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3 years ago

What happens when a main-sequence star exhausts its core hydrogen fuel supply?

Physics

1 answer:

Ivanshal [37]3 years ago

8 0

Explanation:

when a main-sequence star exhausts its core hydrogen fuel supply the core starts to shrink ( lack of fusion reactions) and the rest of the star starts to expands. The fusion reaction leaves the main sequence and begin to fuse helium in a shell outside the core. This mass stars become red supergiant and then evolve to become blue super giant.

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The sun emits energy in the form of electromagnetic waves produced by nuclear reactions deep in the sun's interior. Assume that

tatiyna

Answer:

I = W / 4π R_{s}², P = W / 2π c R_{s}², Io /I_{earth} = 10⁴

Explanation:

The intensity is defined as the ratio between the emitted power and the area of the spherical surface

I = P / A

Since the emitted power is constant and has a value of W for this case, let's look for the area of the sphere on the surface of the sun

A = 4π $R_{s}$ ²

I = W / 4π R_{s}²

.- The radiation pressure for total absorption is

P = S / c

Where S is the Pointer vector that is equal to the intensity

Let's replace

P = W / 2π c R_{s}²

.- We repeat for r = R_{s}/2

I₂ = W / 4π (R_{s}/ 2)²

I₂ = 4 W / 4π R_{s}²

I₂ = 4 Io

I₀ = W / 4piRs2

We calculate the radiation pressure

P₂ = I₂ / c

P₂ = 4 I₀ / c

P₂ = 4 (W / 4pi c Rs2)

.- the relationship between these magnitudes is

I₂ / I₀ = 4

P₂ / P₀ = 4

Let's calculate the intensity on the surface where the Earth is

r = 1.50 10¹¹ m

$I_{earth}$ = W / 4π r²

Io / I_{earth} = r² / $R_{s}$ ²

Io /I_{earth} = (1.5 10¹¹ / 6.96 10⁸) 2

Io /I_{earth} = 4.6 10⁴

Io /I_{earth} = 10⁴

4 0

4 years ago

What is another term for a polar molecule?

Zanzabum

they can be called polar covalents

7 0

3 years ago

Read 2 more answers

What do we call seismic waves that are transmitted along the outside of earth?

crimeas [40]

Surface waves fits that description.

8 0

3 years ago

Two particles are fixed to an x axis: particle 1 of charge −1.50 ✕ 10−7 c at x = 6.00 cm, and particle 2 of charge +1.50 ✕ 10−7

sleet_krkn [62]

Answer : $\underset{E_{R}}{\rightarrow} =-2.44\times10^{5}\ \widehat{i} \ \dfrac{N}{C}$

Explanation :

Given that,

Charge of particle 1 = $-1.50\times10^{-7} c$

Distance x = 6 cm

Charge of particle 2 = $1.50\times10^{-7} c$

Distance x = 27 cm

Total distance = $\dfrac{6+27}{2}$

$r = 16.5\ cm$

Particle 1 is at (6,0) and particle 2 is at (27,0) .

Therefore, midway (16.5, 0)

Now, $r = \dfrac{|6-16.5|}{2} = \dfrac{|27-16.5|}{2} = 10.5\ cm$

Formula of electric field

$E = \dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{q}{r^{2}}$

Now, the the electric field due to particle 1

$\underset{E}{\rightarrow}\ = -\dfrac{9\times10^{9}\times1.50\times10^{-7 }}{10.5}\ \widehat{i} \dfrac{N}{C}$

$\underset{E}{\rightarrow} = \dfrac{13.5\times10^{2}}{(10.5\times10^{-2})^{2}}\widehat{i} \dfrac{N}{C}$

$\underset{E}{\rightarrow} = -1.22\times10^{5}\ \widehat{i} \ \dfrac{N}{C}$

Similarly, the electric field due to particle 2

$\underset{E}{\rightarrow} = -1.22\times10^{5}\ \widehat{i} \ \dfrac{N}{C}$

Resultant Electric field

$\underset{E_{R}}{\rightarrow} = \underset{E_{1}}{\rightarrow} + \underset{E_{2}}{\rightarrow}$

$\underset{E_{R}}{\rightarrow} = -2.44\times10^{5}\ \widehat{i} \ \dfrac{N}{C}$

Hence, this is the required answer.

3 0

3 years ago

Sobre un barco, que se mueve en forma rectilínea, y con velocidad constante de 30 [km/h], se mueve un perro en el mismo sentido

almond37 [142]

Answer:

El observador verá correr al perro sobre la cubierta del barco a una velocidad de 40 kilómetros por hora.

Explanation:

Para determinar la velocidad del perro con respecto al observador sentado desde la playa a través del concepto de velocidad relativa, descrito en la siguiente fórmula:

$v_{P/B} = v_{P} - v_{B}$ (1)

Donde:

$v_{P/B}$ - Velocidad del perro relativo al barco, en kilómetros por hora.

$v_{P}$ - Velocidad del perro con respecto al observador, en kilómetros por hora.

$v_{B}$ - Velocidad del barco con respecto al observador, en kilómetros por hora.

Si sabemos que $v_{B} = 30\,\frac{km}{h}$ y $v_{P/B} = 10\,\frac{km}{h}$ , entonces la velocidad del perro con respecto al observador es:

$v_{P} = v_{B} + v_{P/B}$

$v_{P} = 30\,\frac{km}{h} + 10\,\frac{km}{h}$

$v_{P} = 40\,\frac{km}{h}$

El observador verá correr al perro sobre la cubierta del barco a una velocidad de 40 kilómetros por hora.

6 0

3 years ago