Answer:
50421.6 m³
Explanation:
The river has an average rate of water flow of 59.6 m³/s.
Tributary B accounts for 47% of the rate of water flow. Therefore the rate of water flow through tributary B is:
Flow rate of water through tributary B = 47% of 59.6 m³/s = 0.47 * 59.6 m³/s = 28.012 m³/s
The volume of water that has been discharged through tributary B = Flow rate of water through tributary B * time taken
time = 30 minutes = 30 minutes * 60 seconds / minute = 1800 seconds
The volume of water that has been discharged through tributary B in 30 seconds = 28.012 m³/s * 1800 seconds = 50421.6 m³
(a) The number of vacancies per cubic centimeter is 1.157 X 10²⁰
(b) ρ = n X (AM) / v X Nₐ
<u>Explanation:</u>
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Given-
Lattice parameter of Li = 3.5089 X 10⁻⁸ cm
1 vacancy per 200 unit cells
Vacancy per cell = 1/200
(a)
Number of vacancies per cubic cm = ?
Vacancies/cm³ = vacancy per cell / (lattice parameter)³
Vacancies/cm³ = 1 / 200 X (3.5089 X 10⁻⁸cm)³
Vacancies/cm³ = 1.157 X 10²⁰
Therefore, the number of vacancies per cubic centimeter is 1.157 X 10²⁰
(b)
Density is represented by ρ
ρ = n X (AM) / v X Nₐ
where,
Nₐ = Avogadro number
AM = atomic mass
n = number of atoms
v = volume of unit cell
Answer: 0.95 inches
Explanation:
A direct load on a column is considered or referred to as an axial compressive load. A direct concentric load is considered axial. If the load is off center it is termed eccentric and is no longer axially applied.
The length= 64 inches
Ends are fixed Le= 64/2 = 32 inches
Factor Of Safety (FOS) = 3. 0
E= 10.6× 10^6 ps
σy= 4000ps
The square cross-section= ia^4/12
PE= π^2EI/Le^2
6500= 3.142^2 × 10^6 × a^4/12×32^2
a^4= 0.81 => a=0.81 inches => a=0.95 inches
Given σy= 4000ps
σallowable= σy/3= 40000/3= 13333. 33psi
Load acting= 6500
Area= a^2= 0.95 ×0.95= 0.9025
σactual=6500/0.9025
σ actual < σallowable
The dimension a= 0.95 inches
Maximum shear stress in the pole is 0.
<u>Explanation:</u>
Given-
Outer diameter = 127 mm
Outer radius,
= 127/2 = 63.5 mm
Inner diameter = 115 mm
Inner radius, 
= 115/2 = 57.5 mm
Force, q = 0
Maximum shear stress, τmax = ?
τmax 
If force, q is 0 then τmax is also equal to 0.
Therefore, maximum shear stress in the pole is 0.
