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almond37 [142]
3 years ago
6

When a high-mass main sequence star runs out of both hydrogen and helium in its core, the core begins to fuse

Physics
2 answers:
faust18 [17]3 years ago
7 0

Answer:

The core beings to fuse carbon into heavier elements.

Explanation:

As the star runs out of hydrogen and helium, the energy production in the core stops, and there isn't enough outward pressure to hold the star up. The start contracts, and this heats up the core (pressure increases, temperature increases)  creating temperatures high enough to fuse carbon into heavier elements like oxygen, silicon, neon, sulfur, magnesium, and iron.

But a star can sustain itself on Carbon fusion only for so long. In about 600 years (how long exactly depends on the star ) the star runs out of carbon and what happens next depends on the mass of the star; the star either collapses into a neutron star, or begins fusion heavier elements like neon.

Anastasy [175]3 years ago
4 0

Answer:

In the assignment, the slide has two questions. The first one (what you asked) is carbon. The second answer is supernova.

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You obtain a spectrum of an object in space. The spectrum consists of a number of sharp, bright emission lines. Is this object a
Andreyy89

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4 years ago
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A solid circular shaft and a tubular shaft, both with the same outer radius of c=co = 0.550 in , are being considered for a part
Norma-Jean [14]

Answer:

The power for circular shaft is 7.315 hp and tubular shaft is 6.667 hp

Explanation:

<u>Polar moment of Inertia</u>

(I_p)s = \frac{\pi(0.55)4}2

      = 0.14374 in 4

<u>Maximum sustainable torque on the solid circular shaft</u>

T_{max} = T_{allow} \frac{I_p}{r}

         =(14 \times 10^3) \times (\frac{0.14374}{0.55})

         = 3658.836 lb.in

         = \frac{3658.836}{12} lb.ft

        = 304.9 lb.ft

<u>Maximum sustainable torque on the tubular shaft</u>

T_{max} = T_{allow}( \frac{Ip}{r})

          = (14 \times10^3) \times ( \frac{0.13101}{0.55})

          = 3334.8 lb.in

          = (\frac{3334.8}{12} ) lb.ft

          = 277.9 lb.ft

<u>Maximum sustainable power in the solid circular shaft</u>

P_{max} = 2 \pi f_T

          = 2\pi(2.1) \times 304.9

          = 4023.061 lb. ft/s

          = (\frac{4023.061}{550}) hp

          = 7.315 hp

<u>Maximum sustainable power in the tubular shaft</u>

P _{max,t} = 2\pi f_T

            = 2\pi(2.1) \times 277.9

            = 3666.804 lb.ft /s

            = (\frac{3666.804}{550})hp

            = 6.667 hp

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Electromagnetic force.

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seraphim [82]

Answer:

i think it's a short circuit

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