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tankabanditka [31]
10 months ago
5

the orion nebula (at least the part we can see) is not very old (yet). while several hot, massive stars have had a chance to for

m in the nebula, none of them has been around long enough to go supernova. so relative to supernova remnants, and other areas where the gasses have been 'mixed' with the wreckage from exploded stars, what can you say about the iron levels you'd expect to see in the orion nebula?
Physics
1 answer:
svetoff [14.1K]10 months ago
3 0

One of the brightest nebulae in the night sky, the Orion Nebula may be seen with the unaided eye. The Trapezium is a young open cluster of four main stars in this magnitude 4 interstellar cloud of ionized atomic hydrogen.

<h3>What is the source of the Orion Nebula's crimson glow?</h3>
  • The hydrogen gas in the Orion Nebula, which is powered by radiation from young stars, gives off a crimson tint. The nebula's blue-violet regions are reflecting radiation from bright, blue-white O-type stars while the red areas are emitting light.
  • The Orion Nebula is one of many massive clouds of gas and dust in our Milky Way galaxy, say contemporary astronomers, and is one of the largest. It is approximately 1,300 light years away from Earth. This enormous hazy cocoon, which measures approximately 30 to 40 light-years in diameter, is generating potentially a thousand stars.  

To learn more about Orion nebula refer to:

brainly.com/question/15575332

#SPJ4

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An archer shoots an arrow 75 m distant target; the bull's-eye of the target is at the same height as the release height of the a
Liono4ka [1.6K]

Answer:

16.25^{\circ}

Explanation:

R = Horizontal range of projectile = 75 m

v = Velocity of projectile = 37 m/s

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

Horizontal range is given by

R=\dfrac{v^2\sin2\theta}{g}\\\Rightarrow \theta=\dfrac{\sin^{-1}\dfrac{Rg}{v^2}}{2}\\\Rightarrow \theta=\dfrac{\sin^{-1}\dfrac{75\times 9.81}{37^2}}{2}\\\Rightarrow \theta=16.25^{\circ}

The angle at which the arrow is to be released is 16.25^{\circ}.

4 0
3 years ago
Uniformly charged ring with 180 nC/m and radius R= 58 cm. Find the magnitude of the electric field in KN/C at a point P on the a
raketka [301]

Answer:

3.135 kN/C

Explanation:

The electric field on the axis of a charged ring with radius R and distance z from the axis is E = qz/{4πε₀[√(z² + R²)]³}

Given that R = 58 cm = 0.58 m, z = 116 cm = 1.16m, q = total charge on ring = λl where λ = charge density on ring = 180 nC/m = 180 × 10⁻⁹ C/m and l = length of ring = 2πR. So q = λl = λ2πR = 180 × 10⁻⁹ C/m × 2π(0.58 m) = 208.8π × 10⁻⁹ C and ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m

So, E = qz/{4πε₀[√(z² + R²)]³}

E = 208.8π × 10⁻⁹ C × 1.16 m/{4π8.854 × 10⁻¹² F/m[√((1.16 m)² + (0.58 m)²)]³}

E = 242.208 × 10⁻⁹ Cm/{35.416 × 10⁻¹² F/m[√(1.3456 m² + 0.3364 m²)]³}

E = 242.208 × 10⁻⁹ Cm/35.416 × 10⁻¹² F/m[√(1.682 m²)]³}

E = 6.839 × 10³ Cm²/[1.297 m]³F

E = 6.839 × 10³ Cm²/2.182 m³F

E = 3.135 × 10³ V/m

E = 3.135 × 10³ N/C

E = 3.135 kN/C

3 0
2 years ago
In an application, Germanium is
Zigmanuir [339]

Answer:

produce electronics

Explanation:

The uses of Germanium are recorded beneath: Germanium's principle use is to deliver strong state hardware, semiconductors and fiber optic frameworks. As a phosphor in fluorescent lights.

7 0
3 years ago
Please help I’ll give brainliest
lara [203]
I think the answer is B
6 0
2 years ago
What is the difference between heat exhaustion and heat stroke?
balu736 [363]
I believe the answer is D, Heat exhaustion involves a lack of sweating, while heat stroke involves extreme sweating. Also just to add the on if heat exhaustion is left untreated then it could turn into a heat stroke.
6 0
3 years ago
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