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Illusion [34]
1 year ago
8

irius, the brightest star in the sky, is 2.6 parsecs (8.6 light-years) from Earth, giving it a parallax of 0.379 arcseconds. Ano

ther bright star, Regulus , has a parallax of 0.042 arcseconds. What is its distance in parsecs
Physics
1 answer:
Norma-Jean [14]1 year ago
6 0

The actual distance of Regulus from Earth is 23.81 parsecs.

Given:

Parallax of Regulus, p = 0.042 arc seconds

Calculation:

When an observer changes their position, an apparent change in the object's position takes place. This change can be calculated using the angle ( or semi-angle) made by the observer and object i.e. the angle made between the two lines of observation from the object to the observer.

Thus from the relation of parallax of a celestial body we get:

S = 1/ tan p ≈ 1 / p

where S is the actual distance between the object and the observer

            p is the parallax angle observed

Here for Regulus, we get:

S = 1 / p

  = 1 / (0.042)                                     [ 1 parsecs = 1 arcseconds ]

  = 23.81 parsecs

We know that,

1 parsecs = 3.26 light-years = 206,000 AU

Converting the actual distance into light years we get:

23.81 parsecs = 23.81 × (3.26 light yrs) = 77.658 light-years

Therefore, the actual distance of Regulus from Earth is 23.81 parsecs which is 77.658 in light years.

Learn more about astronomical units here:

<u>brainly.com/question/16471213</u>

#SPJ4

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Arthur conducts a controlled experiment several times. The results of each test show that his hypothesis is not supported. Which
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Hello. You did not inform the experiment that Arthur is conducting, which makes it impossible for your question to be answered accurately. However, I will try to help you in the best possible way.

The hypothesis is an assumption that is made before the experiment is carried out. This hypothesis is formed with the observation of some phenomenon of nature where the researcher believes that two or more elements interact to form a result. In this case, the experiment is carried out to determine whether the assumption, that is, the hypothesis is false or true. In the event that an experiment determines that the hypothesis is false, two things may have occurred: (a) the experiment was set up, or analyzed incorrectly, (b) the elements tested have no relation to the observed phenomenon.

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3 years ago
What will happen when a light bulb is removed from the circuit below, leaving a gap?
Charra [1.4K]

Answer:

The remaining light bulbs will go out.

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3 years ago
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A container of gas is at a pressure of 1.3x10^5 Pa and a volume of 6 m^3. How much work is done by the gas if it expands at a co
Ivenika [448]
In thermodynamics, work of a system at constant pressure conditions is equal to the product of the pressure and the change in volume. It is expressed as follows:

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Hope this answers the question. Have a nice day.
5 0
3 years ago
A snail travels 300 cm in 4 minutes.calculate speed of snail in m/s
LekaFEV [45]

Answer:

\frac{0.0125m}{s}

Explanation:

In order to solve this question we need to know that  speed = \frac{distance}{time}. Then we need to convert 4 minutes into seconds and cm into m. We can do that by multiplying the number of minutes by 60 (because there is 60 seconds in one minute) and dividing the number of cm by 100 (because there is 100 cm in one m). So.......

4min = 4 x 60s = 240s

300cm = 300/100 m = 3m

Now we know that distance = 300m, and that the time = 4min = 240s ⇒

⇒ speed=\frac{distance}{time} = \frac{3m}{240s} = \frac{0.0125m}{s}

5 0
3 years ago
A hot air balloon is moving vertically upwards at a velocity of 3m/s. A sandbag is dropped when the balloon reaches 150m. How lo
gregori [183]

This is a perfect opportunity to stuff all that data into the general equation for the height of an object that has some initial height, and some initial velocity, when it is dropped into free fall.

                       H(t)  =  (H₀)  +  (v₀ T)  +  (1/2 a T²)

 Height at any time 'T' after the drop =

                          (initial height) +

                                              (initial velocity) x (T) +
                                                                 (1/2) x (acceleration) x (T²) .

For the balloon problem ...

-- We have both directions involved here, so we have to define them:

     Upward  = the positive direction

                       Initial height = +150 m
                       Initial velocity = + 3 m/s

     Downward = the negative direction

                     Acceleration (of gravity) = -9.8 m/s²

Height when the bag hits the ground = 0 .

                 H(t)  =  (H₀)  +  (v₀ T)  +  (1/2 a T²)

                  
0    =  (150m) + (3m/s T) + (1/2 x -9.8 m/s² x T²)

                   -4.9 T²  +  3T  + 150  =  0

Use the quadratic equation:

                         T  =  (-1/9.8) [  -3 plus or minus √(9 + 2940)  ]

                             =  (-1/9.8) [  -3  plus or minus  54.305  ]

                             =  (-1/9.8) [ 51.305  or  -57.305 ]

                          T  =  -5.235 seconds    or    5.847 seconds .

(The first solution means that the path of the sandbag is part of
the same path that it would have had if it were launched from the
ground 5.235 seconds before it was actually dropped from balloon
while ascending.)

Concerning the maximum height ... I don't know right now any other
easy way to do that part without differentiating the big equation.
So I hope you've been introduced to a little bit of calculus.

                    H(t)  =  (H₀)  +  (v₀ T)  +  (1/2 a T²)

                  
H'(t)  =  v₀ + a T

The extremes of 'H' (height) correspond to points where h'(t) = 0 .

Set                                  v₀ + a T  =  0

                                      +3  -  9.8 T  =  0

Add 9.8 to each  side:   3               =  9.8 T

Divide each side by  9.8 :   T = 0.306 second

That's the time after the drop when the bag reaches its max altitude.

Oh gosh !  I could have found that without differentiating.

- The bag is released while moving UP at 3 m/s .

- Gravity adds 9.8 m/s of downward speed to that every second.
So the bag reaches the top of its arc, runs out of gas, and starts
falling, after
                       (3 / 9.8) = 0.306 second .

At the beginning of that time, it's moving up at 3 m/s.
At the end of that time, it's moving with zero vertical speed).
Average speed during that 0.306 second = (1/2) (3 + 0) =  1.5 m/s .

Distance climbed during that time = (average speed) x (time)

                                                           =  (1.5 m/s) x (0.306 sec)

                                                           =  0.459 meter  (hardly any at all)

     But it was already up there at 150 m when it was released.

It climbs an additional 0.459 meter, topping out at  150.459 m,
then turns and begins to plummet earthward, where it plummets
to its ultimate final 'plop' precisely  5.847 seconds after its release.  

We can only hope and pray that there's nobody standing at
Ground Zero at the instant of the plop.

I would indeed be remiss if were to neglect, in conclusion,
to express my profound gratitude for the bounty of 5 points
that I shall reap from this work.  The moldy crust and tepid
cloudy water have been delicious, and will not soon be forgotten.

6 0
3 years ago
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