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Mrac [35]
4 years ago
14

Explain the origin of the magnitude designation for determining the brightness of stars. Why does it seem to go backward, with s

maller numbers indicating brighter stars?
Physics
1 answer:
Mashcka [7]4 years ago
5 0

Answer:

Hipparchus was an ancient Greek who classified stars based on the brightness in 129 B.C. He grouped the brightest stars and ranked them 1 (first magnitude) and dimmest stars as 6 (sixth magnitude). Thus, the smaller numbers indicated brighter stars. Now, the scale extends in negative axis as well. More the negative number, brighter is the star. For example, Sun has magnitude -26.74.

This the apparent magnitude which means the classification is based on the brightness of the star as it appears from the Earth.

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Neil pogo sticks to his science class, but stops to pick up his backpack on his way. He travels 8 m east, then 4 m west. what di
Likurg_2 [28]

Answer:

8.9 m NorthEast

Explanation:

a² + b² = c²

8² + 4² = c²

64 + 16 = c²

80 = c²

c = √80

c ≈ 8.9 m NorthEast

8 0
3 years ago
Please help on this one?
Sati [7]
I’m pretty sure it’s A
4 0
4 years ago
An object that is magnetic all of the time is called a.
Novosadov [1.4K]

Answer:

Bar magnets are permanent magnets. This means that their magnetism is there all the time and cannot be turned on or off as it can with electromagnets .

Explanation:

5 0
2 years ago
6. The momentum of a 30.0 g bird with a speed of 12 m.s-1 is 0.36 kg.m.s-1. What will be its momentum 12s later if a constant .0
Keith_Richards [23]

Answer:

Explanation:

initial momentum = .36 kg.m.s⁻¹

negative impulse = force x time = .02 x 12 = .24 kg.m.s⁻¹

final momentum - initial momentum = impulse

final momentum = initial momentum + impulse

= .36 - .24

= .12 kg.m.s⁻¹

7 0
3 years ago
An element has the following natural abundances and isotopic masses: 90.92% abundance with 19.99 amu, 0.26% abundance with 20.99
sashaice [31]

<u>Answer:</u> The average atomic mass of the given element is 20.169 amu.

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of the isotopes each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i     .....(1)

We are given:

  • For isotope 1:

Mass of isotope 1 = 19.99 amu

Percentage abundance of isotope 1 = 90.92 %

Fractional abundance of isotope 1 = 0.9092

  • For isotope 2:

Mass of isotope 2 = 20.99 amu

Percentage abundance of isotope 2 = 0.26%

Fractional abundance of isotope 2 = 0.0026

  • For isotope 3:

Mass of isotope 3 = 21.99 amu

Percentage abundance of isotope 3 = 8.82%

Fractional abundance of isotope 3 = 0.0882  

Putting values in equation 1, we get:

\text{Average atomic mass}=[(19.99\times 0.9092)+(20.99\times 0.0026)+(21.99\times 0.0882)]

\text{Average atomic mass}=20.169amu

Hence, the average atomic mass of the given element is 20.169 amu.

4 0
3 years ago
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