s this statement true or false? Road maps show elevation on Earth’s surface. A. True B. False

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

4 0

3 years ago

After the Curiosity rover photographed some mudstone in an area nicknamed "Yellowknife Bay", it moved on to new outcrops. A scie

pishuonlain [190]

Answer:

The presence of mustones by curiosity rover indicates chances of ancient life.

Explanation:

The Yellowknife bay is a geologic formation of the Gale crater on the planet Mars and NASA named the rover as a curiosity. Its primary goal was to assess the potential habitability and whether the mars are capable to support life.

The presence of the mudstones in the geologic depression area on mars leads to its name as Yellowknife i.e the capital of the northwest territory of Canada.

The findings of the mudstone from the stepped bedded stata indicate the clay minerals that show the presence of water and claims that an ancient creator was found on in this region and these indicators of a potential mafic source of origin of deposition.

4 0

3 years ago

What is the smallest number of distinct points that can define a plane?

stiv31 [10]

The smallest number of distinct points that can define a plane would be 3 points. Below this value, you cannot create a plane, you can only create a line or a point. A plane that has three point is called a triangle. Hope this helps. Have a nice day.

4 0

3 years ago

Which of following areas is best known for its export of cash crops, especially bananas and other fruit?

Gemiola [76]

Answer:

C. Central America.

Explanation:

8 0

3 years ago

Read 2 more answers

Rosy has 16 pencils and 24 erasers. She is putting together packets that will have an equal number of pencils and erasers in eac

NISA [10]

Answer:

She can make 8 packets

Explanation:

3 0

3 years ago