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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 buyer can purchase 70 screwdrivers ten 4-inch length twelve 6 inch length twenty 8-inch length are needed. how many heavy 24-i
jasenka [17]

Answer:

28 , 24-inch screwdrivers

Explanation:

The total number of screwdrivers that can be purchased is = 70

4 - inch length screwdrivers = 10

6- inch length screwdrivers = 12

8- inch length screwdrivers = 20

Total = 20 +12 +10 = 42

Remaining = 70-42 = 28

So, heavy 24-inch screwdrivers = 28

3 0
3 years ago
g A steel water pipe has an inner diameter of 12 in. and a wall thickness of 0.25 in. Determine the longitudinal and hoop stress
zvonat [6]

Answer:

a) \mathbf{\sigma _ 1 = 4800 psi}

     \mathbf{ \sigma _2 = 0}

b)\mathbf{\sigma _ 1 = 6000 psi}

  \mathbf{ \sigma _2 = 3000 psi}

Explanation:

Given that:

diameter d = 12 in

thickness t = 0.25 in

the radius = d/2 = 12 / 2 = 6 in

r/t = 6/0.25 = 24

24 > 10

Using the  thin wall cylinder formula;

The valve A is opened and the flowing water has a pressure P of 200 psi.

So;

\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}

\sigma_{long} = \sigma _2 = 0

\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{200(12)}{2(0.25)}

\mathbf{\sigma _ 1 = 4800 psi}

b)The valve A is closed and the water pressure P is 250 psi.

where P = 250 psi

\sigma_{hoop} = \sigma _ 1 = \frac{Pd}{2t}

\sigma_{long} = \sigma _2 = \frac{Pd}{4t}

\sigma _ 1 = \frac{Pd}{2t} \\ \\ \sigma _ 1 = \frac{250*(12)}{2(0.25)}

\mathbf{\sigma _ 1 = 6000 psi}

\sigma _2 = \frac{Pd}{4t} \\ \\  \sigma _2 = \frac{250(12)}{4(0.25)}

\mathbf{ \sigma _2 = 3000 psi}

The free flow body diagram showing the state of stress on a volume element located on the wall at point B is attached in the diagram below

8 0
3 years ago
Which of the following answer options are your employer's responsibility?
tino4ka555 [31]

Answer:

Develop a written hazard communication program

Implement a hazard communication program

Maintain a written hazard communication program

Explanation:

To find - Which of the following answer options are your employer's responsibility?  Select all that apply.

Develop a written hazard communication program

Implement a hazard communication program

Maintain a written hazard communication program

Solution -

The correct options are -

Develop a written hazard communication program

Implement a hazard communication program

Maintain a written hazard communication program

All are the Responsibilities of an employer

Reason -

The most important duty of the employer is to stay alert and implement a correctly and efficiently written communication program related to hazards of the substances in the workplace.

He also has to maintain the program so that employees do not get affected.

3 0
3 years ago
A flat plate 1.5 m long and 1.0 m wide is towed in water at 20 o C in the direction of its length at a speed of 15 cm/s. Determi
beks73 [17]

Answer:

15.8

0.0944

Explanation:

L = 1.5

B = 1.0

Speed of water = 15cm

Temperature = 20⁰C

At 20⁰C

Specific weight = 9790

Kinematic viscosity v = 1.00x10^-4m²/s

Dynamic viscosity u = 1.00x10^-3

Density p = 998kg/m²

Reynolds number

= 0.15x1.5/1.00x10^-4

= 225000

S = 5

5x1.5/225000^1/2

= 0.0158

= 15.8mm

Resistance on one side of plate

F = 0.664x1x1.0x10^-3x0.15x225000^1/2

= 0.04724N

Total resistance

= 2N

= 2x0.04724

= 0.0944N

3 0
3 years ago
If gain of the critically damped system is increased, the system will behave as a) Under damped b) Over damped c) Critically dam
Ganezh [65]

Answer:

a) Under damped

Explanation:

Given that system is critically damped .And we have to find out the condition when gain is increased.

As we know that damping ratio given as follows

\zeta =\dfrac{C}{C_c}

Where C is the damping coefficient and Cc is the critical damping coefficient.

C_c=2\sqrt{mK}

So from above we can say that

\zeta =\dfrac{C}{2\sqrt{mK}}

\zeta \alpha \dfrac{1}{\sqrt K}

From above relationship we can say when gain (K) is increases then system will become under damped system.

7 0
3 years ago
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