Complete question:
The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.
Answer:
The exit velocity is 629.41 m/s
Explanation:
Given;
initial temperature, T₁ = 1200K
initial pressure, P₁ = 150 kPa
final pressure, P₂ = 80 kPa
specific heat at 300 K, Cp = 1004 J/kgK
k = 1.4
Calculate final temperature;
k = 1.4
Work done is given as;
inlet velocity is negligible;
Therefore, the exit velocity is 629.41 m/s
Anything that can be dissolved/mixed into a solvent. Sugar, salt, acetic acid, etc
The situation presented above is possible because the outlets could be operating in a parallel circuit. <span>Electrical outlets in a house maintain a steady voltage, even when the amount of resistance on them changes because it operates with a parallel circuit wherein voltage is constant even if resistance changes.</span>
Answer:
Explanation:
Conservation of angular momentum
The MGR originally has momentum
L = 100(3.0) = 300 kg•m²/s²
The child can be thought of as a point mass with I = mr²
When she jumps onto the rim of the MGR
300 = (100 + 22(2.0²)ω
ω = 300 / 188 = 1.5957... 1.6 rad/s
As she moves toward the center of the MGR, her moment of inertia goes to zero as her radius goes to zero.
The angular velocity when she reaches the center will again be 3.0 rad/s
