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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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mina [271]
Please mark me brainliest

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3 years ago
What is the only possible value of ml for an electron in an s orbital?
Archy [21]

Answer:

  • zero

Explanation:

m_l     is the magnetic quantum number.

The only possible value for the magnetic quantum number for an electron in an s orbital is 0.

The first three quantun numbers are:

  • n: principal quantum number. It may have positive integer values: 1, 2, 3, 4,5, 6, 7, ...

  • l : Azimuthal or angular momentum quantum number. It may have integer values from 0 to n - 1.

       This quantum number is related to the type (or shape) of the orbital:

        For s orbitals l=0

        For p orbitals l=1

        For d orbitals l=2

         For f orbitals l=3

In this case, it is an s orbital, so we have l=0.

  • m_l , the third quantum number can have integer values  {from-l}   to    {+l}

       Since, for the s orbitals  l=0 , the only possible value for {m_l} is zero.

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3 years ago
An elevator supported by a single cable descends a shaft at a constant speed. The only forces acting on the elevator are the ten
VikaD [51]

Answer:

C. The net work done by the two forces is zero joules.

Explanation:

Work (W) is defined as the product of force F) by the distance (d)the body travels due to this force.  

W= F*d Formula ( 1)

The work is positive (W+) if the force has the same direction of movement of the object.  

The work is negative (W-) if the force has the opposite direction of the movement of the object.

The component of the force that performs work must be parallel to the displacement.  

Calculation of the forces

We apply Newton's second law:

∑F = m*a

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

∑F = m*a  

a = 0 because the elevator moves at constant speed.

∑F=0

T-W = 0

T = W  

Calculation of work done by the forces

W₁ :  work done by the gravity= (+ W*d ) J

W₂ : work done by the tension force=( -T*d) J

Net work done by the two forces(W₁-W₂):

W₁= W₂ .because T = W  

W₁-W₂ =0

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3 years ago
A ball of mass m is thrown into the air in a 45° direction of the horizon, after 3 seconds the ball is seen in a direction 30° f
Rzqust [24]

Answer:

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v_{y}=v_{0y}+a*t=\frac{\sqrt{2} }{2}v_0-9.8t

Now we need to find v_y as a function of v_0. We use the horizontal velocity, which is always the same as follow:

v_x=v_0cos(45\º)=\frac{\sqrt{2} }{2}v_0=v_{t=3}*cos(30\º) \\

We know the angle at 3 seconds:

v_y(t=3)=v_{t=3}*sin(30\º)\\v_{t=3}=\frac{v_y}{sin(30\º)}

Substitute  v_{t=3} in  v_x and then solve for  v_y

\frac{\sqrt{2} }{2}v_0=\frac{v_y*cos(30\º) }{sin(30\º)} \\v_y=\frac{\sqrt{6} }{6}v_0

With this expression we go back to the kinematic equation and solve it for initial speed

\frac{\sqrt{6} }{6} v_0 =\frac{\sqrt{2} }{2}v_0-29.4\\v_0(\frac{\sqrt{6}-3\sqrt{2}}{6} )=-29.4\\v_0=98.37 m/s

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What role does team collaboration play in successfully planning a mission to Mars?
mihalych1998 [28]

Answer:it helps get everyone in the same page

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This is important so everyone knows what they should be doing

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