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3 years ago

Which of the following elements of the CIA triad refers to maintaining and assuring the accuracy of data over its life-cycle?

Engineering

1 answer:

kenny6666 [7]3 years ago

4 0

Answer:

Integrity: involves maintaining and assuring the accuracy of data over its life-cycle

Explanation:

Confidentiality: This is a CIA triad designed to prevent sensitive information from reaching the wrong people, while making sure that the right people have access to it.

Integrity: This is a CIA triad that involves maintaining the consistency, accuracy, and trustworthiness of data over its entire life cycle.

Availability: This is a CIA triad that involves hardware repairs and maintaining a correctly functioning operating system environment that is free of software conflicts.

Authentication:This is a security control that is used to protect the system with regard to the CIA properties.

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Block A hangs by a cord from spring balance D and is submerged in a liquid C contained in beaker B. The mass of the beaker is 1.

nikitadnepr [17]

Answer:

a) $m_e= 3.05 Kg$

b) $\rho=1072.3kg/m^3$

c) $m_e= 3.05 Kg$

Explanation:

From the question we are told that:

Beaker Mass $m_b=1.20$

Liquid Mass $m_l=1.85$

Balance D:

Mass $m_d=3.10$

Balance E:

Mass $m_e=7.50$

Volume $v=4.15*10^{-3}m^3$

Generally the equation for Liquid's density is mathematically given by

$m_e=m_b+m_l+(\rho*v)$

$\rho=\frac{7.50-(1.2+1.85)}{4.15*10^{-3}}$

$\rho=1072.3kg/m^3$

Generally the equation for D's Reading at A pulled is mathematically given by

m_d = mass of block - mass of liquid displaced

$m_d=m- (\rho *v )$

$m=3.10+ (1072.30 *4.15*10^{-3}m^3 )$

$m=18.10kg$

Generally the equation for E's Reading at A pulled is mathematically given by

$m_e=m_b+m_l$

$m_e = 1.20 + 1.85$

$m_e= 3.05 Kg$

6 0

3 years ago

If the tank is designed to withstand a pressure of 5 MPaMPa, determine the required minimum wall thickness to the nearest millim

dmitriy555 [2]

Answer: hello some aspects of your question is missing below is the missing information

The gas tank is made from A-36 steel and has an inner diameter of 1.50 m.

answer:

≈ 22.5 mm

Explanation:

Given data:

Inner diameter = 1.5 m

pressure = 5 MPa

factor of safety = 1.5

<u>Calculate the required minimum wall thickness</u>

maximum-shear-stress theory ( σ allow ) = σγ / FS

= 250(10)^6 / 1.5 = 166.67 (10^6) Pa

given that |σ| = σ allow

3.75 (10^6) / t = 166.67 (10^6)

∴ t ( wall thickness ) = 0.0225 m ≈ 22.5 mm

4 0

2 years ago

Technician A says that carbon monoxide (CO) and hydrocarbon (HC) levels should increase if the air injection reactor (AIR) hoses

lara31 [8.8K]

Answer:Technician A

Explanation:

7 0

3 years ago

Read 2 more answers

Flank wear data were collected in a series of turning tests using a coated carbide tool on hardened alloy steel at a feed of 0.3

Paladinen [302]

Answer:

A) n = 0.6143, c ≈ 640m/min

B) n = 0.6143 , c = 637.53m/min

Explanation:

using the given data

A) A plot of flank wear as a function of time and also A plot for tool when

Flank wear is 0.75 and cutting edge speed is 100m/min, Time of cutting edge is said to be 20.4 min also for cutting edge speed of 155m/min , time for cutting edge is 10 min

is attached below

calculate for the constant N from the second plot

note : the slope will be negative because cutting speed decreases as time of cutting increase

V1 = 100m/min , V2 = 155m/min, T1 = 20.4 min, T2 = 10 min

= - N = $\frac{In(V2) - In(V1)}{In(T2)-ln(T1)}$

therefore - N = $\frac{5.043 - 4.605}{2.302 -3.015}$

= - 0.6143

THEREFORE ( N ) = 0.6143

Determine for the constant C from the second plot as well

note : C is the intercept on the cutting speed axis in 1 min tool life

connecting the two points with a line and extend it to touch the cutting speed axis and measure the value at that point

hence C ≈ 640m/min

B) Calculate the values of N and C in the Taylor equation solving simultaneous equations

using the above cutting speed and time of cutting values we can find the constant N via Taylor tool life equation

Taylor tool life equation = vT = C ------------- equation 1

cutting speed = v = 100m/min and 155m/min

tool life = T = 20.4 min and 10 min

also constant n and c are obtained from the previous plot

back to taylor tool life equation = 100 * 20.4 = C

therefore C = (100)(20.4)^n ---------------- equation 2

also using the second values of v and T

taylor tool life equation = 155 * 10 = C

therefore C = ( 155 )(10)^n ----------------- equation 3

Equate equation 2 and equation 3 and solve simultaneously

(100)(20.4)^n = (155)(10)^n

To find N

take natural log of both sides of the equation

= In ((100)(20.4)^n) = In((155)(10)^n)

= In (100) + nIn(20.4) = In(155) + nIn(10)^n

= n(3.0155) - n (2.3026) = 5.043 - 4.605

= 0.7129 n = 0.438

therefore n = 0.6143

To find C

substitute 0.6143 for n in equation 2

C = (100)(20.4) ^ 0.6143

C = 637.53 m/min

Attached are the two plots for solution A

7 0

3 years ago

A very long pipe of 0.05 m (r1) radius and 0.03 m thickness (r2 - r1) is buried at a depth of 2m (z) to transport liquid nitroge

Elenna [48]

Answer:

The surface temperature of the ground is = 296.946K

Explanation:

Solution

Given

r₁= 0.05m

r₂= 0.08m

Tn =Ti = 77K

Ki = 0.0035 Wm-1K-1

Kg = 1 Wm-1K-1

Z= 2m

Now,

The outer type temperature (Skin temperature pipe)

Q = T₀ -T₁/ln (r2/r1)/2πKi = 2πKi T0 -T1/ln (r2/r1)

Thus,

10 w/m = 2π * 0.0035 = T0 -77/ln 0.08/0.05

⇒ T₀ -77 = 231.72

T₀= 290.72K

The shape factor between the cylinder and he ground

S = 2πL/ln 4z/D

where L = length of pipe

D = outer layer of pipe

S = 2π * 1/4 *2/ 2 * 0.08 = 1.606m

The heat gained in the pipe is = S * Kg * (Tg- T₀)

(10* 1) = 1.606 * 1* (Tg- 290.72)

Tg - 290.72 = 6.2266

Tg = 296.946K

Therefore the surface temperature to the ground is 296.946K

6 0

3 years ago