Answer:
.
Explanation:
The frequency 
of a wave is equal to the number of wave cycles that go through a point on its path in unit time (where "unit time" is typically equal to one second.)
The wave in this question travels at a speed of 
. In other words, the wave would have traveled 
in each second. Consider a point on the path of this wave. If a peak was initially at that point, in one second that peak would be
How many wave cycles can fit into that ? The wavelength of this wave
gives the length of one wave cycle. Therefore:
.
That is: there are 
wave cycles in of this wave.
On the other hand, Because that of this wave goes through that point in each second, that wave cycles will go through that point in the same amount of time. Hence, the frequency of this wave would be
Because one wave cycle per second is equivalent to one Hertz, the frequency of this wave can be written as:
.
The calculations above can be expressed with the formula:
,
where
represents the speed of this wave, and 
represents the wavelength of this wave.
Cool down the Slowest
B. Seawater, because it heats up slower and gives away heat slower than sand.
Read the question wrong the first time
Science is open to change because scientists agree that there are no absolute truths, only the best explanations for the data
there are specifically two cases when science can change: -new data is discovered (for example new skeleton makes us revise the known history of homo species)
-a new model is proposed that fits the data better (such as the heliocentric model of Copernicus)
Answer:
atomic concentration = 2 atoms/unit cell
lattice parameter: a= 3.22 x 10⁻¹⁰ m
atomic radius: r= 1.39 x 10⁻¹⁰m
Explanation:
The atomic concentration is the number of atoms that can fit into a unit cell. It is a known number for each unit cell crystal structure. For a BCC (body-centered cube) crystal structure, atomic concentration is 2 atoms/unit cell because there are a 1/8 part of an atom in each corner of the cube (1/8 x 8= 1 atom) and 1 central atom in the central position of the cube ⇒ n= 1 atom + 1 atom= 2 atoms/unit cell
In order to calculate the lattice parameter a, we introduce the atomic mass 95.94 g/mol and the density 10.22 g/cm³ in the expression for the volume of the cube:
Vc= a³= 
a³= 3.12 x 10⁻²³ m³
⇒ a = ∛(3.12 x 10⁻²³ m³) = 3.22 x 10⁻¹⁰m
Once we know the lattice parameter a, we can calculate the atomic radius r by using the expression of a for a BCC structure:
a= 
⇒ r= a x √3/4= (3.22 x 10⁻¹⁰ m) x √3/4 = 1.39 x 10⁻¹⁰ m
