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Lunna [17]
3 years ago
12

If stellar parallax can be measured to a precision of about 0.01 arcsec using telescopes on the Earth to observe stars, to what

distance does this correspond in space?
Physics
1 answer:
marin [14]3 years ago
7 0

Answer:

It corresponds to a distance of 100 parsecs away from Earth.

Explanation:

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

It is defined in a analytic way as it follows:

       

\tan{p} = \frac{1AU}{d}

Where d is the distance to the star.

p('') = \frac{1}{d} (1)  

Equation (1) can be rewritten in terms of d:

d(pc) = \frac{1}{p('')} (2)

Equation (2) represents the distance in a unit known as parsec (pc).

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 (1AU).

For the case of   (p('') = 0.01):

d(pc) = \frac{1}{0.01}

d(pc) = 100

Hence, it corresponds to a distance of 100 parsecs away from Earth.

<em>Summary:</em>

Notice how a small parallax angle means that the object is farther away.

Key terms:

Parsec: Parallax of arc second

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Suppose that you are headed toward a plateau 55 meters high. If the angle of elevation to the top of the plateau is 40degrees​,
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Answer:

x=65.55m

Explanation:

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From trigonometry properties:

tan(40^{o} )=\frac{55m}{x} \\x=\frac{55m}{tan(40^{o} )} \\x=65.55m

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A box is hanging from two strings. String #1 pulls up and left, making an angle of 50° with the horizontal on the left, and stri
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Answer:

mb = 3.75 kg

Explanation:

System of forces in balance

ΣFx =0  

ΣFy = 0

Forces acting on the box

T₁ : Tension in string 1 ,at angle of 50° with the horizontal on the left

T₂  = 40 N : Tension in string 2, at angle of 75° with the horizontal on the right.

Wb :Weightt of the box (vertical downward)

x-y T₁ and T₂ components

T₁x= T₁cos50°

T₁y= T₁sin50°

T₂x= 30*cos75° = 7.76 N

T₂y= 30*sin75° = 28.98 N

Calculation of the Wb

ΣFx = 0  

T₂x-T₁x = 0

T₂x=T₁x

7.76 = T₁cos50°

T₁ = 7.76 /cos50° = 12.07 N

ΣFy = 0  

T₂y+T₁y-Wb = 0

28.98 + 12.07(cos50°) = Wb

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Calculation of the mb ( mass of the box)

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mb = Wb/g

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On Titan, the largest moon of of Saturn did the Cassini-Huygens probe land in 2004.

To find the answer, we have to know more about the Cassini-Huygens Mission.

<h3>What is Cassini-Huygens mission?</h3>
  • Before arriving at its final destination of Saturn in 2004 and beginning a series of flybys of Saturn's moons, the spacecraft contributed to studies of Jupiter for six months in 2000.
  • In the same year, it launched the Huygens probe to explore Titan's atmosphere and surface makeup on Saturn's moon.
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Thus, we can conclude that, on Titan, the largest moon of of Saturn did the Cassini-Huygens probe land in 2004.

Learn more about the Cassini-Huygens mission here:

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