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3 years ago

In which direction does the sun appear to move across the sky?

Physics

2 answers:

sergiy2304 [10]3 years ago

3 0

Answer:

If we consider a system where the y-axis as the South-North line, and the x-axis as the West-East line (where North and East are the positive sides)

We know that the sun goes from East to West, so in our system, the sun goes from the positive side of the x-axis to the negative side of the x-axis.

Where we would see this if we were standing right in the equator line.

If we where in other point of the planet, the Sun will stil move from East to West, but it will have a little tilt along the path, so we will have a little displacement in the y-axis. This displacement will depend on where we are, if we are at the North of the equator, we will se that the sun seems to go a little towards South as it goes to the West side.

jasenka [17]3 years ago

3 0

Answer:

the sun appears from east to west were it sets

Explanation:

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Answer:

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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

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

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

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

An experiment proposes a possible explanation for a natural phenomenon. A experiment must be stated in a way that it can be test

scZoUnD [109]

Answer:

a) False.

b) True.

c) Flase.

Explanation:

a) An experiment proposes a possible explanation for a natural phenomenon.

False: An experiment doesn't propose possible explanation for a natural phenomenon. Actually, an experiment is designed to test already proposed explanations which normally are first developed in a more <em>theorical </em>context.

b) A experiment must be stated in a way that it can be tested by experiments.

True.

b) An experiment is a theory that has been validated many times by many scientists.

Flase: An experiment is no a theory, but a tool to test theories.

8 0

3 years ago

Consider a uniformly volume‑charged sphere of radius R and charge Q . What is the electric potential on the surface of the spher

VARVARA [1.3K]

To solve this problem we will start by applying the given load ratio, and we will rely on the two types of distances given. Later we will use Gauss's law and through its integrals, in which it is equivalent to the potential we will obtain its value in the center of the sphere. Since it is uniformly charged we have to,

$\frac{Q'}{Q} = \frac{4\pi r^3}{4\pi R^3}$