The process you're fishing for is "polarization", but that's a
misleading description.
Polarization doesn't do anything to change the light waves.
It simply filters out (absorbs, as with a polarizing filter) the
light waves that aren't vibrating in the desired plane, and
allows only those that are to pass.
The intensity of a light beam is always reduced after
polarizing it, because much (most) of the original light
has been removed.
A laser light source may be thought of as an exception,
since everything coming out of the laser is polarized.
Answer:
carbon + oxygen → carbon dioxide
The mass number is the total number of protons and neutrons within an atom and since we know that the unknown element has 6 neutrons, we can simply subtract the number of neutrons from the mass number to get the number of protons.
17 - 6 = 11
There are 11 protons in this unknown element.
Extra:
The number of protons (+) and electrons (-) are equal in a neutral atom so since you know that there are 11 protons you also know that there are 11 electrons. On the periodic table, the element with 11 electrons is Na or Sodium.
Hope this helps! :)
Answer:
v₂ = 0.56 m / s
Explanation:
This exercise can be done using Bernoulli's equation
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
Where points 1 and 2 are on the surface of the glass and the top of the straw
The pressure at the two points is the same because they are open to the atmosphere, if we assume that the surface of the vessel is much sea that the area of the straw the velocity of the surface of the vessel is almost zero v₁ = 0
The difference in height between the level of the glass and the straw is constant and equal to 1.6 cm = 1.6 10⁻² m
We substitute in the equation
+ ρ g y₁ = + ½ ρ v₂² + ρ g y₂
½ v₂² = g (y₂-y₁)
v₂ = √ 2 g (y₂-y₁)
Let's calculate
v₂ = √ (2 9.8 1.6 10⁻²)
v₂ = 0.56 m / s
The acceleration of a body in a gravitational field is independent of its mass. Both the stones will fall with the same acceleration through the same height and hence they will strike the ground simultaneously.
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