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jenyasd209 [6]
1 year ago
5

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

other bright star, Regulus , has a parallax of 0.042 arcseconds. What is its distance in parsecs
Physics
1 answer:
Vikentia [17]1 year ago
6 0

Answer: Sirius, the brightest star in the sky, is 2.6 parsecs (8.6 light-years) from Earth, giving it a parallax of 0.379 arcseconds. Another bright star, Regulus, has a parallax of 0.042 arcseconds. Then, the distance in parsecs will be,23.46.

Explanation: To find the answer, we have to know more about the relation between the distance in parsecs and the parallax.

<h3>What is the relation between the distance in parsecs and the parallax?</h3>
  • Let's consider a star in the sky, is d parsec distance from the earth, and which has some parallax of P amount.
  • Then, the equation connecting parallax and the distance in parsec can be written as,

                                     d=\frac{1}{P}

  • We can say that,

                                    dP=constant.\\thus,\\d_1P_1=d_2P_2

<h3>How to solve the problem?</h3>
  • We have given that,

                                     d_1=2.6 parsecs.\\P_1=0.379arcseconds.\\P_2=0.042 arcseconds.\\d_2=?

  • Thus, we can find the distance in parsecs as,

                                     d_2=\frac{d_1P_1}{P_2} =23.46 parsecs

Thus, we can conclude that, the distance in parsecs will be, 23.46.

Learn more about the relation connecting distance in parsecs and the parallax here: brainly.com/question/28044776

#SPJ4

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The complete text of the problem is:

<em>"Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1000 m/s2 lasting for at least 1 ms will cause injury. Suppose a small child rolls off a bed that is 0.43 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 1.8 mm. If the floor is carpeted, this stopping distance is increased to about 1.1 cm. Calculate the magnitude and duration of the deceleration in both cases, to determine the risk of injury. Assume the child remains horizontal during the fall to the floor. Note that a more complicated fall could result in a head velocity greater or less than the speed you calculate. "</em>

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<u>Solution:</u>

1) Acceleration: -2336 m/s^2 on the hardwood floor, -382 m/s^2 on the carpeted floor

First of all, we need to calculate the speed of the child just before he hits the floor. This can be done by using the equation

v^2 - u^2 = 2ad

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v is the final speed

u = 0 is the initial speed (the child starts from rest)

a = g = 9.8 m/s^2 is the acceleration of gravity

d = 0.43 m is the distance covered by the child as he falls from the bed

Solving for v,

v=\sqrt{2ad}=\sqrt{2(9.8)(0.43)}=2.9 m/s

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So we can find the acceleration by using again the equation

v'^2 - v^2 = 2ad

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a=\frac{v'^2 - v^2}{2d}=\frac{0-2.9^2}{2(0.0018)}=-2336 m/s^2

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therefore the acceleration is

a=\frac{v'^2 - v^2}{2d}=\frac{0-2.9^2}{2(0.011)}=-382 m/s^2

2) Duration: 1.24 ms for the hardwood floor, 7.59 ms for the carpeted floor

We can find the duration of the collision in both cases by using the equation of the acceleration

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a=-2336 m/s^2

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Therefore, the fall on the hardwood floor can result in injury (since the acceleration is greater than 1,000 m/s2 for more than 1 ms), while the fall on the carpeted floor is not dangerous (much less than 1000 m/s^2).

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