The answer for this problem is clarified through this, the
system is absorbing (+). And now see that it uses that the SURROUNDINGS are
doing 84 KJ of work. Any time a system is overshadowing work done on it by the
surroundings the sign will be +. So it's just 12.4 KJ + 4.2 = 16.6 KJ.
Answer:
ºC
Explanation:
First, let's write the energy balance over the duct:

It says that the energy that goes out from the duct (which is in enthalpy of the mass flow) must be equals to the energy that enters in the same way plus the heat that is added to the air. Decompose the enthalpies to the mass flow and specific enthalpies:

The enthalpy change can be calculated as Cp multiplied by the difference of temperature because it is supposed that the pressure drop is not significant.

So, let's isolate
:

The Cp of the air at 27ºC is 1007
(Taken from Keenan, Chao, Keyes, “Gas Tables”, Wiley, 1985.); and the only two unknown are
and Q.
Q can be found knowing that the heat flux is 600W/m2, which is a rate of heat to transfer area; so if we know the transfer area, we could know the heat added.
The heat transfer area is the inner surface area of the duct, which can be found as the perimeter of the cross section multiplied by the length of the duct:
Perimeter:

Surface area:

Then, the heat Q is:

Finally, find the exit temperature:

=27.0000077 ºC
The temperature change so little because:
- The mass flow is so big compared to the heat flux.
- The transfer area is so little, a bigger length would be required.
Answer:
The airliner travels 1.65 km along the runway before coming to a halt.
Explanation:
Given
Resistive forces = (2.90 × 10⁵) N = 290000 N
Mass of the airliner = (1.70 × 10⁵) kg = 170000 kg
Velocity of airliner = 75 m/s
Let the distance over moved by the airliner be equal to d
According to the work-energy theorem, the work done by the resistive forces in stopping the airliner is equal to the travelling kinetic energy of the airliner.
Work done by the resistive forces = (290000) × d = (290,000d) J
Kinetic energy of the airliner = (1/2)(170000)(75²) = 478,125,000 J
290000d = 478,125,000
d = (478,125,000/290,000)
d = 1648.7 m = 1.65 km
Hope this helps!!!
Answer:
a. FTh = 30 N
b. Fw = 30 N
c. a = 200 m/s2
Explanation:
See full explanation in the picture. Please rate as brainliest