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nikdorinn [45]
3 years ago
9

Which of the following is an example of spatial thinking?

Geography
2 answers:
KonstantinChe [14]3 years ago
7 0

Figuring out which cultures approve of sharing bedrooms and which cultures emphasize sleeping alone.

enyata [817]3 years ago
5 0
Describing the colors that you used to paint your bedroom <span>is an example of spatial thinking
</span><span>Geography is a self-professed spatial science.The geographic metaphor, however, is but one of many that incorporates spatial thinking into the rubric of a knowledge base.Because of their essential interest in human-environment relations, geographers have "spatializes" non-spatial data they extract from the real world.In addition to regarding the earth and the people on it as their eminent domain, this allows them to reason spatially about phenomena by representing it in spatial formats, particularly by representing phenomena on maps.But they also spatialize by representing data in graphic formats.For example, they draw population pyramids of the non-spatial demographic data on age structure; 
</span>
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The Canadian Shield, a vast expanse of ancient rock, is centered around which area? A. the Pacific coast of Canada B. the banks
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<span>C. The Hudson Bay
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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 = -<u>0.00515</u>

For galaxy 2:

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

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

z = -<u>0.01707</u>

For galaxy 3:

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

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

z = -<u>0.04008</u>

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 = -<u>0.00517c</u>

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 = -<u>0.01707c</u>

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 = -<u>0.0401c</u>

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 = -<u>21 Mpc </u>

For galaxy 2:

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

d = -5.12 x 10^{3} ÷ 73.8

d = -<u>69.4 Mpc </u>

For galaxy 3:

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

d = -12.03 x 10^{3} ÷ 73.8

d = -<u>163 Mpc</u>

N.B: distance cannot be a negative value

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Answer:

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The earth's basin and range topography is a result of _____.
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The earth's basin and range topography is a result of: c. vertical movement along faults.
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