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.
If he feels like, is interested in it, and is able to grasp it, then why not ? Why not indeed ?
The conservation of energy always holds true even when not clearly observable in machines that are less than 100% efficient. More often than not a machine will suffer energy losses (e.g. consider for a cooling fan: friction between the rotating blades, drag resistance in the air the fan is pushing around, resistance in the wire, and heat radiating/conducting away from the circuitry).
We apply the following equation
T = 2π * sqrt (L/g)
Where g is the gravity = 9.8 m/s^2
L is the longitude of the pendulum (Height of the tower)
T is the period. (T = 18s)
We find L.............> (T /2π)^2 = L/g
L = g*(T /2π)^2...........> L = 80.428 meters
Answer:
The value is 
Explanation:
Generally the moon's radial acceleration is mathematically represented as
Here 
is the angular velocity which is mathematically represented as
substituting 
for T(i.e the period of the moon ) we have
=> 
From the question r(which is the radius of the orbit ) is evaluated as
substitute 
for R and 
H
=> 
So
