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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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Answer: 259.2 KJ

Explanation:

The formula calculate work don in a circuit is given by :-

W=QV, where Q is charge and V is the potential difference.

The formula to calculate charge in circuit :-

Q=It, where I is current and t is time.

Given : Current : I=3A

Potential difference : V=12\ V

Time : t=2\ hr=2(3600)\text{ seconds}=7200\text{ seconds}

Now, Q=3(7200)=21,600\ C

Then,  W=(21600)(12)=259,200\text{ Joules}=259.2\text{ KJ}

Hence, the work done = 259.2 KJ

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Answer:

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3 years ago
Which of the following represent units of capacitance? You may choose more than one correct answer:
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Two uniform, solid cylinders of radius R and total mass M are connected along their common axis by a short, light rod and rest o
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Explanation:

A) To prove the motion of the center of mass of the cylinders is simple harmonic:

System diagram for given situation is shown in attached Fig. 1

We can prove the motion of the center of mass of the cylinders is simple harmonic if

a_{x} = -\omega^{2}  x

where aₓ is acceleration when attached cylinders move in horizontal direction:

<h3>PROOF:</h3>

rotational inertia for cylinders  is given as:

                                  I=\frac{1}{2}MR^{2} -----(1)

Newton's second law for angular motion is:

                                             ∑τ = Iα ------(2)

For linear motion in horizontal direction it is:

                                             ∑Fₓ = Maₓ ------ (3)

By definition of torque:

                                               τ  = RF --------(4)        

Put (4) and (1) in (2)

                                       RF=\frac{1}{2}MR^{2}\alpha

                                       RF=\frac{1}{2}MR^{2}\alpha

from Fig 3 it can be seen that fs is force by which the cylinders roll without slipping as they oscillate

So above equation becomes

                                   f_{s}=\frac{1}{2}MR\alpha------ (5)

As angular acceleration is related to linear by:

                                          a= R\alpha

Eq (5) becomes

                                    f_{s}=\frac{1}{2}Ma_{x}---- (6)

aₓ shows displacement in horizontal direction

From (3)

                                              ∑Fₓ = Maₓ

Fₓ is sum of fs and restoring force that spring exerts:

                                  \sum F_{x} = f_{s} - kx ----(7)

Put (7) in (3)

                                  f_{s} - kx  = Ma_{x}[/tex] -----(8)

Using (6) in (8)

                               \frac{1}{2}Ma_{x} - kx =Ma_{x}

                                     a_{x} = \frac{2k}{3M} x --- (9)

For spring mass system

                                  a= -\omega^{2} x ----- (10)

Equating (9) and (10)

                                  \omega^{2} = \frac{2k}{3M}

\omega = \sqrt{ \frac{2k}{3M}}

then (9) becomes

                                a_{x} = - \omega^{2}x

(The minus sign says that x and  aₓ  have opposite directions as shown in fig 3)

This proves that the motion of the center of mass of the cylinders is simple harmonic.

<h3 /><h3>B) Time Period</h3>

Time period is related to angular frequency as:

                                   T=\frac{2\pi }{\omega}

                                  T = 2\pi \sqrt{\frac{3M}{2k}

                           

 

5 0
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Answer:

Do u have a picture of the graph?

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