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stellarik [79]
4 years ago
9

The hot glowing surfaces of stars emit energy in the form of electromagnetic radiation. It is a good approximation to assume e =

1 for these surfaces. Find the radii of the following stars (assumed to be spherical):________
(a) Rigel, the bright blue star in the constellation Orion, which radiates energy at a rate of 2.7×1032W and has surface temperature 11,000 K;
(b) Procyon B (visible only using a telescope), which radiates energy at a rate of 2.1×1023W and has surface temperature 10,000 K.
(c) Compare your answers to the radius of the earth, the radius of the sun, and the distance between the earth and the sun. (Rigel is an example of a supergiant star, and Procyon B is an example of a white dwarf star.)
Physics
1 answer:
Svetllana [295]4 years ago
7 0

Answer:

Explanation:

(a) Rigel, 2.7x10^32W, T = 11,000K

But L = 4pR²sT⁴

L = 2.7x10^32W, T = 11,000K, s= 5.67 x 10^-8, R= radius in meters

Rigel parallax, p = 0.00378 arc sec

Substituting the various values and making R the subject of the formula

R² = L/(4psT⁴)

R² = 2.7x10^32/(4 x 0.003878 x 5.67x10^-8 x (11,000)⁴)

R² = 2.7x10^32/1.2877x10^7

R² = 2.096761668 x 10^25

R = 4.579041021 x 10^12meters

(b)

Procyon B, 2.1x10^23W, T = 10,000K

But L = 4pR²sT⁴

L = 2.1x10^23W, T = 10,000K, s= 5.67 x 10^-8, R= radius in meters

Procyon B parallax, p = 0.00284 arc sec

R² = 2.1x10^23/(4 x 0.00284 x 5.67x10^-8 x (10,000)⁴)

R² = 2.1x10^23/(6.441 x 10^6)

R² = 3.26036 x 10^16

R = 1.80565 x 10^8 meters

(c) The radius of Rigel is given as 4.579041021 x 10^12meters and the radius of Procyon B is given by 1.80565 x 10^8 meters shows the remarkable difference between a super-giant star(Rigel) and a white dwarf star (Procyon B)

The radius of the sun a red star is 6.96 x 10^8meters which shows a certain level of resemblance with the size of a dwarf white star Procyon B.

The sun is larger than Procyon B as estimated above.

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Answer:

a) 0.167 μC/m^2

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Explanation:

Hello!

First let's find the surface charge density:

a)

Since thesatellite is metallic, the accumalted charge will be uniformly distribuited on its surface. Therefore the charge density σ will be:

σ = Q/A

Where A is the area of the satellite, which is:

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Now let's calculate the electric field

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Just outside the surface of the satellite the elctric field will be:

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what is the angular speed w of the system immediately after the collision in terms f the sstem parameters and I
denis23 [38]

Answer: hello some part of your question is missing attached below is the missing detail

answer :

<em>w</em>f = M( v cos∅ )D / I

Explanation:

The Angular speed <em>wf  </em>of the system after collision in terms of the system parameters  and I can be expressed as

considering angular momentum conservation

Li = Lf

M( v cos∅ ) D = ( ML^2 / 3  + mD^2 ) <em>w</em>f

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In terms of system parameters and I

<em>w</em>f = M( v cos∅ )D / I

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