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Svetradugi [14.3K]
3 years ago
11

Explain why the air we breathe and solids, such as steel and bronze, are solutions or homogeneous mixtures, just like salt water

is ?
Physics
2 answers:
eimsori [14]3 years ago
7 0

Answer:

Explanation:

There are two major types of mixtures; homogeneous and heterogeneous mixtures.

Homogeneous mixture is a mixture in which its constituent particles are evenly distributed throughout the mixture. Examples are steel, bronze and air (as stated in the question). While heterogeneous mixture is a mixture in which it's constituent particles are not evenly distributed throughout the mixture. Example includes sand and water solution.

Salt water is an homogeneous mixture because the percentage of salt in a sample of salt from a particular mixture is the same in every other sample (of the same volume) of the same mixture.

From the above, we can deduce that air is an homogeneous mixture because a sample of air (in a particular area) will contain the same constituent by proportion as another sample of air (in the same area). This means that the percentage of oxygen, nitrogen, carbon dioxide, water vapour, argon and other substances in one sample of the air will be equal to the percentage of this constituents in the other sample of the air.

Same can be said about steel and bronze. For steel, each gram of the same steel will contain the same percentage of carbon and iron. And also, each gram of the same bronze will contain the same percentage of copper, tin and other metals present.

For better understanding, let's consider an heterogeneous mixture. A mixture of sand and water solution cannot have the same proportion throughout, this is because sand particles in some samples (of a particular volume) of a particular mixture will be different from sand particles in some other samples (of the same volume) of the same mixture.

svet-max [94.6K]3 years ago
3 0
Because you can't physically see the difference.
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Two vectors are presented as a=3.0i +5.0j and b=2.0i+4.0j find (a) a x b, ab (c) (a+b)b and (d) the component of a along the dir
Svet_ta [14]
Let's ask this question step by step:
 Part A) 
 a x b = (3.0i + 5.0j) x (2.0i + 4.0j) = (12-10) k = 2k
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 Part (c)
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 comp (ba) = (a.b) / lbl
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A) 750 m

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c=3.0\cdot 10^8 m/s is the speed of light

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\lambda=\frac{c}{f}=\frac{3\cdot 10^8 m/s}{12\cdot 10^9 Hz}=0.025 m

Now we can use the formula of the single-slit diffraction to find the radius of aperture of the beam:

y=\frac{m\lambda D}{a}

where

m = 1 since we are interested only in the central fringe

D = 30 km = 30,000 m

a = 2.0 m is the aperture of the antenna (which corresponds to the width of the slit)

Substituting, we find

y=\frac{(1)(0.025 m)(30000 m)}{2.0 m}=375 m

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B) 0.23 W/m^2

First we calculate the area of the surface of the microwave at a distance of 30 km. Since the diameter of the circle is 750 m, the radius is

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7 0
3 years ago
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