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LiRa [457]
3 years ago
6

1. What is the wave speed of a wave that has a frequency of 100 Hz and a wavelength of 0.30 m?

Chemistry
2 answers:
poizon [28]3 years ago
8 0

<u>Answer:</u> The wave speed of the wave is 30 m/s

<u>Explanation:</u>

Wave speed is defined as the product of its wavelength and frequency.

Mathematically,

\text{Wave speed}=\lambda\times \nu

where,

\lambda = wavelength of the wave = 0.30 m

\nu = frequency of the wave = 100 Hz = 100s^{-1}

Putting values in above equation, we get:

\text{Wave speed}=0.30m\times 100s^{-1}\\\\\text{Wave speed}=30m/s

Hence, the wave speed of the wave is 30 m/s

Sav [38]3 years ago
5 0
Wave speed = frequency x wavelength 
                     =   100 x 0.3
                     =       30 m/s.

Hope this helps!
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Molybdenum (Mo) has a body centered cubic unit cell. The density of Mo is 10.28 g/cm3. Determine (a) the edge length of the unit
ser-zykov [4K]

<u>Answer:</u>

<u>For a:</u> The edge length of the unit cell is 314 pm

<u>For b:</u> The radius of the molybdenum atom is 135.9 pm

<u>Explanation:</u>

  • <u>For a:</u>

To calculate the edge length for given density of metal, we use the equation:

\rho=\frac{Z\times M}{N_{A}\times a^{3}}

where,

\rho = density = 10.28g/cm^3

Z = number of atom in unit cell = 2  (BCC)

M = atomic mass of metal (molybdenum) = 95.94 g/mol

N_{A} = Avogadro's number = 6.022\times 10^{23}

a = edge length of unit cell =?

Putting values in above equation, we get:

10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm

Conversion factor used:  1cm=10^{10}pm  

Hence, the edge length of the unit cell is 314 pm

  • <u>For b:</u>

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

R=\frac{\sqrt{3}a}{4}

where,

R = radius of the lattice = ?

a = edge length = 314 pm

Putting values in above equation, we get:

R=\frac{\sqrt{3}\times 314}{4}=135.9pm

Hence, the radius of the molybdenum atom is 135.9 pm

4 0
3 years ago
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