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Simora [160]
3 years ago
13

What is stellar parallax?a.It describes the fact that stars are actually moving relative to one another, even though to our eyes

the stars appear fixed in the constellations. b.It is the change in the set of constellations that we see at different times of year in the evening sky. c.It is the slight back-and-forth shifting of star positions that occurs as we view the stars from different positions in Earth's orbit of the Sun. d.It is the daily rise and set of the stars.

Physics
1 answer:
bagirrra123 [75]3 years ago
3 0

Answer:

Option C

C. It is the slight back-and-forth shifting of star positions that occurs as we view the stars from different positions in Earth's orbit of the Sun.

Explanation:

The angle due to the change in position of a nearby object against the background stars it is known as parallax.          

The parallax angle can be used to find out the distance by means of triangulation. Making a triangle between the nearby star, the Sun and the Earth(as is shown in the image below), knowing that the distance between the Earth and the Sun (150000000 Km) is defined as 1 astronomical unit.

d(pc) = \frac{1}{p('')}  (1)

Equation (1) represents the distance in a unit known as parsec (pc).

Key terms:

Parsec: Parallax of arc second      

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Find the cube roots of 27(cos 327° + i sin 327° ). Write the answer in trigonometric form.
Sati [7]

Answer:

z^{\frac{1}{3} }= -0.978 + i\cdot 2.836, z^{\frac{1}{3} }= -1.967 - i\cdot 2.265, z^{\frac{1}{3} }= 2.945 - i\cdot 0.571

Explanation:

The cube root of the complex number can determined by the following De Moivre's Formula:

z^{\frac{1}{n} } = r^{\frac{1}{n} }\cdot \left[\cos\left(\frac{x + 2\pi\cdot k}{n} \right) + i\cdot \sin\left(\frac{x+2\pi\cdot k}{n} \right)\right]

Where angles are measured in radians and k represents an integer between 0 and n - 1.

The magnitude of the complex number is 27 and the equivalent angular value is 1.817\pi. The set of cubic roots are, respectively:

k = 0

z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{1.817\pi}{3} \right)+i\cdot \sin\left(\frac{1.817\pi}{3} \right)]

z^{\frac{1}{3} }= -0.978 + i\cdot 2.836

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z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{3.817\pi}{3} \right)+i\cdot \sin\left(\frac{3.817\pi}{3} \right)]

z^{\frac{1}{3} }= -1.967 - i\cdot 2.265

k = 2

z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{5.817\pi}{3} \right)+i\cdot \sin\left(\frac{5.817\pi}{3} \right)]

z^{\frac{1}{3} }= 2.945 - i\cdot 0.571

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effort is the answer

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