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Akimi4 [234]
3 years ago
8

A conglomerate is a rock that forms as a result of _____. . . Choices:Intense heat and pressure. Compaction and cementation. Rap

id cooling. Slow cooling.
Physics
2 answers:
sergeinik [125]3 years ago
7 0

A conglomerate is a rock that forms as a result of compaction and cementation. The correct answer between all the choices given is the first choice or letter A. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

12345 [234]3 years ago
5 0

A conglomerate is a rock that forms as a result of compaction and cementation.

Conglomerate rock is a type of sedimentary rock which are formed as a result of compaction and cementation.

<h2>Further Explanation</h2><h3>Rock </h3>
  • A rock is a naturally occurring substances that is a solid mixture of one or more minerals together with organic matter.
  • Rocks are classified based on the chemical composition, texture and the way they are formed.  
<h3>Rock cycle</h3>
  • One type of rock may change from one form to another through a rock cycle.
  • Rock cycle is the process through each various rocks change from one form to another, normally an interchange between the three major types of rocks.
<h3>Major classification of rocks:</h3><h3>Sedimentary rocks </h3>
  • Sedimentary rocks are types of rocks that are formed through accumulation of sediments at low temperatures in tectonic layers and sinks. The process involved include compaction and cementation of sediments.
  • These sediments includes; pebbles, shells, sand and other material fragments.  
  • The sediments accumulates in layers and then harden into rocks over a period of time.
  • Examples of sedimentary rocks include; limestone and conglomerate
<h3>Metamorphic rocks</h3>
  • These are types of rocks that are formed as a result of changes that occurs due to intense heat and pressure under the surface of the earth. They result from action of heat and pressure on other rocks that pre-existed.
  • These types of rocks are characterized by shiny crystals, ribbon-like layers among other features.
  • Examples of metamorphic rocks are marble and gneiss
<h3>Igneous rocks </h3>
  • These are types of rocks that are formed as a result hardening and cooling of magma from volcanic eruptions. Magma may cool inside the earth or when on the surface of the earth as a result of volcanic eruptions. The lava from this eruptions cools and hardens to form metamorphic rocks.
  • Igneous rocks are glass-like and shiny with no crystals. They may also have tiny spaces and holes due to gas bubbles trapped during the cooling process.
  • Examples of igneous rocks include obsidian and basalt.

Key words: Rocks, chemical composition, sedimentary rocks, conglomerate.

<h3>Learn more about;</h3>
  • Rocks and rock types; brainly.com/question/2817889
  • Sedimentary rocks; brainly.com/question/2817889
  • Igneous rocks; brainly.com/question/2817889
  • Metamorphic rocks; brainly.com/question/2817889

Level; High school

Subject: Geography

Topic:  Rocks

sub-topic: classification of rocks

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A) v = 6.93 m/s

B) v = 4.9 m/s

C) x_m = 0.015m

D) v_max = 5.2 m/s

Explanation:

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x = 6 cm = 0.06 m

k = 400 N

m = 0.03 kg

F = 6N

A) from work energy law, work dome by the spring on ball which now became a kinetic energy is;

Ws = K.E = ½kx²

Similarly, kinetic energy of ball is;

K.E = ½mv²

So, equating both equations, we have;

½kx² = ½mv²

Making v the subject gives;

v = √(kx²/m)

Plugging in the relevant values to give;

v = √((400 × 0.06²)/0.03)

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B) If there is friction, the total work is;

Ws = ½kx² - - - (1)

Work of the ball is;

Wb = KE + Wf

So, Wb = ½mv² + fx - - - (2)

Combining both equations, we have;

½mv² + fx = ½kx²

Plugging in the relevant values, we have;

(½ × 0.03 × v²) + (6 × 0.06) = ½ × 400 × 0.06²

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0.015v² = 0.72 - 0.36

v² = 0.36/0.015

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v = 4.9 m/s

C) The speed is greatest where the acceleration stops i.e. where the net force on the ball is zero. (ie spring force matches 6.0N friction)

So, from F = Kx;

(x is measured into barrel from end where F = 0)

Thus; 6.0 = 400x

x_m = 6/400

x_m = 0.015m from the end after traveling 0.045m

D) Initial force on ball = (Kx - F) =

[(400 x 0.06) - 6.0] = 18N

Final force on ball = 0N

Mean Net force on ball = ½(18 + 0)

Mean met force, F_m = 9N

Net Work Done on ball = KE = 9N x 0.045m = 0.405 J

Thus;

½m(v_max)² = 0.405J

(v_max)² = 2 x 0.405/0.03

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v_max = 5.2 m/s

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Answer:

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