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.
Answer:
The number of oxygen molecules in the left container greater than the number of hydrogen molecules in the right container.
Explanation:
Given:
Molar mass of oxygen, 
Molar mass of hydrogen, 
We know ideal gas law as:
where:
P = pressure of the gas
V = volume of the gas
n= no. of moles of the gas molecules
R = universal gs constant
T = temperature of the gas
∵
where:
m = mass of gas in grams
M = molecular mass of the gas
∴Eq. (1) can be written as:
as: 
So,
Now, according to given we have T,P,R same for both the gases.
∴The molecules of oxygen are more densely packed than the molecules of hydrogen in the same volume at the same temperature and pressure. So, <em>the number of oxygen molecules in the left container greater than the number of hydrogen molecules in the right container.</em>
It's b, because the more force an object it is given the harder it will be for it to slow down.
Here,
Load distance (Ld) = 30 cm
Effort distance (Ed) = 60 cm
Load (L) = 200N
Effort (E) = ?
Now, By using formula,
or, E * Ed = L * Ld
or, E * 60 = 200 * 30
or, E = 6000/60
◆ E = 100N
This is a Right answer...
I hope you understand...
Answer: The period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.
Explanation:
Given: Mass = 5 kg
Spring constant = 6 N/m
Formula used to calculate period is as follows.
where,
T = period
m = mass
k = spring constant
Substitute the values into above formula as follows.
Thus, we can conclude that the period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.
