Explanation:
Recrystallization: contact pressure causing grains to "fuse" together
Cementation
: precipitation of bonding agents between grains
Compaction
: increase in density due to weight of overburden
Lithification is the process by which sediments are converted into sedimentary rocks. During this process, recrystallication, compaction and cementation of mineral grains occur.
The process starts with the compaction of sediments. The over burden weight of new sediments in the basin adds to the one originally deposited. This compresses the sediment. The volume of reduced and the density increases.
Recrystallization follows suit as the contact pressure of grains makes them fuse together. It is more like reworking of sediments. In this process, cementing materials can precipitate and cause sediments to be more fused together.
This is why most sediment are made up of clasts in a matrix of cementing materials.
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Answer:
A constant value everywhere in the universe.
Explanation:
The speed of light in a vacuum is a constant value. It is not affected by change in frequency or wavelength of the light.
Mathematically the speed of light is given as:
c = λf
where λ = wavelength and f - frequency
The speed of light is the constant of proportionality between frequency and wavelength. In order words, wavelength and frequency are inversely proportional. As the wavelength increases, frequency decreases and vice versa.
While the change in wavelength and frequency of light affect the energy of the light, its speed is a constant value as long as the medium is a vacuum.
The speed of light is also not dependent on the manner with which the light wave is moving.
Answer:
a) t = 2.0 s, b) x_f = - 24.56 m, Δx = 16.56 m
Explanation:
This is an exercise in kinematics, the relationship of position and time is indicated
x = 5 t³ - 9t² -24 t - 8
a) ask when the velocity is zero
speed is defined by
v =
let's perform the derivative
v = 15 t² - 18t - 24
0 = 15 t² - 18t - 24
let's solve the quadratic equation
t =
t1 = -0.8 s
t2 = 2.0 s
the time has to be positive therefore the correct answer is t = 2.0 s
b) the position and distance traveled for a = 0
acceleration is defined by
a = dv / dt
a = 30 t - 18
a = 0
30 t = 18
t = 18/30
t = 0.6 s
we substitute this time in the expression of the position
x = 5 0.6³ - 9 0.6² - 24 0.6 - 8
x = 1.08 - 3.24 - 14.4 - 8
x = -24.56 m
we see that all the movement is in one dimension so the distance traveled is the change in position between t = 0 and t = 0.6 s
the position for t = 0
x₀ = -8 m
the position for t = 0.6 s
x_f = - 24.56 m
the distance
ΔX = x_f - x₀
Δx = | -24.56 -(-8) |
Δx = 16.56 m
Answer:
v = 16.11 m / s
Explanation:
For this exercise we must use the principle of conservation of energy. We set a reference system on the part of the platform without elongation
starting point. When the spring is compressed
Em₀ = K_e + U = ½ k x² + m g x ’
final point. The point where it leaves the platform
Em_f = K = ½ m v²
energy is conserved
Em₀ = Em_f
½ k x² + m g x ’= ½ m v²
v² =
x² + g x
let's calculate
v² =
1.25² + 9.8 1.25
v² = 247.159 + 12.25 = 259.409
v = 16.11 m / s