What is the primary function of NCEES?

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Answer 1

Answer:

Answer:

A primary function of NCEES is to prepare standardized, confidential examinations that are used by the state and territorial boards to help determine the competency of individuals seeking to become licensed to practice as professional engineers and land surveyors.

Explanation:

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web application 3: where's the beef? provide a screenshot confirming that you successfully completed this exploit: [place screenshot here] write two or three sentences outlining mitigation strategies for this vulnerability: [enter answer here]

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Web application 3’s “Where’s the Beef” exploit can be seen in the screenshot provided. In order to mitigate this vulnerability, developers should take measures to ensure that input validation is performed on all data.

Additionally, developers should enforce strict rules on which characters are allowed in user inputs and reject all requests that don’t follow these rules. Finally, developers should implement best practices such as password hashing to ensure that user data is secure.

Developers should use secure coding techniques, such as sanitizing user input and properly escaping HTML output, to protect against injection-based attacks. Also, developers should implement authentication and authorization techniques to ensure that only authorized users have access to sensitive data. Furthermore, developers should use an up-to-date web application firewall to protect against known exploits, and use secure protocols such as HTTPS to protect data in transit. Finally, developers should ensure that software is kept up-to-date and patched to prevent exploitation of known vulnerabilities.

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For a steel alloy it has been determined that a carburizing heat treatment of 11-h duration will raise the carbon concentration to 0.38 wt% at a point 2.3 mm from the surface. Estimate the time (in h) necessary to achieve the same concentration at a 5.3 mm position for an identical steel and at the same carburizing temperature.

Answers

Answer:

Time =t2=58.4 h

Explanation:

Since temperature is the same hence using condition

x^2/Dt=constant

where t is the time as temperature so D also remains constant

hence

x^2/t=constant

2.3^2/11=5.3^2/t2

time=t^2=58.4 h

three string are attached to a small metal ring, two of the strings make and angle of 35° with the vertical and each is pulled with a force of 7Newton. What force must be applied to the third string to keep the rings stationary?

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No clue sorry man I would help but I need help too

You are analyzing an open-return wind tunnel that intakes air at 20 m/s and 320K. When the flow exits the wind tunnel it is moving at a speed of 250 m/s. What is the temperature of the flow exiting that wind tunnel

Answers

The solution is in the attachment

Answer:

please find attached.

Explanation:

As a means of preventing ice formation on the wings of a small, private aircraft, it is proposed that electric resistance heating elements be installed within the wings. To determine representative power requirements, consider nominal flight conditions for which the plane moves at 100 m/s in air that is at a temperature of -23 degree C. If the characteristic length of the airfoil is L = 2 m and wind tunnel measurements indicate an average friction coefficient of of C_f = 0.0025 for the nominal conditions, what is the average heat flux needed to maintain a surface temperature of T_s = 5 degree C?

Answers

Answer:

Average heat flux=3729.82 W/ $m^(2)$

Explanation:

An environmental engineer is considering three methods for disposing of a nonhazardous chemical sludge: land application, fluidized-bed incineration, and private disposal contract. The estimates for each method are below. (a) Determine which has the least cost on the basis of an annual worth comparison at 10% per year. (b) Determine the equivalent present worth value of each alternative using its AW value

Answers

Answer:

Explanation:

The annual worth of land application = $ 110,000 ( A/P , 10% , 5 years ) + $ 95,000 - $ 15,000 (A/F , 10% , 5 years )

The annual worth of land application = $ 110,000 X 0.263797 + $ 95,000 - $ 15,000 X 0.163797

The annual worth of land application = $ 29,017.54 + $ 95,000 - $ 2,456.95

The annual worth of land application = $ 121,560.59

The annual worth of land Incineration = $ 800,000 ( A/P , 10% , 6 years ) + $ 60,000 - $ 250,000 X (A/F , 10% , 6 years )

The annual worth of land Incineration = $ 800,000 X 0.229607 + $ 60,000 - $ 250,000 X 0.129607

The annual worth of land Incineration = $ 211,283.85

The annual worth of contract = $ 190,000

The land application has the least cost , hence it is preferred .

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