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Lilit [14]
1 year ago
6

the most likely places where stars and planetary systems are forming in the universe are- the most likely places where stars and

planetary systems are forming in the universe are- in nebulae composed of gas and dust. in regions surrounding quasars. in the rarified space between galaxies. in the centers of galaxies.
Physics
1 answer:
igomit [66]1 year ago
8 0

The most likely places where stars and planetary systems are forming in the universe are in nebulae composed of gas and dust.

Stars are massive, hot, luminous bodies of gas that emit large amounts of radiation and derive their energy from nuclear fusion. A planetary system comprises of a star and all the objects that revolve around it.

Stars are born in nebulae. They are a cluster of gas and dust scattered throughout a galaxy. Gravitational attraction between the gas and dust particles causes turbulence, which gives rise to knots. When these knots acquire sufficient mass, they collapse, and when the reach a certain temperature, nuclear fusion of hydrogen atoms into helium atoms begins. At this point, star has been born.

Planets can be said to be formed as a by-product of star formation, through a process called accretion. Smaller objects stick together through gravity, forming bigger objects. Gravitational pull from the nearby star and the object's own momentum causes the object to revolve round the star. At this point, a planet has been born.

Thus, the most likely places where stars and planetary systems are forming in the universe are in nebulae composed of gas and dust.

Learn more about planetary systems here:

brainly.com/question/13092561

#SPJ4

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8 0
3 years ago
A scaffold of mass 57 kg and length 6.5 m is supported in a horizontal position by a vertical cable at each end. A window washer
Mrac [35]

Explanation:

The given data is as follows.

     m_{1} = 57 kg,        m_{2} = 79 kg

      l_{1} = 6.5 m,         l_{2} = (6.5 - 1.9) m = 4.6 m

(a)  The sum of torque ends about far end is as follows.

    m_{1}g \frac{l_{1}}{2} + m_{2}g \times l_{2} - T \times l_{1} = 0

     57 \times 9.81 \times \frac{6.5}{2} + 79 \times 9.81 \times (6.5 - 1.9) - T \times 6.5 = 0

                     T = 828 N

Therefore, 828 N is the tension in the cable closer to the painter.

(b)  Now, we will calculate the sum about close ends as follows.

m_{1}g \frac{l_{1}}{2} + m_{2}g \times (1.9) - T \times l_{1}    

  57 \times 9.81 \times \frac{6.5}{2} + 79 \times 9.81 \times (1.9) - T \times 6.5 = 0

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Therefore, 506 N is the tension in the cable further from the painter.  

8 0
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The eye is actually a multiple-lens system, but we can approximate it with a single-lens system for most of our purposes. When t
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Explanation:

Given that,

The optical power of the equivalent single lens is 45.4 diopters.

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or

f = 2.2 cm

(b) We need to find how far in front of the retina is this "equivalent lens" located. It is given by using lens formula as :

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So, at 2.2 cm in front of the retina is this "equivalent lens" located.

Hence, this is the required solution.

5 0
4 years ago
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