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Reika [66]
3 years ago
14

The Milky Way is often considered to be an intermediately wound, barred spiral, which would be type ________ according to Hubble

.
Physics
1 answer:
barxatty [35]3 years ago
7 0

Answer: SBb

On 1930 the astronomer Edwin Hubble classified the galaxies based on their visual appearance into elliptical, spiral and irregular, being the first two classes the most frequent.  

So, according to this classification, the Milky Way is a barred spiral galaxy (SBb in Hubble's notation system) because it has a central bar-shaped structure of bright stars that spans from one side of the galaxy to the other. In addition, its spiral arms seem to emerge from the end of this "bar".

Scientifics considered this, after measuring the the disk and central bulge region of the galaxy, and the conclusion is the Milky Way fulfills these conditions, because is a galaxy that orbits on its same axis and with this rotation its arms are twisted in opposite directions around the mentioned axis.

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A solenoid of length 0.35 m and diameter 0.040 m carries a current of 5.0 A through its windings. If the magnetic field in the c
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A solenoid of length 0.35 m and diameter 0.040 m carries a current of 5.0 A through its windings. If the magnetic field in the center of the solenoid is 2.8 x 10⁻² T, what is the number of turns per meter for this solenoid?

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the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.

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length of solenoid, L= 0.35 m

diameter of the solenoid, d = 0.04 m

current through the solenoid, I = 5.0 A

magnetic field in the center of the solenoid, 2.8 x 10⁻² T

The number of turns per meter for the solenoid is calculated as follows;

B =\mu_o  I(\frac{N}{L} )\\\\B =  \mu_o  I(n)\\\\n = \frac{B}{\mu_o I} \\\\n = \frac{2.8 \times 10^{-2}}{4 \pi \times 10^{-7} \times 5.0} \\\\n = 4.5 \times 10^3 \ turns/m

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A heat engine with 0.300 mol of a monatomic ideal gas initially fills a 1000 cm3 cylinder at 500 K . The gas goes through the fo
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Complete Question:

A heat engine with 0.300 mol of a monatomic ideal gas initially fills a 1000 cm3 cylinder at 500 K . The gas goes through the following closed cycle: - Isothermal expansion to 5000 cm3. - Isochoric cooling to 400 K . - Isothermal compression to 1000 cm3. - Isochoric heating to 500 K .

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b) what is the thermal efficiency

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a) Work done for 1 cycle = 402.13

b) Thermal efficiency = 20%

Explanation:

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Temperature, T = 500 K

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R = 8.314 J/ mol.K

Work done, W = nRT ln(V₂/V₁)

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W = 1247.1 * ln5

W₁ = 2007.13 J

Isochoric cooling

In an Isochoric process, volume is constant i.e. V₂ = V₁ = V

W = nRT ln(V/V)

But  ln(V/V) = ln 1 = 0

Work done, W₂ = 0 Joules

Isothermal Compression to 1000 cm³

V₂ = 1000 cm³

V₁ = 5000 cm³

W = nRT ln(V₂/V₁)

W = 0.3 * 8.314 * 400 ln(1000/5000)

W₃ = -1605 J

Isochoric heating to 500 K

Since there is no change in volume, no work is done

W₄ = 0 J

a) Work done for 1 cycle

W = W₁ + W₂ + W₃ + W₄

W = 2007.13 + 0 + 0 -1605+0

W = 402.13 Joules

b) Thermal efficiency

Thermal efficiency = (Net workdone for 1 cycle)/(Heat absorbed)

Heat absorbed = Work done due to thermal expansion = 2007.13 J

Thermal efficiency = 402.13/2007.13

Thermal efficiency = 0.2

Thermal efficiency = 0.2 * 100% = 20 %

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