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The lowest constant acceleration needed for takeoff from a 1.80 km runway is 2.8 m/s².
To find the answer, we need to know about the Newton's equation of motion.
<h3>What's the Newton's equation of motion to find the acceleration in term of initial velocity, final velocity and distance?</h3>
- The Newton's equation of motion that connects velocity, distance and acceleration is V² - U²= 2aS
- V= final velocity, U= initial velocity, S= distance and a= acceleration
<h3>What's the acceleration, if the initial velocity, final velocity and distance are 0 m/s, 360km/h and 1.8 km respectively?</h3>
- Here, S= 1.8 km or 1800 m, V= 360km/h or 100m/s , U= 0 m/s
- So, 100²-0= 2×a×1800
=> 10000= 3600a
=> a= 10000/3600 = 2.8 m/s²
Thus, we can conclude that the lowest constant acceleration needed for takeoff from a 1.80 km runway is 2.8 m/s².
Learn more about the Newton's equation of motion here:
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Answer:

Explanation:
Given that
T₁ = 290 K
P₁ = 100 KPa
Power P =5.5 KW
mass flow rate

Lets take the exit temperature = T₂
We know that


If we assume that process inside the compressor is adiabatic then we can say that





That is why the exit pressure will be 4091 KPa.