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Vikki [24]
3 years ago
14

Scientists want to protect a beach from erosion. what can they do

Geography
2 answers:
Pepsi [2]3 years ago
8 0
They can prevent erosion by adding a sea wall.
Pepsi [2]3 years ago
4 0
Sand can be trucked in from other sources and filtered for sediments to eventually make way into the air and water. They can also build seawalls, revetments, and jetties along the shoreline. :)
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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:

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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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