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vodka [1.7K]
3 years ago
12

In Ch. 1.6, the authors point out that interstellar space is not actually as empty as it seems. There is actually a lot of diffu

se gas, like good old H hydrogen, and ________.
Physics
1 answer:
wel3 years ago
6 0

Answer:

very small solid particles called interstellar dust.

Explanation:

In the space between the stars there is gas and dust, which represent at least 20% of the mass of our galaxy. In the Milky Way it is considered that there is a gas density of approximately 0.2 to 0.5 atoms / cm3 in the surroundings of the Sun; with respect to the dust an average of 1 g / cm3 is estimated.

Gas is about atoms and molecules, mainly hydrogen; In order of abundance, helium, carbon, oxygen, nitrogen and iron follow. On the other hand, the dust is tiny particles, generally smaller than 10 microns; the dust does not shine and therefore it is only distinguished when it is projected on bright regions (nebulae or clusters).

Interstellar matter is mainly concentrated towards the plane of the galaxy, in the strip corresponding to the Milky Way; there you can see bright nebulas of diffuse character called nebulas. These nebulae are classified according to three types: (a) bright or emission nebulae, (b) reflection nebulae and (c) planetary nebulae.

Hydrogen appears both ionized and neutral; The bright nebulae are composed of ionized hydrogen and other ionized elements. Non-ionized (neutral) hydrogen is found in the spiral arms of the Milky Way and can be detected through radio waves.

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8. A turtle crawls along a straight line, which we will call the x-axis with the positive direction to the right. The equation f
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(a) The turtle's initial velocity is 4 m/s, initial position of the turtle is 5 cm, and initial acceleration is -1.25 m/s².

(b) The time when the velocity of the turtle is zero is 3.2 s.

(c) The time taken for the turtle to return to its starting point is 6.4 s.

(d) The time taken for the turtle to travel 30 cm is 0.08 s.

<h3>Initial velocity of the turtle</h3>

The initial velocity of the turtle is calculated as follows;

v = \frac{dx}{dt} \\\\x = 5 + 4t -0.625t^2\\\\v(t) = 4 - 1.25t\\\\v(0) = 4-0\\\\v(0) = 4 \ m/s

<h3>Initial acceleration of the turtle</h3>

The initial acceleration of the turtle is calculated as follows;

a = \frac{dv}{dt} \\\\v(t) = 4 - 1.25t\\\\a = -1.25\ m/s^2

<h3>Initial position of the turtle</h3>

x(t) = 5 + 4t - 0.625t²

x(0) = 5 cm

<h3>Time when the velocity becomes zero</h3>

v(t) = 4 - 1.25t

0 = 4 - 1.25t

1.25t = 4

t = 4/1.25

t = 3.2 s

<h3>Time taken to return to starting point</h3>

The total distance traveled is calculated as follows

v² = u² + 2ad

0 = (4)² + 2(-1.25)d

0 = 16 - 2.5d

2.5d = 16

d = 16/2.5

d = 6.4 m

Time to travel the given distance;

d = ut + ¹/₂at²

6.4 = (4)t + ¹/₂(-1.25)t²

6.4 = 4t - 0.625t²

0.625t² - 4t + 6.4 = 0

solve the quadratic equation using formula method;

t = 3.2 s

The time travel the distance two times, = 2 x 3.2 s = 6.4 s

<h3>Time taken for the turtle to travel 30 cm</h3>

d = ut + ¹/₂at²

0.3 = (4)t + ¹/₂(-1.25)t²

0.3 = 4t - 0.625t²

0.625t² - 4t + 0.3 = 0

solve the quadratic equation using formula method;

t = 0.08 s

Learn more about velocity here: brainly.com/question/6504879

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