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miskamm [114]
3 years ago
8

In the 5 Code of Federal Regulations (C.F.R.), it is recommended that an individual has security awareness training before s/he

can access information. The C.F.R is unusual in that it requires all users to receive broad training in system/application life cycle management, security planning and system/application security management, risk management, and contingency planning.A. TrueB. false
Engineering
2 answers:
GarryVolchara [31]3 years ago
6 0

Answer: A it is true.

Explanation: It is true under Subpart C—Information Security Responsibilities for Employees who Manage or Use Federal Information Systems

Number 3 and 4

(3) Program and functional managers must receive training in information security basics; management and implementation level training in security planning and system/application security management; and management and implementation level training in system/application life cycle management, risk management, and contingency planning.

(4) Chief Information Officers (CIOs), IT security program managers, auditors, and other security-oriented personnel (e.g., system and network administrators, and system/application security officers) must receive training in information security basics and broad training in security planning, system and application security management, system/application life cycle management, risk management, and contingency planning.

Helen [10]3 years ago
5 0

Answer:

B. False

Explanation:

The Code of Federal Regulations (C.F.R.) also called the administrative law refers to the codes of conducts that governs and sets the boundaries for administrative agencies in the United States.

The C.F.R. code stipulates that every individual who wants to access the Federal Information System be exposed to a certain basic level of security awareness before doing so. However, the C.F.R. does not stipulate that all users undergo broad training in system/application life cycle management, security planning and system/application security management, risk management, and contingency planning.

According to the C.F.R. code, training is made available to individuals based on their roles and responsibilities; Executives receive a basic security training, Program managers receive management training as well as basic security training, Chief Information Officers & other security-oriented personnels are the ones that receive broad training in system/application life cycle management, security planning and system/application security management, risk management, and contingency planning.

The C.F.R. code also stipulates that new employees be introduced & acquainted with security training depending on their roles/positions before granting them access to the systems, that current employees be refreshed often with security training and to make security training available to employees when there is any significant change in the agency's information system procedure or if an employee is given a new role that demands additional training.

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A rigid tank contains 1 kg of oxygen (O2) at p1 = 35 bar, T1 = 180 K. The gas is cooled until the temperature drops to 150 K. De
andreyandreev [35.5K]

Answer:

a. Volume = 13.36 x 10^-3 m³ Pressure = 29.17 bar  b. Volume = 14.06 x 10^-3 m³ Pressure = 22.5 bar

Explanation:

Mass of O₂ = 1kg, Pressure (P1) = 35bar, T1= 180K, T2= 150k Molecular weight of O₂ = 32kg/Kmol

Volume of tank and final pressure using a)Ideal Gas Equation and b) Redlich - Kwong Equation

a. PV=mRT

V = {1 x (8314/32) x 180}/(35 x 10⁵) = 13.36 x 10^-3

Since it is a rigid tank the volume of the tank must remain constant and hnece we can say

T2/T1 = P2/P1, solving for P2

P2 = (150/180) x 35 = 29.17bar

b. P1 = {RT1/(v1-b)} - {a/v1(v1+b)(√T1)}

where R, a and b are constants with the values of, R = 0.08314bar.m³/kmol.K, a = 17.22(m³/kmol)√k, b = 0.02197m³/kmol

solving for v1

35 = {(0.08314 x 180)/(v1 - 0.02197)} - {17.22/(v1)(v1 + 0.02197)(√180)}

35 = {14.96542/(v1-0.02197)} - {1.2835/v1(v1 + 0.02197)}

Using Trial method to find v1

for v1 = 0.5

Right hand side becomes =  {14.96542/(0.5-0.02197)} - {1.2835/0.5(0.5 + 0.02197)} = 31.30 ≠ Left hand side

for v1 = 0.4

Right hand side becomes =  {14.96542/(0.4-0.02197)} - {1.2835/0.4(0.4 + 0.02197)} = 39.58 ≠ Left hand side

for v1 = 0.45

Right hand side becomes =  {14.96542/(0.45-0.02197)} - {1.2835/0.45(0.45 + 0.02197)} = 34.96 ≅ 35

Specific Volume = 35 m³/kmol

V = m x Vspecific/M = (1 x 0.45)/32 = 14.06 x 10^-3 m³

For Pressure P2, we know that v2= v1

P2 = {RT2/(v2-b)} - {a/v2(v2+b)(√T2)} = {(0.08314 x 150)/(0.45 - 0.02197)} - {17.22/(0.45)(0.45 + 0.02197)(√150)} = 22.5 bar

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3 years ago
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Answer:

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3 0
3 years ago
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Why is the uniaxial tension test commonly used to evaluate the mechanical properties of metals?
inn [45]

Answer:

Because with this test you can determine complex material parameters like Young’s modulus, yield strength, ultimate strength and elongation at break. This is important because it provides us with the factor of safety  that needs to be built-in the products using these materials.

Explanation:

This test consists in place the material between to tweezers to subdued the material into a stress-strain test. The figure shows the procedure.

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3 years ago
“In a List of Positive Integers, Set MINIMUM to 1. For each number X in the list L, compare it to MINIMUM. If X is smaller, set
artcher [175]

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7 0
2 years ago
Calculate the maximum load that a 7075 series aluminum alloy bar (with a T6 temper heat treatment) can support without permanent
Aleksandr [31]

Answer:

The maximum load the bar can withstand = 35.43 KN

Explanation:

Ultimate tensile strength of the given aluminium bar \sigma = 540 M pa

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We know that the ultimate strength of the bar is calculated from

\sigma = \frac{P_{max} }{A}

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P_{max} = 540 × 65.61

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Therefore the maximum load the bar can withstand = 35.43 KN

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3 years ago
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