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.
Answer:
in the acceleration process the quantity α and w must increase
the deceleration process the alpha quantity must constant a direction opposite to the angular velocity
Explanation:
Acceleration and angular velocity are related to linear
v = w xr
a = αx r
The bold letters indicate vectors and the cross is a vector product, therefore if
we can see that the relationship between linear and angular variables is direct
therefore in the acceleration process the quantity α and w must increase as well as their linear counterparts
in the deceleration process the alpha quantity must constant as the linear acceleration and must have a direction opposite to the angular velocity
<span>6.6 degrees C
Let's model the student as a 125 w furnace that's been operating for 11 minutes. So
125 w * 11 min = 125 kg*m^2/s^3 * 11 min * 60 s/min = 82500 kg*m^2/s^2 = 82500 Joule
So the average kinetic energy increase of each gas molecule is
82500 J / 6.0x10^26 = 1.38x10^-22 J
Now the equation that relates kinetic energy to temperature is:
E = (3/2)Kb*Tk
E = average kinetic energy of the gas particles
Kb = Boltzmann constant (1.3806504Ă—10^-23 J/K)
Tk = Kinetic temperature in Kelvins
Notice the the energy level of the gas particles is linear with respect to temperature. So we don't care what the original temperature is, we just need to know by how much the average energy of the gas particles has increased by.
So let's substitute the known values and solve for Tk
E = (3/2)Kb*Tk
1.38x10^-22 J = (3/2)1.3806504Ă—10^-23 J/K * Tk
1.38x10^-22 J = 2.0709756x10^-23 J/K * Tk
6.64 K = Tk
Rounding to 2 significant digits gives 6.6K. So the temperature in the room will increase by 6.6 degrees K or 6.6 degrees C, or 11.9 degrees F.</span>
Answer:
v₂ = 5.7 m/s
Explanation:
We will apply the law of conservation of momentum here:
where,
Total Initial Momentum = 340 kg.m/s
m₁ = mass of bike
v₁ = final speed of bike = 0 m/s
m₂ = mass of Sheila = 60 kg
v₂ = final speed of Sheila = ?
Therefore,
<u>v₂ = 5.7 m/s </u>
