A systems analyst is concerned that the current authentication system may not provide the appropriate level of security. The company has integrated WAYF within its federation system and implemented a mandatory two-step authentication system. Some accounts are still becoming compromised via phishing attacks that redirect users to a fake portal, which is automatically collecting and replaying the stolen credentials.
Which of the following is a technical solution that would BEST reduce the risk of similar compromises?
A. Security awareness training
B. Push-based authentication
C. Software-based TOPT
D. OAuth tokens
E. Shibboleth

Answer: The tracer lathe is a roughing operation for the output shaft on rear wheel drive transmissions.

Explanation:

Answer:

Hey mate......

Explanation:

This is ur answer.....

Hope it helps!

Brainliest pls!

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Answer: Create lessons learned at the end of the project.

Explanation:

Lessons learned are the experiences that are gotten from a project which should be taken into account for the future projects. Lesson learned are created at the end of the project.

The main objective of the lessons learned is that they show both the positive experience and the negative experience of a project and this will help the future projects that will be undertaken.

Answer:

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Answer:

  5.9 watts

Explanation:

The secondary voltage is the primary voltage multiplied by the turns ratio:

  (120 V)(41) = 4920 V

The power is the product of voltage and current:

  (4920 V)(1.2·10^-3 A) = (4.92)(1.2) W = 5.904 W

The power consumed is about 5.9 watts.

The causes of low compression in an engine cylinder and leakage past the exhaust can be identified by performing mechanical tests on the engine

The correct technician is Tech B; Blown head gasket would generally leak past the exhaust valve

The reason for arriving at the above selection is as follows:

A cylinder leakage test is generally performed after there is an indication of low compression from one or more cylinders. The test allows the identification of the part of the cylinder that is leaking.

The cylinder that shows sign of low compression is often the location that the cylinder leakage test is performed, although, the test can be performed on every cylinder

A cylinder leakage test is valid only when

Therefore, the positioning of the piston in the bottom dead center during a leakage test as mentioned by Tech A  is not correct

The route of coolant in the engine are sealed by the cylinder head gasket, where there is a blown head gasket, the coolant will be allowed to leak into the cylinders and into the exhaust, causing white smoke and sweet odor of antifreeze

Therefore, Tech B is correct, coolant leakage into the exhaust is an indication of a blown head gasket

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Answer:

[tex]Q_2 = 32[/tex] mL/s

Explanation:

Given :

The flow is incompressible viscous flow.

The initial flow rate, [tex]Q_1[/tex] = 1 mL/s

Initial diameter, [tex]D_1= D_0[/tex]

Initial length, [tex]L_1=L_0[/tex]

The initial pressure difference to maintain the flow, [tex]P_1=P_0[/tex]

We know for a viscous flow,

[tex]$\Delta P = \frac{32 \mu V L}{D^2}$[/tex]

[tex]$\Delta P = \frac{32 \mu Q L}{\frac{\pi}{4}D^4}$[/tex]

[tex]$Q \propto \Delta P \times D^4$[/tex]

[tex]$\frac{Q_1}{Q_2}= \frac{P_1}{P_2} \times \left( \frac{D_1}{D_2} \right)^4$[/tex]

[tex]$\frac{1}{Q_2}= \frac{P_0}{2P_0} \times \left( \frac{D_0}{2D_0} \right)^4$[/tex]

[tex]$\frac{1}{Q_2}= \frac{1}{2} \times \left( \frac{1}{2} \right)^4$[/tex]

[tex]$\frac{1}{Q_2}= \frac{1}{32}$[/tex]

∴ [tex]Q_2 = 32[/tex] mL/s

The flow rate if the pressure difference was increased to 2P0 and the diameter was increased to 2D0 is; Q2 = 32 mL/s

We are given;

Initial flow rate; Q1 = 1 mL/s

Initial uniform diameter; D0

Initial Length; L0

Initial Pressure difference; P0

Relationship between pressure, flow rate and diameter for vicious flow is given by;

Q1/Q2 = (P1/P2) × (D1/D2)⁴

Where;

Q1 is initial flow rate

Q2 is final flow rate

P1 is initial pressure difference

P2 is final pressure difference

D1 is initial diameter

D2 is final diameter

We are told that the pressure difference was increased to 2P0 and the diameter was increased to 2D0. Thus;

P2 = 2P0

D2 = 2D0

Thus;

1/Q2 = (P0/2P0) × (D0/2D0)⁴

>> 1/Q2 = ½ × (½)⁴

1/Q2 = 1/32

Q2 = 32 mL/s

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Answer:

Explanation:

Be bop

Answer:

not understand language

Answer:

80% of the people that were killed weren't wearing a safety flotation device ( in correct terminology Personal Flotation Device, or PFD )

Explanation:

Hence they drowned due to the lack of safety.

The condition present in over 80% of paddling fatalities is that the victims were said to be not wearing a personal flotation device (PFD).

The act of paddling is known to be a sport or something one can do for fun or for games.

Conclusively, a lot of paddling fatalities 1995-2000 was due to the fact that the people who were engage in the sport or act were not putting on  a personal flotation device (PFD) and this can lead to death or drowning.

