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Gwar [14]
2 years ago
12

If you wanted to see a star behind an interstellar dust cloud, what "colour" of light should you look for?

Physics
1 answer:
vladimir1956 [14]2 years ago
5 0

RED colour of light you should look for if you wanted to see a star behind the interstellar cloud.

In our galaxy and other galaxies, an interstellar cloud is often a buildup of gas, plasma, and dust. In other words, an interstellar cloud is a portion of the interstellar medium (ISM), the matter and radiation that is present in the space between star systems in a galaxy, that is denser than typical.

Interstellar dust has an extremely saturated orange to brownish-red tint that turns saturated red when there is a modest quantity of hydrogen emission.

Learn more about interstellar cloud here: brainly.com/question/28138302

#SPJ4

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If 478 watts of power are used in 14 seconds,how much work was done
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Answer:

6692J

Explanation:

Power is defined as the rate at which work is being done.

So,

    Power  = \frac{workdone}{time }  

  Work done  = Power x time

Given parameters:

Power  = 478watts

Time  = 14s

So;

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A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.
ivann1987 [24]

Answer:

v₂ = 176.24 m/s

Explanation:

given,

angle of projectile = 45°

speed = v₁ = 150 m/s

for second trail

speed = v₂ = ?

angle of projectile = 37°

maximum height attained formula,

H_{max}= \dfrac{v^2 sin^2(\theta)}{g}

now,

H_{max}= \dfrac{v_1^2 sin^2(\theta_1)}{g}

H_{max}= \dfrac{v_2^2 sin^2(\theta_2)}{g}

now, equating both the equations

\dfrac{v_2^2}{v_1^2}=\dfrac{sin^2(\theta_1)}{sin^2(\theta_2)}

\dfrac{v_2^2}{150^2}=\dfrac{sin^2(45^0)}{sin^2(37^0)}

   v₂² = 31061.79

   v₂ = 176.24 m/s

velocity of projectile would be equal to v₂ = 176.24 m/s

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