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QUESTION: A pure jet engine propels and aircraft at 340 m/s through air at 45 kPa and -13C. The inlet diameter of this engine is 1.6 m, the compressor pressure ratio is 13, and the temperature at the turbine inlet is 557C. Determine the velocity at the exit of this engines nozzle and the thrust produced.
ANSWER: Due to the propulsion from the inlet diameter of this engine bring 1.6 m allows the compressor rations to radiate allowing thrust propultion above all velocitic rebisomes.
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Answer:
applying 1st eq of motion vf=vi+at we have to find a=vf-vi/t here a=50-30/2=10 so we got a=10m/s²
Answer:
Maximum altitude above the ground = 1,540,224 m = 1540.2 km
Explanation:
Using the equations of motion
u = initial velocity of the projectile = 5.5 km/s = 5500 m/s
v = final velocity of the projectile at maximum height reached = 0 m/s
g = acceleration due to gravity = (GM/R²) (from the gravitational law)
g = (6.674 × 10⁻¹¹ × 5.97 × 10²⁴)/(6370000²)
g = -9.82 m/s² (minus because of the direction in which it is directed)
y = vertical distance covered by the projectile = ?
v² = u² + 2gy
0² = 5500² + 2(-9.82)(y)
19.64y = 5500²
y = 1,540,224 m = 1540.2 km
Hope this Helps!!!
