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QveST [7]
3 years ago
7

Describe the main distinguishing features of spiral, elliptical, and irregular galaxies.

Physics
1 answer:
hram777 [196]3 years ago
5 0

Answer:

Spiral galaxies consist of a flat, rotating disk of stars, gas and dust, and a central concentration of stars known as the bulge. These are surrounded by a much fainter halo of stars, many of which reside in globular clusters.

Elliptical galaxies have smooth, featureless light-profiles and range in shape from nearly spherical to highly flattened, and in size from hundreds of millions to over one trillion stars. In the outer regions, many stars are grouped into globular clusters. Most elliptical galaxies are composed of older, low-mass stars, with a sparse interstellar medium and minimal star formation activity They are often chaotic in appearance, with neither a nuclear bulge nor any trace of spiral arm structure. Collectively they are thought to make up about a quarter of all galaxies.

irregular galaxies were once spiral or elliptical galaxies but were deformed by gravitational action. they are shapeless.

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Convert 43 km/h to m/s.​
SIZIF [17.4K]

Answer:

43km/h to m/s = 11.9444

Explanation:

1 km = 1000 m; 1 hr = 3600 sec. To convert km/hr into m/sec, multiply the number by 5 and then divide it by 18.

6 0
2 years ago
Where is a trench most likely to occur?
Dafna1 [17]
D. convergent plate boundary involving an oceanic plate
7 0
3 years ago
Read 2 more answers
3. How can we determine the volume of a regular object?
JulsSmile [24]
Length•Width•Height is the answer
4 0
3 years ago
3. A certain wire, 3 m long, stretches by 1.2 mm when under tension of 200 N. By how much does
nikitadnepr [17]

Answer:

The extension of the second wire is   e_2 = 0.0024 \  m =  2.4 mm

Explanation:

From the question we are told that

    The length of the wire is L  = 3 \ m

     The elongation of the wire is  e =  1.2mm =  \frac{1.2}{1000} =  0.0012 m

        The tension is F  =  200 \ N

       The length of the second wire is  L_2   =  6 \ m

     

Generally the Young's modulus(Y) of this material is  

        Y  = \frac{stress}{strain }

Where stress =  \frac{F}{A}

    Where A is the area which is evaluated as  

           A = \pi r^2

  and   strain = \frac{extention}{length} =  \frac{e}{L}

   So

        Y  = \frac{\frac{F}{\pi r^2 } }{ \frac{e}{L}  }

Since the wire are of the same material Young's modulus(Y)  is constant

So we have  

              \frac{F * L }{r^2 e}  =  \pi * Y = constant

              F * L   =  constant   * r^2 e

Now the ration between the first and the second wire is

         \frac{F_1}{F_2}  * \frac{L_1}{L_2} =  \frac{r*2_1}{r^2}  *  \frac{e_1}{e_2}

Since tension , radius are constant

   We have

           \frac{L_1}{L_2} =   \frac{e_1}{e_2}

substituting values

          \frac{3}{6} =   \frac{0.0012}{e_2}

          0.5 e_2 =  0.0012

         e_2 = \frac{ 0.0012  }{0.5}

          e_2 = 0.0024 \  m =  2.4 mm

3 0
3 years ago
A proton is moved from a position where the electric potential is 125 V to a position where the electric potential is 275 V. The
kakasveta [241]
We determine the electric potential energy of the proton by multiplying the net electric potential to the charge of the proton. The net electric potential is the difference of the final state to the that of the initial state. So, it would be 275 - 125 = 150 V.

electric potential energy = 150 (<span>1.602 × 10-19) = 2.4x10^-17 J</span>
7 0
3 years ago
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