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Naddika [18.5K]
3 years ago
13

2. The lifting condensation level for the atmosphere is 1000m. The wet adiabatic lapse rate is 6.5 oC/km. An air parcel is emitt

ed from the ground surface with a temperature of 30 oC. The ground air temperature is 20 oC and the atmospheric lapse rate 8 oC/km. To what level will the air parcel rise

Geography
1 answer:
Nikitich [7]3 years ago
3 0

Answer:

2.855km

Explanation: see attachment

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If you are talking about settlers who went to America...

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(Btw, I typed this as my sister blasted the Star Spangled banner for no reason lol)

Hope this helps! Sorry if it doesn't.

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One of the organizations of the United Nations is the International Monetary Fund or IMF.

Explanation:

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Why do people in the United States vote the way they do?
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D

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Read 2 more answers
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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Differences in timings across different counties across the world. The need for conversion of timings in order to convey time.
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