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Viktor [21]
3 years ago
8

[OU.04]What type of galaxy is described as having a circular shape with most of the objects concentrated around its center? spir

al elliptical irregular barred spiral
Physics
2 answers:
Annette [7]3 years ago
8 0

Answer:

The correct answer would be a barred spiral as it mentions an elongated spiral, so it is not a regular spiral galaxy. So the correct answer would be D

Explanation:

4.04 notes

GalinKa [24]3 years ago
5 0

Answer:

spiral

Explanation:

A galaxy may be defined as the huge collection of celestial bodies like the stars, planets, dark matters, dust and gases and the solar systems. It is derived from the Greek work, 'Galaxias' , which means "milky". There are many types of galaxies. They are : Spiral galaxy, Elliptical galaxy, Irregular galaxy and Barred spiral galaxy.

A spiral galaxy is one which contains a sphere shaped bulge which consists of old stars and it is devoid of dust and gas. Its circular shape composes a disk. It is of circular shape having most of the objects concentrated in its center.

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If a 2,000-kg car hits a tree with 500 n of force over a time of 0.5 seconds, what is the magnitude of its impulse?
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The correct answer is:
<span>The rate at which a waves energy flows through a given unit of area

In fact, light intensity is defined as the light power per unit of area:
</span>I= \frac{P}{A}
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The interior space of large box is kept at 30 C. The walls of the box are 3 m high and have a ‘sandwich’ construction consisting
White raven [17]

Answer:

\frac{\dot Q}{A} =20.129\ W.m^{-2}

T_1=27.58\ ^{\circ}C & T_2=2.41875\ ^{\circ}C

Explanation:

Given:

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  • height of the walls of box, h=3\ m
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  • thermal conductivity of plywood, k_p=0.104\ W.m^{-1}.K^{-1}
  • thickness of sandwiched Styrofoam, x_s=5\ cm=0.05\ m
  • thermal conductivity of Styrofoam, k_s=0.04\ W.m^{-1}.K^{-1}
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<u>From the Fourier's law of conduction:</u>

\dot Q=\frac{dT}{(\frac{x}{kA}) }

\dot Q=\frac{dT}{R_{th} } ....................................(1)

<u>Now calculating the equivalent thermal resistance for conductivity using electrical analogy:</u>

R_{th}=R_p+R_s+R_p

R_{th}=\frac{x_p}{k_p.A}+\frac{x_s}{k_s.A}+\frac{x_p}{k_p.A}

R_{th}=\frac{1}{A} (\frac{x_p}{k_p}+\frac{x_s}{k_s}+\frac{x_p}{k_p})

R_{th}=\frac{1}{A} (\frac{0.0125}{0.104}+\frac{0.05}{0.04}+\frac{0.0125}{0.104})

R_{th}=\frac{1.4904}{A} .....................(2)

Putting the value from (2) into (1):

\dot Q=\frac{30-0}{\frac{1.4904}{A} }

\dot Q=\frac{30\ A}{1.4904}

\frac{\dot Q}{A} =20.129\ W.m^{-2} is the heat per unit area of the wall.

The heat flux remains constant because the area is constant.

<u>For plywood-Styrofoam interface from inside:</u>

\frac{\dot Q}{A} =k_p.\frac{T_i-T_1}{x_p}

20.129=0.104\times \frac{30-T_1}{0.0125}

T_1=27.58\ ^{\circ}C

&<u>For Styrofoam-plywood interface from inside:</u>

\frac{\dot Q}{A} =k_s.\frac{T_1-T_2}{x_s}

20.129=0.04\times \frac{27.58-T_2}{0.05}

T_2=2.41875\ ^{\circ}C

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