Answer:
T = 20.42 N
Explanation:
given data
standard altitude = 30,000 ft
velocity Ca = 500 mph = 0.4 m/s
inlet areas Aa = 7 ft² = 0.65 m²
exit areas Aj = 4.5 ft² = 0.42 m²
velocity at exit Cj = 1600 ft/s = 487.68 m/s
pressure exit 
j = 640 lb/ft² = 0.3 bar
solution
we get here thrust of the turbojet that is express as
thrust of the turbojet T = Mg × Cj - Ma × Ca + ( j Aj - a Ag ) .............1
here Ma = Mg
Ma = a × Ca Aa = 0.042 kg/s
put value in equation 1 we get
T = 0.042 × (487.68 -0.14) + ( 0.3 × - 0.3 × 0.65 )
T = 20.42 N
Answer:
A) True. I hope this helps
Answer:
The heat transferred to water equals 1600 kJ
Explanation:
By the conservation of energy we have
All the kinetic energy of the moving vehicle is converted into thermal energy
We know that kinetic energy of a object of mass 'm' moving with a speed of 'v' is given by
Thus
Thus the heat transferred to water equals 
Answer:
Fatigue factor of safety is 2.0267
Explanation:
Solution is attached below.
Answer:
747464646647477565756t6565665664674646465765
