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Answer:
hello your question is incomplete attached below is the missing equation related to the question
answer : 40.389° , 38.987° , 38° , 39.869° , 40.265°
Explanation:
<u>Determine the friction angle at each depth</u>
attached below is the detailed solution
To calculate the vertical stress = depth * unit weight of sand
also inverse of Tan = Tan^-1
also qc is in Mpa while σ0 is in kPa
Friction angle at each depth
2 meters = 40.389°
3.5 meters = 38.987°
5 meters = 38.022°
6.5 meters = 39.869°
8 meters = 40.265°
Answer:
ΔQ = 4930.37 BTu
Explanation:
given data
height h = 8ft
Δt = 8 hours
length L = 24 feet
R value = 16.2 hr⋅°F⋅ft² /Btu
inside temperature t1 = 68°F
outside temperature t2 = 16°F
to find out
number of Btu conducted
solution
we get here number of Btu conducted by this expression that s
......................1
here A is area that is = h × L = 8 × 24 = 1492 ft²
put here value we get
solve it we get
ΔQ = 4930.37 BTu
Answer: the theoretical density is 4.1109 g/cm³
Explanation:
first the image of one set of ZnS bonding in the crystal structure, we calculate the value of angle θ
θ + ∅ + 90° = 180°
θ = 90° - ∅
θ = 90° - ( 109.5° / 2 )
θ = 35.25°
next we calculate the value of x from the geometry
given that; distance angle d = 0.234
x = dsinθ
= 0.234 × sin35.25°)
= 0.135 nm = 0.135 × 10⁻⁷ cm
next we calculate the length of the unit cell
a = 4x
a = 4(0.135)
a = 0.54 nm = 0.54 × 10⁻⁷ cm
next we calculate number of formula units
n' = (no of corner atoms in unit ell × contribution of each corner atom in unit cell) + ( no of face center atom in a unit cell × contribution of each face center atom in a unit cell)
n' = 8 × 1/8) + ( 6 × 1/2)
= 1 + 3
= 4
next we calculate the theoretical density using this equation
P = [n'∑(Ac + AA)] / [Vc.NA]
= [n'∑(Ac + AA)] / [(a)³NA]
where the ∑Ac is sum of atomic weights of all cations in the formula unit( 65.41 g/mol)
∑AA is the sum of weights of all anions in the formula unit( 32.06 g/mol)
Na is the Avogadro’s number( 6.023 × 10²³ units/mole)
so we substitute
P = [4( 65.41 + 32.06)] / [ ( 0.54 × 10⁻⁷ )³ × (6.023 × 10²³)]
= 389.88 / 94.84
= 4.1109 g/cm³
therefore the theoretical density is 4.1109 g/cm³
