If i wanted as few birthdays as possible which planet would i want to live on?

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

Where do scientists fit into the process of making environmental laws?

4vir4ik [10]

hello there

Andrea's point is probably the answer you're looking for, however, you should keep in mind that many (most?) environmental laws are driven more by politics than by the available data. Scientists do research on any number of topics (environmental or not) that have implications for our society. We (scientists) like to think that policy makers take our data and results into consideration when drafting legislation, but I would say that in my experience this rarely happens. For one, there have rarely been any federal legislators who understand science. Even the few physicians who've gone into politics seem largely to fail to understand very basic tenets of science, or at least they don't often demonstrate it with their political actions. Leaving environmental issues, for example, if politicians took science into account, there would be no discussion in any school district in the country about whether or not evolution should be taught in schools, as there are simply *NO* scientists who are qualified to have an opinion who would suggest it shouldn't. Environmental issues are not really any different.

hope that helped bye

8 0

3 years ago

Chemical weathering in most rapid in what climate

Veseljchak [2.6K]

These chemical processes need water , and occur more rapidly at the higher temperature ,so warm ,damp climates are best.

6 0

3 years ago

WRITE A PARAGRAPH (minimum of 5 SENTENCES) EXPLAINING

spayn [35]

Answer:

Explanation:

hi

5 0

2 years ago

Gideon noted that three inches of rain had fallen on his patio over night , the following day, scientists measured and found tha

EleoNora [17]

Answer:

The water has left the system before if could be measured.

Explanation:

Some of the water has flowed quickly along as surface runoff and through the river channel. Scientists would then missed the chance to measure this rainfall. Some water will have been absorbed by nearby vegetation. The vegetation then releases the water in a vapor state, to the atmosphere. This process is known as transpiration. Heat in the atmosphere causes evaporation and the water never joins the soil of groundwater to be measured as part of the water table.

8 0

3 years ago