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drek231 [11]
3 years ago
9

The luminous star Alnilam in the Orion belt is 1,340 light-years away from Earth. Use the conversion factor 1 parsec = 3.262 lig

ht-years to find the closest conversion of this distance from light-years to parsecs (pc).

Physics
2 answers:
svlad2 [7]3 years ago
8 0

The answer is 410.8 pc.

Angelina_Jolie [31]3 years ago
6 0

Answer:

The distance to the star Alnilam in the Orion belt is 410.8 Pc

Explanation:

The angle due to the change in position of a nearby object against the background stars is known as parallax. This angle is gotten when the position of the object is measured in January and then in July according to the configuration of the Earth with respect to the Sun in those months (see the image below).

       

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:

\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), which is the distance of a star with a parallax angle of 1 arc second.

For the case of the star Alnilam in the Orion belt ( 1340 light-years):

Since 1 parsec is equivalent to 3.262 light-years, therefore:

d = 1340 ly \cdot \frac{1pc}{3.262 ly} ⇒ $410.8 Pc

Notice how both , Light years and parsecs are units for distance.          

Key terms:

Parsec: Parallax of arc second.

Light-year: Distance traveled by light in one year.

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Answer:

VB -  VA  =  - 33.4

Explanation:

Generally the workdone in moving the proton is mathematically represented as

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    KE_i  =  \frac{1}{2} m v_a^2

Here v_a is the velocity at A with value  50 m/s

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    KE_i  =  \frac{1}{2} (1.67*10^{-27}) * 50^2

    KE_i  = 2.09 *10^{-24} \  J

Also  

     KE_f  =  \frac{1}{2} m v_b^2

Here v_a is the velocity at A with value 80 km/s = 80000 m/s

=>   KE_f  =  \frac{1}{2} (1.67*10^{-27}) * 80000^2

=>   KE_f  = 5.34 *10^{-18} \  J

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    W  =   5.34 *10^{-18}  - 2.09 *10^{-24}

     W  =   5.34 *10^{-18}  m/s

Now this workdone is also mathematically represented as

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    q *  V =   5.34 *10^{-18}

Here  q =  1.60*10^{-19} C

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        V =   \frac{5.34 *10^{-18} }{1.60*10^{-19}}

         V =   33.4 \  V

Generally proton movement is in the direction of the electric field it means that  VA>VB

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2 years ago
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Goryan [66]
Yeah that’s is correct
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3 years ago
A man has 887.5 J of kinetic energy while running with a velocity of 5 m/s. What is his mass?
monitta

Answer:

The mass of the man is 71 kg

Explanation:

Given;

kinetic energy of the man, K.E = 887.5 J

velocity of the man, v = 5 m/s

The mass of the man is calculated as follows;

K.E = ¹/₂mv²

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2K.E = mv²

m = 2K.E / v²

m = (2 x 887.5) / (5)²

m = 71 kg

Therefore, the mass of the man is 71 kg

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