Answer:
i think the Ohio state university Wexner medical center
Explanation:
Answer:
The three major network categories are LAN (Local Area Network), MAN (Metropolitan Area Network) and WAN (Wide Area Network)
Explanation:
A Local Area Network (LAN) is the simplest form of a network. It is usually limited to a geographical area and consists of a group of computers in the same organisation linked together. In most cases, the same technology is used for all computers in this network (eg. ethernet).
A Metropolitan Area Network (MAN) can connect multiple LANs as long as they are close to each other (geographically). MANs allow high speed remote connection as if the devices were on the same LAN.
A Wide Area Network (WAN) can connect multiple LANs across large geographical distances. Unlike MANs, the speed might differ based on factors such as distance. The internet is the most popular example of a WAN.
The scenarios whereby licensure plays a role is:
- Jorge opens his own consulting business after he has gained enough experience working for other firms.
- Employers in Sarah's area require formal verification that she meets all the standards and possesses all the skills required of a professional architect.
<h3>What is the meaning of licensure?</h3>
Licensure is defined as a limited practice or the use of a professional title that requires a license. Thus, prior to the engagement in specific professional activities, such as driving a vehicle on public roads, a license issued under a practice act is required.
Therefore, from the given situations, the perfect option that requires a licensure is seen in Option A and Option D.
Hence, a license issued under this title act prevents licensees from using a certain occupational title, however anybody can execute the activity using a less limited title.
Learn more about Licensure here;
brainly.com/question/26686639
Answer:
D
Explanation:
the way vertices are connected may be different so having same number of edges do not mean that total degree will also be same.
Answer:
a) 15.37 mm
b)
c) 5.7186 W/m². K
d) 0.60 m
Explanation:
Given that :
The surface temperature = 130°C = ( 130+ 273 ) K = 473 K
suspended in quiescent air at 25°C = ( 25 + 273 ) K = 298 K
![T_f = \frac{T_s + T \infty}{2 } \\ \\ T_f = {403+298}{2} \\ \\ T_f = \frac{701}{2} \\ \\ T_f = 350 \ K](https://tex.z-dn.net/?f=T_f%20%3D%20%5Cfrac%7BT_s%20%2B%20T%20%5Cinfty%7D%7B2%20%7D%20%5C%5C%20%5C%5C%20T_f%20%3D%20%7B403%2B298%7D%7B2%7D%20%5C%5C%20%5C%5C%20T_f%20%3D%20%5Cfrac%7B701%7D%7B2%7D%20%20%5C%5C%20%5C%5C%20T_f%20%3D%20350%20%5C%20K)
Atmospheric Pressure = 1 atm
The properties obtained from Table A - 4 include :
v = ![20.92 *10^{-6} \ \ m^2 /s](https://tex.z-dn.net/?f=20.92%20%2A10%5E%7B-6%7D%20%5C%20%5C%20m%5E2%20%2Fs)
k = 0.03 W/m K
Pr = 0.700
η = 5
![Gr_x = 9.8[T_s - T \infty] \frac{x^3}{v^2}](https://tex.z-dn.net/?f=Gr_x%20%3D%209.8%5BT_s%20-%20T%20%5Cinfty%5D%20%5Cfrac%7Bx%5E3%7D%7Bv%5E2%7D)
![Gr_x = 9.8*\frac{1}{350}[130-25] \frac{x^3}{20.92*10^{-6}^2}](https://tex.z-dn.net/?f=Gr_x%20%3D%209.8%2A%5Cfrac%7B1%7D%7B350%7D%5B130-25%5D%20%5Cfrac%7Bx%5E3%7D%7B20.92%2A10%5E%7B-6%7D%5E2%7D)
![Gr_x =6.718*10^9 \ x^3](https://tex.z-dn.net/?f=Gr_x%20%3D6.718%2A10%5E9%20%5C%20x%5E3)
![\gamma = 5(0.15)((6.718*10^9)(0.15)^{3}/4)^{-1/4}](https://tex.z-dn.net/?f=%5Cgamma%20%3D%205%280.15%29%28%286.718%2A10%5E9%29%280.15%29%5E%7B3%7D%2F4%29%5E%7B-1%2F4%7D)
![\gamma = 0.01537 \ m](https://tex.z-dn.net/?f=%5Cgamma%20%3D%200.01537%20%5C%20m)
![\gamma = 15.37 \ mm](https://tex.z-dn.net/?f=%5Cgamma%20%3D%2015.37%20%5C%20mm)
Hence, the boundary layer thickness at a location 0.15 m measured from the lower edge is 15.37 mm
b) The maximum velocity in the boundary layer
with f'(n) = 0.275
![u = \frac{2v}{x} Gr_x^{1/2} f'(n)](https://tex.z-dn.net/?f=u%20%3D%20%5Cfrac%7B2v%7D%7Bx%7D%20%20Gr_x%5E%7B1%2F2%7D%20f%27%28n%29)
![u= \frac{2*20.92*10^{-6}}{0.75} [6.718*10^9(0.15)^3]^{1/2} * 0.275](https://tex.z-dn.net/?f=u%3D%20%5Cfrac%7B2%2A20.92%2A10%5E%7B-6%7D%7D%7B0.75%7D%20%20%5B6.718%2A10%5E9%280.15%29%5E3%5D%5E%7B1%2F2%7D%20%2A%200.275)
u = 0.3659 m/s
the maximum velocity in the boundary layer at this location is 0.3659 m/s
the position in the boundary layer where the maximum occur is calculated as:
![y_{max} = \frac{1}{5}(15.37 \ mm)](https://tex.z-dn.net/?f=y_%7Bmax%7D%20%3D%20%5Cfrac%7B1%7D%7B5%7D%2815.37%20%5C%20mm%29)
3.074 mm
c) Using the similarity solution result, , determine the heat transfer coefficient 0.15 m from the lower edge.
we know that:
![Nu_x = \frac{h_x *x }{K} = (Gr_x/4)^{1/4} \ g (pr)](https://tex.z-dn.net/?f=Nu_x%20%3D%20%5Cfrac%7Bh_x%20%2Ax%20%7D%7BK%7D%20%3D%20%28Gr_x%2F4%29%5E%7B1%2F4%7D%20%20%5C%20g%20%28pr%29)
= ![(6.718*10^9(0.15)^3/4)^{1/4} *0.586](https://tex.z-dn.net/?f=%286.718%2A10%5E9%280.15%29%5E3%2F4%29%5E%7B1%2F4%7D%20%2A0.586)
= 28.593
Making
the subject from the above formula:
![h_x = \frac{Nu_xK}{x}](https://tex.z-dn.net/?f=h_x%20%3D%20%5Cfrac%7BNu_xK%7D%7Bx%7D)
![h_x= \frac{28.593*0.03}{0.15}](https://tex.z-dn.net/?f=h_x%3D%20%5Cfrac%7B28.593%2A0.03%7D%7B0.15%7D)
= 5.7186 W/m². K
d) to determine the location on the plate that the boundary layer we become turbulent ; we have the following:
![Ra _ {x,c} = Gr_x,c^{pr} \approx 10^9](https://tex.z-dn.net/?f=Ra%20_%20%7Bx%2Cc%7D%20%3D%20Gr_x%2Cc%5E%7Bpr%7D%20%5Capprox%2010%5E9)
![x_c = [10^9/6.718*10^9(0.7)]^{1/3}](https://tex.z-dn.net/?f=x_c%20%3D%20%5B10%5E9%2F6.718%2A10%5E9%280.7%29%5D%5E%7B1%2F3%7D)
0.60 m