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3 years ago

On the map above, what is the name of the island to the west of Letter B?

Geography

2 answers:

JulijaS [17]3 years ago

8 0

The correct answer is option A. Kilimanjaro

Kilimanjaro or Mount Kilimanjaro has three volcanic cones.

The names of these cones are Kibo, Mawenzi, and Shira. Kilimanjaro is a dormant volcano in Tanzania.

In addition, Kilimanjaro is the highest mountain in Africa.

The height of the mountain is about 4,900 metres i.e.16,100 ft from its base, and 5,895 metres i.e. 19,341 ft above sea level.

Hope this helps!

Please mark brainliest! ^^

Yanka [14]3 years ago

5 0

Answer:

A. Kilimanjaro

Explanation:

hope this helps

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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

For galaxy 3:

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

d = -12.03 x $10^{3}$ ÷ 73.8

d = -163 Mpc

N.B: distance cannot be a negative value

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iogann1982 [59]

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