. It is directly dependent on the mineral's crystallographic structure.

Slaty cleavage refers to the cleavage that is commonly found in the low grade metamorphic rock slate. Here, the fine grained minerals are so closely spaced and the lines of weakness are all parallel to each other, that allows the minerals to break in fine and thin layers. During the time of formation of slate, the smaller fractures are formed due to the increasing pressure. When groundwater moves through these fractures it forms a weak zone due to which these slaty cleavage forms.

When the rocks undergoes deformation and changes in the pressure and temperature condition then the cleavages are formed on the mineral's surface. The calcite contains rhombohedral cleavage are considered to be the perfect cleavage. Whereas the cleavage in amphibole is uneven and comprised of two directions that intersect each other at 56° and 124°.


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Discuss three Main approaches in map analysis ?

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Answer:

"Three main approaches for conducting research using concept maps as a tool emerged: relational, cluster, and word frequency. These approaches are included in the different phases of research process such as data collection, analysis, and presentation."

source from: www.tandfonline.com

Is this a good answer?

1. Calculate the angle of the noon sun above the horizon right here at Placerita Canyon, which is at ~34N. Use the analemma on the next page to find the declination of the sun for today’s date. Make sure you identify today’s date! SA = 90º - AD SA is the solar altitude AD is the arc distance between your latitude and that of today’s declination of the sun

Answers

Final answer:

To calculate the noon sun's angle above the horizon at Placerita Canyon, identify today's date and find the corresponding solar declination using the analemma. Next, calculate the arc distance (AD) by subtracting this declination from 34°N, and finally subtract the AD from 90° to get the solar altitude (SA). This gives you the sun's angle above the horizon at noon.

Explanation:

To calculate the angle of the noon sun above the horizon at a location of approximately 34°N latitude, given that the solar altitude (SA) is 90º minus the arc distance (AD). Start with the latitude of 34°N. Identify today's date and find the corresponding declination of the sun using the analemma. Subtract this value from 34°N (or vice versa depending on which is greater) for the AD. Finally, subtract the AD from 90° to get the SA.

For example, if today's declination of the sun was 23°N: AD = 34°N - 23°N = 11°, then SA = 90º - 11º = 79º. So, at noon the sun would be 79 degrees above the horizon at Placerita Canyon if today's date corresponds to a declination of 23°N.

Learn more about Solar Altitude Calculation here:

brainly.com/question/33603439

#SPJ2

Answer:

The latitude at which the noon sun is directly overhead is defined as the sun's declination. For example, if the sun appears directly overhead at 20°N latitude, the sun's declination is thus 20°N. Figure 7-3, known as analemma, shows the declination of the sun throughout the year. Thus, if you would like to know where the sun will be directly overhead on September 20, you can look on the analemma and find that it will be at 1°N. Given the declination, the noon sun angle at a particular latitude on any given day can be calculated.

Figure 7-3. The analemma, a graph illustrating the latitude of the overhead noon sun throughout the year.

Figure 7-4 shows that for each degree of latitude that the place is away from the latitude where the noon sun is overhead (declination), the angle of noon sun becomes one degree lower from being vertical (or 90°) above the horizon. Noon sun angle for any given latitude on any given day can be calculated using the following equation:

Noon Sun Angle = 90° - (Latitude +/- Declination)

In this equation, use " - " when the latitude and declination are on the same hemisphere and use " + " when they are on the opposite hemispheres. For example, on April 11, the declination is 8°N, the noon sun angle at Springfield, Illinois (40N) on April 11, therefore, should be 58° (90° - (40°N - 8°N).

Figure 7-4. Noon sun angle calculation. The example in this figure illustrates how to calculate the noon sun angle for a city 40° north of the equator (like Springfield, Illinois) on December 22.

Questions

Based on Figure 7-3, the sun shifts its position between which two important latitudes (find the terminologies for these two latitudes in Fig. 7-4). (2 points)

Using the analemma (Fig. 7-3), determine the declinations on each of the following dates. (6 points)

August 10 _________________

September 21 _______________

June 21 ______________

October 15 ____________

December 22 ______________

March 21 _________________

Calculate the noon sun angle in Grayslake, Illinois on each of the dates listed below. (6 points)

Noon Sun Angle in Grayslake on

August 10 _________________

September 21 _______________

June 21 ______________

October 15 ____________

December 22 ______________

March 21 _________________

Obtain the average temperature in Grayslake on each of the dates listed below using a weather station online. (6 points)

Average Temperature in Grayslake on:

August 10 ________________

September 21 _______________

June 21 _____________

October 15 ____________

December 22 ______________

March 21 _________________

Plot the noon sun angle and temperature on each of the dates you obtained in Questions 3-4 on the graph provided below and connect your temperature and noon sun angle plots using smooth lines. Discuss the relationship between noon sun angles and average temperature in Grayslake throughout the year based on your graph. Use the right vertical axis to show temperature variation from -30° F at bottom and +105° F on top (label temperature value at equal amount interval in between: 2 units for every 5°F). Use the left vertical axis to show noon sun angle, with 0° at the bottom and 90°F on top (label noon sun angle value at equal amount intervals in between: every 3 units per 10°). (4 points)

J F M A M J J A S O N D

Months

Figure 7-5. Variation of temperature and noon sun angle in Grayslake.

brief paragraph summarizing the yearly movement of the overhead noon sun and how the intensity of insolation aries in Grayslake throughout the year. (2 points)

Calculate the maximum, minimum, and average noon sun angles at the place in Grayslake. Show your calculations. (3 points)

Maximum noon sun angle:

Minimum noon sun angle:

Average noon sun angle:

When the noon sun is directly overhead a ship in the Pacific Ocean on October 16, the ship's latitude is _____________. Explain your answer. (2 points)

What is HL? Trapezoid

Answers

Answer:

Please post question under mathematics category

Explanation:

Does not belong in geography.

The Pacific garbage patch is a(n) Question 14 options: dead zone south of Hawaii. ocean space, twice the size of Texas, where schools of stingray feed on detritus. continuous, expanding array of floating plastics in the north pacific gyre. landfill for marine wastes located on Midway atoll.

Answers

Answer:ocean space

Explanation:

1) The Sun has an average surface temperature of approximately 6000 K. Calculate the power per unit area released by the surface of the Sun using Stefan-Boltzmann Law and report the value below. Units are a must. (2 points)Power per unit area released by the Sun=

Answers

Answer:

Power per unit area released by the Sun is 73,488,384 Watt per meter square.

Explanation:

The Sun has an average surface temperature of approximately,T= 6000 K

The power per unit area released by the surface of the Sun be P'

Stefan-Boltzmann Law  is given as:

P'=(Power)/(Area)=\sigma T^4

where

\sigma =5.6704* 10^(-8) Watt/m^2K^4 Stefan-Boltzmann constant

T = Absolute temperature of the body

Power per unit area released by the Sun:

=5.6704* 10^(-8) Watt/m^2K^4* (6000 K)^4=73,488,384 Watt/m^2

What role did geography play in how the greeks obtained food

Answers

I would say that geography meant that Greece was very rocky so agriculture was difficult so Greece would have to rely on trade for some of its food. On the other hand, Greece has many islands in the Mediterranean which probably allowed the residents to catch fish and supplement their diet with that and perhaps trade fish for agricultural products.
Greece was really rocky and supported growth of only a few crops,so they did trading to obtain food.