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3 years ago

The Law of Superposition places the youngest rock on top of older rock, except when what two rock formations are present?

Geography

1 answer:

Nookie1986 [14]3 years ago

7 0

Answer:

The Law of Superposition places the youngest rock on top of older rock, except when faults and igneous rock are present (option 1).

Explanation:

The law of superposition of strata -proposed by Nicolás Steno based on the observations of Avicenna- states that the rock strata observed in the earth's crust are the product of the deposit of rocks progressively over time, leaving the newest rocks in the surface and the oldest in lower strata.

In successive studies and observations made by Steno, this scientist could appreciate the existence of some exceptions to his principle of superposition:

The presence of geological faults could cause an alteration in the arrangement of the layers, different from the order they should have.
Magma activity is a factor that would lead to the deposit of igneous rocks between the layers of already formed rocks.

The Law of Superposition is, in geology, the basis on which the discipline of sedimentary stratigraphy is based.

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Imagine that you have obtained spectra for several galaxies and have measured the observed wavelength of a hydrogen emission lin

Lerok [7]

Answer:

A. z (galaxy 1) = -0.00515, z (galaxy 2) = -0.01707, z (galaxy 3) = -0.04008

B. v (galaxy 1) = -0.00517c, v (galaxy 2) = -0.01707c, v (galaxy 3) = -0.0401c

C. d (galaxy 1) = -21 Mpc,d (galaxy 2) = -69.4 Mpc, d (galaxy 3) = -163 Mpc

Explanation:

We begin by listing out the parameters we were given:

λ (obsv) = 656.3 nm, λ (emit 1) = 659.7 nm, λ (emit 2) = 667.7 nm,

λ (emit 3) = 683.7 nm

A) Using the Redshift formula, we have:

z = [λ (obsv) - λ (emit)] ÷ λ (emit)

For galaxy 1:

z = [λ (obsv) - λ (emit 1)] ÷ λ (emit 1)

z = (656.3 - 659.7) ÷ 659.7 = -0.00515

z = -0.00515

For galaxy 2:

z = [λ (obsv) - λ (emit 2)] ÷ λ (emit 2)

z = (656.3 - 667.7) ÷ 667.7 = -0.01707

z = -0.01707

For galaxy 3:

z = [λ (obsv) - λ (emit 3)] ÷ λ (emit 3)

z = (656.3 - 683.7) ÷ 683.7 = -0.04008

z = -0.04008

B) Using the Doppler formula, we have:

(Δλ ÷ λ) = v ÷ c

v = c * (Δλ ÷ λ)

but, z = (Δλ ÷ λ)

⇒ v = c * z

speed of light (c) = 3 x $10^{8}$ m/s

For galaxy 1:

v = c * z

Substitute z into the equation calculated from A) above

v = 3 x $10^{8}$ * (-0.00515)

v = -1.55 x $10^{6}$ m/s

v = -0.00517c

For galaxy 2:

v = c * z

Substitute z into the equation calculated from A) above

v = 3 x $10^{8}$ * (-0.01707)

v = -5.12 x $10^{6}$ m/s

v = -0.01707c

For galaxy 3:

v = c * z

Substitute z into the equation calculated from A) above

v = 3 x $10^{8}$ * (-0.04008)

v = -12.03 x $10^{6}$ m/s

v = -0.0401c

N.B: the negative value of velocity connotes that the galaxies are moving away from us (not towards us)

C) Using Hubbleʹs law, we have:

v = H · d

where:

v = velocity of a galaxy (km/s), d = distance (Mpc),

H = Hubble's constant (km/s/Mpc) = 73.8 km/sec

d = v ÷ H

We use the velocities calculated in B) above

For galaxy 1:

v = -1.55 x $10^{3}$ km/s

d = -1.55 x $10^{3}$ ÷ 73.8

d = -21 Mpc

For galaxy 2:

v = -5.12 x $10^{3}$ km/s

d = -5.12 x $10^{3}$ ÷ 73.8

d = -69.4 Mpc

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v = -12.03 x $10^{3}$ km/s

d = -12.03 x $10^{3}$ ÷ 73.8

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