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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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Consider the following two well-mixed, isothermal gas-phase batch reactors for the elementary and irreversible decomposition of
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Answer:

attached below

Explanation:

8 0
3 years ago
If 1,490 compacted cubic yards of in-place soil is required for a project, how many loads of import will be required? The import
trasher [3.6K]

Answer:

149 loads of import will be required for the project.

Explanation:

bank cubic yards = bcy

Loose cubic yards = lcy

Calculate for the required bank cubic yards.

Required bank cubic yards = required in place cubic yards/shrinkage

Substitute

required in place cubic yards = 1,490 ccy

shrinkage percentage = 95% = 0.95

Required bank cubic yards = 1,490/0.95 = 1,568.42 bcy

Determine the volume of the material to be transported

Loose cubic yards = bank cubic yards * (1 + swell percentage)

swell percentage = 14% = 0.14

Loose cubic yards = 1,568.42*(1+0.14) = 1,788 lcy

loads of import that will be required = Loose cubic yards/truck haul capacity

loads of import that will be required = 1,788/12 = 149 load

3 0
3 years ago
Read 2 more answers
A spherical gas container made of steel has a(n) 17-ft outer diameter and a wall thickness of 0.375 in. Knowing that the interna
Arte-miy333 [17]

Answer:

Maximum Normal Stress σ = 8.16 Ksi

Maximum Shearing Stress τ = 4.08 Ksi

Explanation:

Outer diameter of spherical container D = 17 ft

Convert feet to inches D = 17 x 12 in = 204 inches

Wall thickness t = 0.375 in

Internal Pressure P = 60 Psi

Maximum Normal Stress σ = PD / 4t

σ = PD / 4t

σ = (60 psi x 204 in) / (4 x 0.375 in)

σ = 12,240 / 1.5

σ = 8,160 P/in

σ = 8.16 Ksi

Maximum Shearing Stress τ = PD / 8t

τ = PD / 8t

τ = (60 psi x 204 in) / (8 x 0.375 in)

τ = 12,240 / 3

τ = 4,080 P/in

τ = 4.08 Ksi

7 0
3 years ago
Steam flows through a copper pipe (kc=150Wm∙K⁄) that has an inner diameter of 19 mm and an outer diameter of 21 mm. The convecti
olchik [2.2K]

Answer:

a) The temperature on the outer surface of the pipe is approximately 179.97 °C

b) The thickness of the insulation  is approximately 0.857 m

Explanation:

We have;

\dfrac{1}{U} = \dfrac{1}{\alpha _A}

αA = 200 W/(m²·K)

\dot q =  (T₂ - T₁) × U

\dot q =  (200 - 180) × 200 = 4,000

For the pipe, we have;

\dfrac{1}{U} =\dfrac{x}{kc }

\dot q/U=  (T₂ - T₁)

∴ 4000×\dfrac{0.001}{150} = (180 - T₂)

T₂ ≈ 179.97 °C

The temperature on the outer surface of the pipe, T₂ ≈ 179.97 °C

b) For the insulation, we have;

\dfrac{1}{U} = \dfrac{x}{ki } = \dfrac{x}{0.03}

T₂ - T₃ = 179.97 °C - 40°C ≈ 139.97°C

\dot q/U=  (T₂ - T₃)

x = \dfrac{\dot q \cdot kc}{T_2 - T_3} = \dfrac{4,000 \times 0.03}{139.97} \approx 0.857

The thickness of the insulation, x ≈ 0.857 m

7 0
3 years ago
You wish to lift a 12,000 lb stone by a vertical distance of 15 ft. Unfortunately, you can only generate a maximum pushing force
natali 33 [55]

Answer:

A) 180000 lbft

B) 6

C) 90 ft

Explanation:

work to be imputed is

Wi = F x d

= 12,000 lb x 15 ft

= 180000 lbft of work

Work you can produce is

Wa = F x d

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Mechanical advantage that will be needed = Wi/Wa = 180000/30000

= 6

Lenght of ramp = 6 x 15 ft = 90 ft

7 0
3 years ago
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