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svet-max [94.6K]
3 years ago
11

Interstellar matter is quite evenly distributed throughout the milky way galaxy. true or false

Physics
1 answer:
melomori [17]3 years ago
4 0

Answer:

False

Explanation:

In addition to stars, our galaxy contains abundant diffuse matter that is distributed throughout its volume and constitutes what we call the interstellar medium. This medium plays a fundamental role in the life cycle of the stars, since it is where the matter from which they are born resides, and it is the place to which it returns when the stars expel their outer layers at death.

The interstellar medium is a complex environment. <u>Its matter is </u><u>not </u><u>distributed uniformly</u>, but consists of different phases with temperatures ranging from a few degrees Kelvin (near absolute zero) in the areas of star formation to the millions of degrees Kelvin observed in supernova remnants. The densities of interstellar matter also vary orders of magnitude according to the phase, but they are always so low that they rival those that can be achieved in the best vacuum chambers of terrestrial laboratories. Depending on the density and temperature conditions, interstellar matter is in a molecular, atomic, or ionized state, although the state is not permanent, since matter circulates between the different phases in a continuous cycle of evolution on a galactic scale.

Due to the very different characteristics of its multiple phases, the interstellar medium has to be studied using various observational techniques and different types of telescopes. The coldest components of the interstellar medium do not emit visible light, and require the observation of telescopes sensitive to the weak emission of radio waves that this material produces. Using different radio telescopes, such as the 40-meter diameter of the Yebes Observatory, which the Institute of Radio Astronomy Millimeter, to which the IGN belongs, has in Grenoble and Granada, or the recently opened Atacama Large Millimeter / submillimeter Array in the Atacama desert in Chile, astronomers from the National Astronomical Observatory contribute to characterize the physical and chemical properties of the molecular clouds where stars are born and of the circumestellar shells produced by the stars in the last stages of their lives . The study of these regions is helping to complete our knowledge of the most unknown phases of the complex life cycle of stars.

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g The electric power needs of a community are to be met by windmills with 40-m-diameter rotors. The windmills are to be located
Ksenya-84 [330]

Answer:

Explanation:

Given Data

The diameter of the wind mills is d = 40m

Velocity of the air is V = 6 m / s

Required power output is:  P ₀ = 2100 k W

Expression to calculate the exergy of the air is

E = V ² / 2

Substitute the value in above expression

E = ( 6 m / s ) ² / 2

E = 18 m ² / s ² x (1kJ/kg / 1000m²/s²)

E = 0.018 k J / k g

Expression to calculate the density of the air is

P v =m R T

m /v = P  /RT ⋯ ⋯( I )

Here  

m  is the mass of the air,  

v  is the volume of the air,  

P  is the atmospheric pressure,  

T  is the standard temperature at the atmospheric pressure and  

R  is the gas constant

As the density is

ρ = m /V

Substitute the value in expression (I)

ρ = 101  kP a /( 0.287 k J / k g ⋅ K ) ( 298 K )

ρ = 1.180 k g / m ²

Expression to calculate the mass flow rate is

m = ρ A V ⋯ ⋯ ( I I )

Here  A  is the area of the windmill

Expression to calculate the  A  is

A = π /4  d ²

Substitute the value in above expression

A = π /4 ( 40 m ²)

A = 1256.63 m ²

Substitute the value in expression (II)

m = ( 1.180 k g / m ³) ( 1256.63 m ²) ( 6 m / s )

m = 8896.94  k g / s

Expression to calculate the maximum power available to the windmill is

P w = m ( V ² /2 )

Substitute the value in above expression

P w = 8896.94  k g / s ( (6m/s)²/2 )

P w = 160144.92 W  × ( 1 W /1000 k W )

P w = 160.144 k W

Expression to calculate the number of windmills required is

n = P o /P w

Substitute the value in above expression

n=2100kw/160.144kw

n=13.11

8 0
3 years ago
Energy transformation in wound spring of a toy car?<br>give your own answrr​
BaLLatris [955]

Answer:

The work done in winding the spring gets stored in the wound up spring in the form of elastic potential energy (i.e potential energy due to change in shape). ... During this process, the potential energy stored in it gets converted to kinetic energy. This turns the wheels of the toy car.

Explanation:

7 0
2 years ago
Read 2 more answers
The volume of a fixed mass of gas is directly proportional to its
Leni [432]

Answer: Charles's law

Explanation:

Charles's law is one of the gas laws, and it explains the effect of temperature changes on the volume of a given mass of gas at a constant pressure. Usually, the volume of a gas decreases as the temperature decreases and increases as the temperature also increases.

Mathematically, Charles's law can be expressed as:

V ∝ T

V = kT or (V/T) = k

where v is volume, T is temperature in Kelvin, and a k is a constant.

7 0
3 years ago
a cone is constructed by cutting a sector from a circular sheet of metal with radius . the cut sheet is then folded up and welde
Svet_ta [14]

The expression for the radius and height of the cone can be obtained from

the property of a function at the maximum point.

  • The \ radius, \ of \ the \ base \ of \ the \ cone \ is \ \sqrt{ \dfrac{3}{4}} \times radius \ of \ circular \ sheet \ metal
  • The height of the cone is half the length of the radius of the circular sheet metal.

Reasons:

The part used to form the cone = A sector of a circle

The length of the arc of the sector = The perimeter of the circle formed by the base of the cone.

Volume \ of \ a \ cone = \dfrac{1}{3} \cdot  \pi \cdot r^3 \cdot h

  • Volume \ of \ a \ cone, \, V = \dfrac{1}{3} \cdot  \pi \cdot r^3 \cdot \sqrt{(s^2- r^2)}

θ/360·2·π·s = 2·π·r

Where;

s = The radius of he circular sheet metal

h = s² - r²

  • \dfrac{dV}{dr} = \dfrac{d}{dr}  \left(\dfrac{1}{3} \cdot  \pi \cdot r^3 \cdot \sqrt{(s^2- r^2)}\right) = \dfrac{\pi \cdot (3 \cdot r^2 \cdot  s^2 - 4 \cdot r^4)}{\sqrt{(s^2- r^2)}} = 0

3·r²·s² - 4·r⁴ = 0

3·r²·s² = 4·r⁴

3·s² = 4·r²

\underline{\left  \right. The \ radius, \, r =\sqrt{ \dfrac{3}{4}} \cdot s}

\underline{The \ height, \, h =\sqrt{s^2 - \dfrac{3}{4}\cdot s^2} = \dfrac{s}{2}}}

Learn more here:

brainly.com/question/14466080

3 0
2 years ago
A ball is dropped from the top of a 77 m building. With what speed does the ball hit the ground? _________ m/s
vitfil [10]

Answer:

38.87 m/s

Explanation:

Given that the ball is dropped from a height = 77 m

u = 0 m/s

s = 77 m

a = g = 9.81 m/s²

Applying the expression as:

v^2-u^2=2as

Applying values as:

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 77+0^2}\\\Rightarrow v=38.87\ m/s

<u>The speed with which the ball hit the ground = 38.87 m/s</u>

3 0
3 years ago
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