Answer:
The break force that must be applied to hold the plane stationary is 12597.4 N
Explanation:
p₁ = p₂, T₁ = T₂
The heat supplied = 
× Heating value of jet fuel
The heat supplied = 0.5 kg/s × 42,700 kJ/kg = 21,350 kJ/s
The heat supplied = 
· 
= 20 kg/s
The heat supplied = 20* = 21,350 kJ/s
= 1.15 kJ/kg
T₃ = 21,350/(1.15*20) + 485.03 = 1413.3 K
p₂ = p₁ × p₂/p₁ = 95×9 = 855 kPa
p₃ = p₂ = 855 kPa
T₃ - T₄ = T₂ - T₁ = 485.03 - 280.15 = 204.88 K
T₄ = 1413.3 - 204.88 = 1208.42 K
T₅ = 1208.42*(2/2.333) = 1035.94 K
= √(1.333*287.3*1035.94) = 629.87 m/s
The total thrust = × 
= 20*629.87 = 12597.4 N
Therefore;
The break force that must be applied to hold the plane stationary = 12597.4 N.
I honestly don’t know because I honestly don’t know what I mean
Answer:
a. Fraction of Atom = 2.41E-5 when T = 600K
b. Fraction of Atom = 5.03E-10 when T = 298K
Explanation:
a.
Given
T = Temperature = 600K
Qv = Energy for formation = 0.55eV/atom
To calculate the fraction of atom sites, we make use of the following formula
Nv/N = exp(-Qv/kT)
Where k = Boltzmann Constant = 8.62E-5eV/K
Nv/N = exp(-0.55/(8.62E-5 * 600))
Nv/N = 0.000024078672493307
Nv/N = 2.41E-5
b. When T = 298K
Nv/N = exp(-0.55/(8.62E-5 * 298))
Nv/N = 5.026591237904E−10
Nv/N = 5.03E-10 ----- Approximated
