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Aleks [24]
3 years ago
5

Which one of the following activities is not exempt from licensure pursuant to Chapter 471, F.S.? A person practicing engineerin

g on property owned by him or her. A full time electrical engineer of Progress Energy Corp. A civil engineer employed full time by the U.S. Army Corps of Engineers. An Independent consultant working on the design of an electrical distribution system project for Progress Energy Corp. Applicants for licensure with degrees from foreign institutions are required to document "substantial equivalency to ABET criteria to the FBPE. They can do this by: Providing a transcript from their institution to the Board. Providing a notarized certification that they have completed the requisite college credit hours set forth in Rules 61615-20.007(2)(a) thru (2)(d), F.A.C. Getting the evaluation of substantial equivalency from a provider of the service that is approved by the FBPE Passing the Principles & Practice examination In order to verify an applicant's experience, the FBPE: Follows guidelines set forth in Rule 61615 20.002, F.A.C. Relies on information obtained from personal references Requires evidence of employment from employers or supervisors who are employed in the engineering profession. • All of the above.
Engineering
1 answer:
elixir [45]3 years ago
6 0

Answer:

Your questions are three in one, so the answers are below;

1) An independent consultant working on design of electrical distribution is not included in exemption in 471.003

2) Applicants from foreign are required to document substantial equivalency as per ABET criteria.

3)  Requires the evidence of employment from employers or supervisor who are employed in engineering profession

Explanation:

1) An independent consultant working on design of electrical distribution is not included in exemption in 471.003

2) Applicants from foreign are required to document substantial equivalency as per ABET criteria.

3)  Requires the evidence of employment from employers or supervisor who are employed in engineering profession

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A pressure gage connected to a tank reads 50 psi at a location where the barometric reading is 29.1 inches Hg. Determine the abs
Effectus [21]

Answer:

Absolute pressure , P(abs)= 433.31 KPa

Explanation:

Given that

Gauge pressure P(gauge)=  50 psi

We know that barometer reads atmospheric pressure

Atmospheric pressure P(atm) = 29.1 inches of Hg

We know that

1 psi = 6.89 KPa

So 50 psi = 6.89 x 50 KPa

P(gauge)=  50 psi =344.72 KPa

We know that

1 inch = 0.0254 m

29.1 inches = 0.739 m

Atmospheric pressure P(atm) = 0.739 m of Hg

We know that density of Hg =13.6\times 10^3\ kg/m^3

P = ρ g h

P(atm) = 13.6 x 1000 x 9.81 x 0.739 Pa

P(atm) = 13.6  x 9.81 x 0.739 KPa

P(atm) =98.54 KPa

Now

Absolute pressure = Gauge pressure + Atmospheric pressure

P(abs)=P(gauge) + P(atm)

P(abs)= 344.72 KPa + 98.54 KPa

P(abs)= 433.31 KPa

3 0
3 years ago
How can feeding plant crops to animals be considered an efficient use of those crops?
stira [4]

Answer:

Most of Our Grain is Being Consumed by Animals, Not Humans. How the Planting of Crops Used to Feed Livestock is Contributing to Habitat Destruction.

Explanation:

The major feed grains are corn, sorghum, barley, and oats.

7 0
3 years ago
. Consider the single-engine light plane described in Prob. 2. If the specific fuel consumption is 0.42 lb of fuel per horsepowe
Trava [24]

Answer:

Hence the Range and Endurance of single engine plane is given by

650.644 miles and 5.3528 hrs at standard sea level.

Explanation:

Given :

A single engine light plane with ,

Specific fuel consumption 0.42lb/hr/hp.

Fuel capacity =44 gal.

Gross weight =3400 lb.

To find :

Range and Endurance of the plane.

Solution:

Consider  all standard measures of standard single engine propeller plane

as

Wing span =35.8 fts.

Wing swing area=174 sq ft

parasite drag coefficient  =Cd.o.=0.025

Oswald's eff. factor= 0.8

ρ=0.002377= corresponds to standard sea level constant.

Now

Formula for Range is given by, Breguent formula.

R=(η/c)  *(Cl/Cd)*ln(W1/W0)

here η is Oswald's constant,

Now calculating lift(Cl) and drag coefficient (Cd)

Cl=W/(1/2*ρ*v^2*S)

W=Gross weight

ρ=0.002377

Assume v=200 ft/sec normally,

S=174 Sq .ft.

CI=3400/(1/2*0.002377*200*200*174)

=6800/16543.9

=0.4110

Now calculating drag constant,

AR=(wing span)^2/wing swing area

=(35.8)^2/174

=7.37

Now

Drag Coefficient

Cd=Cd.o.+ (Cl^2)/(pie*e*AR)

=0.025+(0.4110)^2/(3.142*0.8*7.36)

=0.0342

Given that 44 gal fuel capacity and in Aviation weight of fuel is 5.64 lb/gal

hence weight of fuel=W1=3400- (44*5.64)

=3151.84

Now

for specific fuel consumption=0.42  lb/hp/hr

=0.42  lb*(1/550 ft)*(1/3600)sec

=2.12 *10^-7 lb/ft/sec

Now further calculating range

R=(η/c)  *(Cl/Cd)*ln(W1/W0)

={0.8/(2.12*10^-7)}*(0.4110/0.0342)*ln(3151.84/3400)

=0.024908/0.072504

=0.34354*10^7

=3.4353 *10^6 fts.

1mi =5280 ft

=(3.4353/5280)*10^6

=650.644 miles

Now

For Endurance

E=(η/c)*{(Cl^3/2)/Cd}*(2*ρ*S)^1/2*[1/(W1)^1/2  -1/(W0)^1/2].

=(0.8/2.12*10^-7)*{(0.4110^3/2)/0.0342}*(2*0.002377*174)^1/2*[1/(3151.84)^1/2  -1/(3400)^1/2]

=3.7735*10^6*7.7043*0.8272*0.0006629

=0.01927*10^6

=1.927*10^4 sec

here 1hr =3600 sec

E=(1.927/3600)*10^4

=5.3528 hrs

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3 years ago
List the parts of a manual transmission <br><br> List the parts of a typical clutch assembly?
True [87]

Answer:

Explanation: Clutch Plate.

Clutch Cover.

Clutch Bearing (Release bearing)

Release Fork (clutch fork)

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2 years ago
The purpose of daytime running lights (DRLs) is to _____.
balu736 [363]
D make your car easy to spot in daytime because they do not illuminate a lot
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