Answer:

Explanation:
Previous concepts
Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:

Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =
MO, where MO is the moment of the force F about point O. The equation expressing the rate of change of angular momentum is this one:
MO = H˙ O
Principle of Angular Impulse and Momentum
The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:

Solution to the problem
For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is
".
If we analyze the staritning point we see that the initial velocity can be founded like this:

And if we look the figure attached we can use the point A as a reference to calculate the angular impulse and momentum equation, like this:

](https://tex.z-dn.net/?f=0%2B%5Csum%20%5Cint_%7B0%7D%5E%7B4%7D%2020t%20%280.15m%29%20dt%20%3D0.46875%20%5Comega%20%2B%2030kg%5B%5Comega%280.15m%29%5D%280.15m%29)
And if we integrate the left part and we simplify the right part we have

And if we solve for
we got:

It is habahi Yw with yuuuuuy I am a little more confused about
Answer:
a. true
Explanation:
Firstly, we need to understand what takes places during the compression process in a quasi-equilibrium process. A quasi-equilibrium process is a process in during which the system remains very close to a state of equilibrium at all times. When a compression process is quasi-equilibrium, the work done during the compression is returned to the surroundings during expansion, no exchange of heat, and then the system and the surroundings return to their initial states. Thus a reversible process.
While for a non-quasi equilibrium process, it takes more work to move the piston against this high-pressure region.
Answer:
1) q=18414.93 W
2) C=12920$
Explanation:
Given data:
pipe length L=25m
pipe diameter D=100mm =0.1 m
air temperature
=
=25
°C.....= 298.15k
pipe surface temp
=150
°C.....=423.15k
surface emissivity e= 0.8
boiler efficiency η=0.90
natural gas price Cg=$0.02 per MJ
1) Total heat loss and radiation heat loss combined
q=
q=
б(
^4-
^4)]....... (1)
б=5.67×10^-8 W/m^2K^4 (boltzmann constant)
area A =L.Dπ=25×0.1π=7.85 m^2
putting all these values in eq (1)
q=18414.93 W
2) suppose boiler is operating non stop annual energy loss will be
E=q.t
=18414.93.3600.24.365
=5.81×10^11 J
to find furnace energy consumption
Ef =E/η
=6.46×10^5 MJ
annual cost
C=Cg. Ef
=12920$
I would honestly select every one of the given options. Gor a company evaluating this new material it would be very valuable to hit each of these factors.