Learn more about  paddling fatalities from

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Answer:

Logging while drilling (LWD) is a technique of conveying well logging tools into the well borehole downhole as part of the bottom hole assembly (BHA). ... In these situations, the LWD measurement ensures that some measurement of the subsurface is captured in the event that wireline operations are not possible

Answer:

As per my experience Egor Popov is one the best book for mechanics of materials.

Explanation:

hope this helps

Given :

Power, P = 20 kW

Speed, N = 430 rpm

Allowable shear stress, τ = 65 MPa

Torque in the shaft is given by :

[tex]$P=\frac{2 \pi NT}{60}$[/tex]

[tex]$T=\frac{60 \times 20 \times 10^3}{2 \pi \times 430}$[/tex]

T = 444.37 N.m

Diameter of the solid shaft is

[tex]$d=\sqrt[3]{\frac{16 T}{\pi \tau}}[/tex]

[tex]$d=\sqrt[3]{\frac{16 \times 444.37}{3.14 \times 65}}[/tex]

[tex]$d=\sqrt[3]{34.83} $[/tex]

d = 3.265 m

d = 326.5 mm

Internal diameter of the hollow shaft is :

[tex]$\frac{T}{\frac{\pi}{32} \left( d_0^4 - d_i^4 \right)}=\frac{\tau}{d_0/2}$[/tex]

[tex]$\frac{444.37}{\frac{3.14}{32} \left( 0.036^4 - d_i^4 \right)}=\frac{65 \times 10^6}{0.036/2}$[/tex]

[tex]$\frac{444.37}{0.09 \left( 1.6 \times 10^{-6} - d_i^4 \right)}=\frac{65 \times 10^6}{0.018}$[/tex]

[tex]$\frac{7.99}{ \left( 1.6 \times 10^{-6} - d_i^4 \right)}=5850000$[/tex]

[tex]$1.3\times 10^{-6} = 1.6 \times 10^{-6} - d_i^4 \right)}$[/tex]

[tex]$d_i^4=300000$[/tex]

[tex]$d_i = 23.40$[/tex] mm

Percentage savings in the weight is given by :

Percentage saving = [tex]$\frac{W_{solid}-W_{hollow}}{W_{solid}}\times100$[/tex]

                                 [tex]$=\frac{V_{solid}-V_{hollow}}{V_{solid}}\times100$[/tex]

                                 [tex]$=\frac{d^2 - (d_0^2 - d_i^2)}{d^2} \times 100$[/tex]

                               [tex]$=\frac{(326.5)^2 - (0.036^2 - (32.40)^2)}{(326.5)^2} \times 100$[/tex]

                                 [tex]$=\frac{106602 - \left(1.29 \times 10^{-3} - 1049.76 \right)}{106602} \times 100$[/tex]

                                  [tex]$=\frac{106602 - 1049 }{106602} \times 100$[/tex]

                                  [tex]$=\frac{105553 }{106602} \times 100$[/tex]

                                  = 99.01 %

Answer:

Check it out here

Explanation:

Answer:

No it is not permitted

Explanation:

It is not permitted  because as per NEC 410.21 policy no other conductor is allowed to be passed through integral junction box luminaries unless such conductor supply recessed luminaries.

The marking will show that the Luminaries is of the right construction or right installation to ensure that the the conductors ( in the outer boxes ) will not be exposed to temperatures greater than the conductor rating, hence the lack of marking makes it not to be permitted.

Answer:

Non-functional requirements when defined and executed well will help to make the system easy to use and enhance the performance

Explanation:

Answer:

X = 1 (1st digit in the code)

Y = 0 (2nd digit)

Z = 4 (3rd multiplier digit)

104 → 10 × 10^4 Ω

→ 10 × 10000Ω

→ 100 kΩ

resistors are marked 104, 105, 205, 751, and 754. The resistor marked with 104 should be 100kΩ (10x10^4), 105 would be 1MΩ (10x10^5), and 205 is 2MΩ (20x10^5). 751 is 750Ω (75x10^1), and 754 is 750kΩ (75x10^4).

Here we need to understand how a code in a resistor gives us information on the resistor. Here we will see that the code means that the resistance is 100,000 Ω.

When we use numbers, let's assume that we have 3 single-digit numbers abc.

So if the code in our resistor is abc, this will mean that the resistance of the resistor is:

ab×10^c Ω

Using this general rule we can see that if the code is 104, then the resistance will be:

r = 10×10^4 Ω

 = 100,000 Ω

Then we can conclude that the resistive value of this resistor is  100,000 Ω

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Answer:

The main sections included in the bill of quantities are Form of Tender, Information, Requirements, Pricing schedule, Provisional sums, and Day works.

