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svet-max [94.6K]
4 years ago
11

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

Physics
1 answer:
melomori [17]4 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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A straight wire of length 0.53 m carries a conventional current of 0.2 amperes. What is the magnitude of the magnetic field made
olga55 [171]

Explanation:

It is given that,

Length of wire, l = 0.53 m

Current, I = 0.2 A

(1.) Approximate formula:

We need to find the magnitude of the magnetic field made by the current at a location 2.0 cm from the wire, r = 2 cm = 0.02 m

The formula for magnetic field at some distance from the wire is given by :

B=\dfrac{\mu_oI}{2\pi r}

B=\dfrac{4\pi \times 10^{-7}\times 0.2\ A}{2\pi \times 0.02\ m}

B = 0.000002 T

B=10^{-5}\ T

(2) Exact formula:

B=\dfrac{\mu_oI}{2\pi r}\dfrac{l}{\sqrt{l^2+4r^2} }

B=\dfrac{\mu_o\times 0.2\ A}{2\pi \times 0.02\ m}\times \dfrac{0.53\ m}{\sqrt{(0.53\ m)^2+4(0.02\ m)^2} }

B = 0.00000199 T

or

B = 0.000002 T

Hence, this is the required solution.

4 0
3 years ago
Two hockey players are about to collide on the ice, One player has a mass of
Furkat [3]

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19 kg x m/s East

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Two masses are connected by a string which passes over a pulley with negligible mass and friction. One mass hangs vertically and
monitta

Answer:

a=2,5m/s^2

Explanation:

From the question we are told that:

Coefficient of kinetic friction \mu= 0.200

Vertical Mass M_v=3kg

Horizontal mass M_h=3.00kg  

Generally the equation for kinetic force F_k is mathematically given by

F_k=\mu N\\F_k=0.2*3\\F_k=0.6

Generally the equation for T is mathematically given by

For M_v=3kg3g-T=3a

For M_h=3kg

T=M_v V+F_k\\T=3.0a+0.6

Therefore substituting

3-3a-0.6=3a\\2.4g=6a

a=2,5m/s^2

3 0
3 years ago
Three carts of masses 4.0 kg, 10kg, and 3.0 kg move on a frictionless track with speeds of v1 = 5.0m/s, v2=3.0m/s, and v3=-3.6 m
gogolik [260]

2.24 m/s is the calculated velocity.

Initial velocity (u) squared plus two times the acceleration (a) times the displacement equals final velocity (v) squared (s). Final velocity (v) is equal to the square root of initial velocity (u) squared plus two times the acceleration (a) times displacement when v is the variable being solved for (s).

The cart's masses and speeds are known.

M1 = 4.00 kg, M2 = 10.0 kg, M3 = 3.00 kg, etc.

v1 = 5.00 m/s = 5.00 m/s, v2 = 3.00 m/s = 3.00 m/s, v3 = -4.00 m/s = 4.00 m/s, and m1v1+m2v2+m3v3 = (m1+m2+m3) v=d frac m 1v 1+m 2v 2+m 3v 3, where (m1+m2 + m3) is the product of (v1 v 1+m2v2+m3v3).

"m 1+m 2+m 3" is equivalent to "m 1+m2+m3/m1v 1+m2v2 +m3v3"

the three carts' final velocities are calculated as follows: v=d frac

{4.00kg\sdot5.00m/s+10.0kg\sdot3.00m/s-3.00kg\sdot4.00m/s} 4.24m/s = 4.50kg+10.0kg+3.00kg vs. 4.50kg+10.0kg+3.00kg

5.00m/s/4.00kg/5.00m/s+10.0m/s/3.00m/s/4.00m/s =2.24m/s.

2.24 m/s is the calculated final velocity.

Learn more about velocity here-

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