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VashaNatasha [74]
2 years ago
7

To what is the Enviromana?​

Geography
1 answer:
yuradex [85]2 years ago
3 0

Answer:

c

Explanation:

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St. Petersburg has been described as the most "western" Russian city. (a). true (b).false
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It is true because people did described him as the most western Russian in the city back in the days
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Imagine that you have obtained spectra for several galaxies and have measured the observed wavelength of a hydrogen emission lin
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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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Where on Earth would you experience the highest pressure?
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Answer:

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