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storchak [24]
4 years ago
6

Why is the curve between 1950 and 1980 relatively flat and centered around zero degrees difference from the baseline? (Hint: how

is the temperature change being compared over time?)

Physics
1 answer:
zimovet [89]4 years ago
3 0

Look at the title of the graph, in small print under it.

Each point is "compared to 1950-1980 baseline". So the set of data for those years is being compared to itself. No wonder it matches up pretty close !

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1.524777636594663 cm^3
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Ted is a high school student. He is very good at math, but he is not a very good English student. He likes all of his teachers;
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4 years ago
What happens to waves near the shore?
Lady_Fox [76]

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Explanation:

8 0
3 years ago
The star Sirius has an apparent magnitude of -1.46 and appears 95-times brighter compared to the more distant star Tau Ceti, whi
dimaraw [331]

Answer:

(a) Apparent magnitude is the perceived brightness of an astronomical object

Absolute magnitude is the luminosity based on viewing an object from a 32.6 light-years distance

Bolometric magnitude is the total emitted radiation of a star

(b) The apparent magnitude of the star Tau Ceti = 3.51

(c) The distance between the Earth and Tau Ceti is 1.13 × 10¹⁴ km

Explanation:

(a) Apparent magnitude is an estimate of an astronomical objects' brightness as the object is perceived from the Earth

The absolute magnitude  is the magnitude an object appears to have when viewed from a 32.6 light-years distance while having constant transfer of its luminosity that is not affected by cosmic dust and objects present in the line of sight

The bolometric magnitude of a star is the sum total of the star's radiation released over all electromagnetic spectrum wavelengths

(b) The apparent magnitude of the star Tau Ceti is found using the following equation;

m_{2}-m_{1} = -2.512\times log\left (\dfrac{B_{2}}{B_{1}}  \right )

Where:

m₁ = Apparent magnitude of Tau Ceti

m₂ = Apparent magnitude of  Sirius = -1.46

B₁ = Brightness of Tau Ceti

B₂ = Brightness of Sirius

\left \dfrac{B_{2}}{B_{1}}  \right  = 95

Hence we have;

-1.46-m_{1} = -2.512\times log\left (95 \right )

m₁ = -1.46 + 2.512 × log(95) = 3.51

The apparent magnitude of the star Tau Ceti = 3.51

(c) The distance between the Earth and Tau Ceti is found using the following equation;

m-M = 5\times log\left (\dfrac{d}{10}  \right )

Where:

m = Apparent magnitude of Tau Ceti

M = Absolute magnitude of Tau Ceti = 5.69

d = The distance between the Earth and Tau Ceti

Which gives;

3.51-5.69 = 5\times log\left (\dfrac{d}{10}  \right )

\therefore \dfrac{d}{10} = 10^{-0.436} = 0.3664

d = 10 × 0.3664 = 3.664 parsecs = 3.664 × 3.0857 × 10¹⁶ m

d = 1.13 × 10¹⁴ km.

4 0
3 years ago
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