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Vaselesa [24]
3 years ago
10

How are sky maps made?

Physics
2 answers:
Andre45 [30]3 years ago
5 0
A star chart or star map is a map of night sky
r-ruslan [8.4K]3 years ago
4 0
A star chart<span> or </span>star map<span> is a </span>map<span> of the night </span>sky <span>Astronomers divide these into grids to use them more easil They are used to identify and locate astronomical objects such as stars constellations and galaxie it is in my text book </span>
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A track begins at 0 meters and has a total distance of 100 meters. Juliet starts at the 10-meter mark while practicing for a rac
shusha [124]

Answer : The correct option is (D).

Explanation :

Given that,

A track begins at 0 meters and has a total distance of 100 meters. Juliet starts at the 10-meter mark while practicing for a race.

We have to find her position after she runs 45 meters.

From the attached figure,

Let A is the position of Juliet. O is the initial point such that OA = 10 m, AB = 45 m and OP = 100 m.

So, using simple mathematics, it is clear that the position of Juliet after running 45 meters will be 55 m. It is OB in the figure.

So, the correct option is (D) " 55 meters ".

3 0
3 years ago
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Which if the following statements is true? A. Instantaneous acceleration is always changing. B. An object at rest has an instant
ycow [4]
B is the right one

hope this helps you
3 0
3 years ago
Remy wonders if the height of the mountain has anything to do with the eventual size of the tsunami wave. How should Remy test t
Anna71 [15]

Answer:

<em>The answer is B</em>

Explanation:

<em>I got this from study island</em>

5 0
3 years ago
which statement best describes how changes behave a.opposites charges repel against each other and like charges attract to each
Anon25 [30]

I think it is B hope this helps

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7 0
2 years ago
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A parallel-plate capacitor in air has a plate separation of 1.76 cm and a plate area of
Monica [59]

Answer:

Explanation:

Plate separation, d = 1.76 cm = 0.0176 m

Area of plates, A = 25 cm^2 = 0.0025 m^2

V = 255 V

(a) Capacitance of capacitor

C = \frac{\epsilon _0A}{d}

C = \frac{8.854\times 10^{-12}\times 0.0025}{0.0176}

C = 1.258 x 10^-12 F

charge is same before and after immersion as the battery is disconnected

q = C V

q = 1.258 x 10^-12 x 255 = 3.2 x 10^-10 C

(b)

Capacitance before, C = 1.258 x 10^-12 C

capacitance after, C' = k x C = 80 x 1.258 x 10^-12 = 100.64 x 10^-12 C

Where, k is the dielectric constant of water = 80

Potential difference after immersion, V' = V / k = 255 / 80 = 3.1875 V

(c) initial energy,

U = \frac{q^{2}}{2C}

U = \frac{(3.2\times 10^{-10})^{2}}{2\times 1.258\times 10^{-12 }}=4.07\times 10^{-8}J

Final energy

U' = \frac{q^{2}}{2C'}

U' = \frac{(3.2\times 10^{-10})^{2}}{2\times 100.64\times 10^{-12}}=5.08\times 10^{-10}J

6 0
3 years ago
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