Answer:
This might reduce outage probability because multiple base stations are able to receive a given mobile signal at a time which lead to signal outage decrease or weakness.
Answer:
Work = 651,1011 kJ
Explanation:
Let´s take the car as a system in order to apply the first law of thermodynamics as follows:

Where

And considering that there is no mass transfer and that the only energy flows that interact with the system are the heat losses and the work needed to move the car we have:

Regarding the energy system we have the following:

By doing the calculations we have:
![E_{system,final}- E_{system,initial}=[0,1*900]_{internal}+[0,5*900(30^2-10^2)/1000)_{kinetic}+(900*10*(20-0)/1000)_{potential}\\E_{system,final}- E_{system,initial}=90+360+180=630kJ](https://tex.z-dn.net/?f=E_%7Bsystem%2Cfinal%7D-%20E_%7Bsystem%2Cinitial%7D%3D%5B0%2C1%2A900%5D_%7Binternal%7D%2B%5B0%2C5%2A900%2830%5E2-10%5E2%29%2F1000%29_%7Bkinetic%7D%2B%28900%2A10%2A%2820-0%29%2F1000%29_%7Bpotential%7D%5C%5CE_%7Bsystem%2Cfinal%7D-%20E_%7Bsystem%2Cinitial%7D%3D90%2B360%2B180%3D630kJ)
Consider that in the previous calculation, the kinetic and potential energy terms were divided by 1.000 to change the units from J to kJ.
Finally, the work needed to move the car under the required conditions is calculated as follows:

Consider that in the previous calculation, the heat loss was changed previously from BTU to kJ.
Answer:
"Tempering Process" seems to be the appropriate choice.
Explanation:
- Tempering seems to be a method of heat preparation which is mostly used in completely hard materials to increase consistency, strength, durability, and also some decreasing brittleness.
- The tempering method is used to examine good functionality as well as flexural by reducing stiffness again after the substance has indeed been quenched towards its toughest state.
Answer:
2800 [MPa]
Explanation:
In fracture mechanics, whenever a crack has the shape of a hole, and the stress is perpendicular to the orientation of such, we can use a simple formula to calculate the maximum stress at the crack tip

Where
is the magnitude of he maximum stress at the tip of the crack,
is the magnitude of the tensile stress,
is
the length of the internal crack, and
is the radius of curvature of the crack.
We have:
![r_{c}=1.9*10^{-4} [mm]](https://tex.z-dn.net/?f=r_%7Bc%7D%3D1.9%2A10%5E%7B-4%7D%20%5Bmm%5D)
![l_{c}=3.8*10^{-2} [mm]](https://tex.z-dn.net/?f=l_%7Bc%7D%3D3.8%2A10%5E%7B-2%7D%20%5Bmm%5D)
![\sigma_{c}=140 [MPa]](https://tex.z-dn.net/?f=%5Csigma_%7Bc%7D%3D140%20%5BMPa%5D)
We replace:
![\sigma_{m} = 2*(140 [MPa])*(\frac{\frac{3.8*10^{-2} [mm]}{2}}{1.9*10^{-4} [mm]})^{0.5}](https://tex.z-dn.net/?f=%5Csigma_%7Bm%7D%20%3D%202%2A%28140%20%5BMPa%5D%29%2A%28%5Cfrac%7B%5Cfrac%7B3.8%2A10%5E%7B-2%7D%20%5Bmm%5D%7D%7B2%7D%7D%7B1.9%2A10%5E%7B-4%7D%20%5Bmm%5D%7D%29%5E%7B0.5%7D)
We get:
![\sigma_{m} = 2*(140 [MPa])*(\frac{\frac{3.8*10^{-2} [mm]}{2}}{1.9*10^{-4} [mm]})^{0.5}=2800 [MPa]](https://tex.z-dn.net/?f=%5Csigma_%7Bm%7D%20%3D%202%2A%28140%20%5BMPa%5D%29%2A%28%5Cfrac%7B%5Cfrac%7B3.8%2A10%5E%7B-2%7D%20%5Bmm%5D%7D%7B2%7D%7D%7B1.9%2A10%5E%7B-4%7D%20%5Bmm%5D%7D%29%5E%7B0.5%7D%3D2800%20%5BMPa%5D)
Answer:
asking questions, drawing conclusions, and gathering information.
Explanation:
hope this helps!