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Sever21 [200]
3 years ago
14

50 POINTS

Geography
1 answer:
fomenos3 years ago
6 0

Well there are Primary waves, Secondary waves,and Surface waves

  • <u><em>Primary waves:</em></u>

<u><em>These waves can travel through fluids and solids and are longitudinal - this means they transfer their energy through compression, like a slinky forming compressed areas when you push one end (sound is also an example of a longitudinal wave).</em></u>

<u><em>This also means that they transfer energy parallel to the direction of the wave, so if a wave is travelling north to south, the energy will be transferred in this direction.</em></u>

<u><em>P-waves are the fastest of the three seismic wave</em></u>

  • <u><em>Secondary waves:</em></u>

<u><em>S-waves cannot travel through air or water, only through solids, but they have a larger amplitude (this is the height of a wave, measured from the highest point to the middle line) so are more destructive in the case of an earthquake.</em></u>

<u><em>They are transverse waves, meaning they transfer energy perpendicular (at right angles) to the direction of the wave - like a rope being shaken up and down (this is like the classic up and down wave associated with light).</em></u>

<u><em>S-waves are slower than P-waves.</em></u>

  • <u><em>Surface waves:</em></u>

<u><em>The final type of seismic wave occurs along the boundary between two different substances (e.g. rock and air). They can be either longitudinal (Rayleigh) or transverse (Love and Rayleigh).</em></u>

<u><em>These waves travel slower than both S and P waves, but have a higher amplitude and so can be the most destructive of all the seismic waves.</em></u>

<u><em>All three types of seismic wave are generated in an earthquake, and we can monitor them to find out about the nature of one. For example, by measuring the difference in arrival at a monitor between the fast P waves and slower S waves to find the location of an earthquake's focus.</em></u>


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The correct answers are as follows:
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2. The type of cloud that usually brings precipitation is NIMBUS. The Nimbus cloud usually form dark colored, wet looking layers, that are typically associated with precipitation. The precipitation can be light or moderate. The cloud can also result in continuous rain or snow.

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Cirrus is an high level cloud, altocumulus is a middle level cloud, stratus is a low level cloud and the fog is the closest to the earth surface. 

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1. At the location of any star, west is defined as the direction of the apparent motion, and east is the opposite direction. Is
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Answer and Explanation:

1. In order to determine the direction whether west is toward left or right, let us suppose that we are facing towards the  south direction.

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2. The vast majority of the stars seem to move in a similar way at both the north and south poles but not all the stars.

Since stars in northern half of the globe seem to turn clockwise however some are not seen because of going under the skyline and which is valid for southern side of the equator too.

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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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Answer:

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