A project is meant to achieve clearly defined goals within present constraints

The correct option for the source of work in a project which is NOT a source of authorized work to be performed is option (a)

a. Scope creep

The reason for the selecting the above option is as follows;

In project management, alongside work planned in a work package, authorized work to be performed can come from;

A scope creep, however, is generally is due to the inclusion of the end

users of the project at a time that is behind schedule, insufficient analysis of

the requirements, and project complexity underestimation. and consists of

changes made after a project has commenced

A scope creep therefore, changes the amount of work to be done which

therefore alters the project timeline, and due to project, budgetary, and

end user constraints, CANNOT be a source of authorized work to be

performed

The correct option for the source that is NOT a authorized work to be performed is option a. Scope creep

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Answer: hello the complete question is attached below

answer:

A) Group symbol = SW

B) Group name = well graded sand , fine to coarse sand

C) It is not a clean sand given that ≤ 50% particles are retained on No 200

Explanation:

A) Classifying the soil according to USCS system

 ( using 2nd image attached below )

description of sand :

The soil is a coarse sand since  ≤ 50% particles are retained on No 200 sieve, also

The soil is a sand given that more than 50% particles passed from No 4 sieve

The soil can be a clean sand given that fines ≤ 12%

The soil can be said to be a well graded sand because the percentage of particles passing through decreases gradually over time

Group symbol as per the 2nd image attached below = SW

B) Group name = well graded sand , fine to coarse sand

C) It is not a clean sand given that ≤ 50% particles are retained on No 200

Answer:

B

Explanation:

Anxiety is very common especially nowadays but it's especially common in psychological disorders

Answer:

348 g

Explanation:

The heat gained by the ice equals the heat lost by the hot tea.

Also, since we require a very, very small amount of ice and some water, to still have some ice, the temperature of the hot tea has to decrease to the melting point of ice which is 0°C. So, the final temperature of the mixture is 0 °C.

So, mc(T - T') + mL = -MC(T - T") where m = mass of ice, c =  specific heat of ice = 2090 J/kgK., T = final temperature of mixture = 0 °C, T' = initial temperature of ice = -12 °C, L = heat of fusion of H2O = 3.33 × 10⁵ J/kg, M = mass of hot tea = ρV where ρ = density of tea = 1.00 g/mL and V = volume of hot tea = 350 mL. So, M = ρV = 1.00 g/mL × 350 mL = 350 g = 0.350 kg, C = specific heat of tea = 4190 J/kgK and T" = initial temperature of tea = 85 °C.

Making m subject of the formula, we have

mc(T - T') + mL = -MC(T - T")

m[c(T - T') + L] = -MC(T - T")  

m = -MC(T - T")/[c(T - T') + L]

substituting the values of the variables into the equation, we have

m = -MC(T - T")/[c(T - T') + L]

m = -0.350 kg × 4190 J/kgK(0 °C - 85 °C.)/[2090 J/kgK(0 °C. - (-12 °C)) + 3.33 × 10⁵ J/kg]

m = -1466.5 J/kK(- 85 °C)/[2090 J/kgK(0 °C + 12 °C)) + 3.33 × 10⁵ J/kg]

m = 124652.5 J/[2090 J/kgK(12 °C) + 3.33 × 10⁵ J/kg]

m = 124652.5 J/[25080 J/kg + 3.33 × 10⁵ J/kg]

m = 124652.5 J/358080 J/kg

m = 0.3481 kg

m = 348.1 g

m ≅ 348 g

So, the minimum amount of ice to be added is 348 g.

Complete question:

A steel component with a tensile strength of 800 MPa and fracture toughness Kic=20 MPa Nm is known to contain internal cracks (also called tunnel cracks) with the maximum length of 1.4 mm. This steel is being considered for use in a cyclic loading application for which the designed stresses vary from 0 to 420 MPa. Would you recommend using this steel in this application?

a. Not sure. Because cyclic loading is applied. Fatigue test is needed in order to make the recommendation.

b. Yes, this because the tensile strength of steel is much higher than the applied highest stress of 420 MPa.

c. Yes, this because the calculated critical stress to fracture for the cracks is higher than the highest applied stress of 420 MPa and the steel can withstand the stress of 420 MPa.

d. No. Although the calculated critical stress to fracture for the cracks is slightly higher than the highest applied stress of 420 MPa and the steel may withstand the static stress of 420 MPa, the cyclic loading may cause rapid fatigue fracture.

Answer:

A. Not sure. Because cyclic loading is applied. Fatigue test is needed in order to make the recommendation.

Explanation:

we are not sure if to recommend this alloy for this application given that this material has already been left to experience fatigue degradation. the cyclic load application brings about a growth in the crack. We know that cyclic loading is continuous loading that is useful for the testing of fatigue. Therefore the answer to this question is option a. We cannot make recommendations except fatigue testing has been carried out.

thank you!

Answer:

I don't understand the question

Answer:

Explanation:

Increased by the constant. Take a very simple case.

4  +  5 +  6 = 15

The mean is 5  (obtained by dividing the total (15) by the number of terms (3).

Now add a constant say 6

4 + 6 = 10

5 + 6 = 11

6 + 6 = 12

Total = 33/3 = 11

So the mean 5 is increased by the constant 6.

Now do the same thing more symbolically.

4 + c

5 + c

6 + c

Total = 15 + 3c

Divide by 3 you get 5 + c

If you want a more formal proof involving n terms, leave a note.