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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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The design for a new cementless hip implant is to be studied using an instrumented implant and a fixed simulated femur.
OlgaM077 [116]

Answer:

a) the velocity of the implant immediately after impact is 20 m/s

b) the average resistance of the implant is 40000 N

Explanation:

a) The impulse momentum is:

mv1 + ∑Imp(1---->2) = mv2

According the exercise:

v1=0

∑Imp(1---->2) = F(t2-t1)

m=0.2 kg

Replacing:

0+F(t_{2} -t_{1} )=0.2v_{2}

if F=2 kN and t2-t1=2x10^-3 s. Replacing

0+2x10^{-3} (2x10^{-3} )=0.2v_{2} \\v_{2} =\frac{4}{0.2} =20m/s

b) Work and energy in the system is:

T2 - U(2----->3) = T3

where T2 and T3 are the kinetic energy and U(2----->3) is the work.

T_{2} =\frac{1}{2} mv_{2}^{2}  \\T_{3} =0\\U_{2---3} =-F_{res} x

Replacing:

\frac{1}{2} *0.2*20^{2} -F_{res} *0.001=0\\F_{res} =40000N

3 0
3 years ago
For the pipe-fl ow-reducing section of Fig. P3.54, D 1 5 8 cm, D 2 5 5 cm, and p 2 5 1 atm. All fl uids are at 20 8 C. If V 1 5
bonufazy [111]

Answer:

The total force resisted by the flange bolts is  163.98 N

Explanation:

Solution

The first step is to find  the pipe cross section at the inlet section

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A₁ = π /4 D₁²

D₁ =  diameter of the pipe at the inlet section

Now we insert 8 cm for D₁ which gives us A₁ = π /4 D (8)²

=50.265 cm² * ( 1 m²/100² cm²)

= 5.0265 * 10^⁻³ m²

Secondly, we find cross section area of  the pipe at the inlet section

A₂ = π /4 D₂²

D₂ =  diameter of the pipe at the inlet section

Now we insert 5 cm for D₁ which gives us A₁ = π /4 D (5)²

= 19.63 cm² * ( 1 m²/100² cm²)

= 1.963 * 10^⁻³ m²

Now,

we write down the conversation mass relation which is stated as follows:

Q₁ = Q₂

Where Q₁ and Q₂ are both the flow rate at the exist and inlet.

We now insert A₁V₁ for Q₁ and A₂V₂ for Q₂

So,

V₁ and V₂ are defined as the velocities at the inlet and exit

We now insert 5.0265 * 10^⁻³ m² for A₁ 5 m/s for V₁ and 1.963 * 10^⁻³ m² for A₂

= 5.0265  * 5 = 1.963 * V₂

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8 0
3 years ago
A highway reconstruction project is being undertaken to reduce crash rates. The reconstruction involves a major realignment of t
CaHeK987 [17]

Answer:

The provided length of the vertical curve is satisfactory for the reconstruction design speed of 60 mi/h

Explanation:

The explanation is shown on the first uploaded image

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3 years ago
How can I solve 23.5 million Nona meters to millimeters using no calculator because I have to show my work
katrin2010 [14]

Express it in standard form and apply the basic indices laws to simplify

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3 years ago
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A hydraulic cylinder is to be used to move a workpiece in a manufacturing operation through a distance of 50 mm in 10 s. A force
svetlana [45]

Answer:

The answer to this question is 1273885.3 ∅

Explanation:

<em>The first step is to determine the required  hydraulic flow rate liquid if working pressure and  if a cylinder with a piston diameter of 100 mm is available.</em>

<em>Given that,</em>

<em>The distance = 50mm</em>

<em>The time t =10 seconds</em>

<em>The force F = 10kN</em>

<em>The piston diameter is = 100mm</em>

<em>The pressure = F/A</em>

<em> 10 * 10^3/Δ/Δ </em>

<em> P = 1273885.3503 pa</em>

<em>Then</em>

<em>Power = work/time  = Force * distance /time</em>

<em> = 10 * 1000 * 0.050/10</em>

<em>which is  =50 watt</em>

<em>Power =∅ΔP</em>

<em>50 = 1273885.3 ∅</em>

5 0
3 years ago
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