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2 years ago

Cisco what three best practices can help defend against social engineering attacks?

Engineering

1 answer:

SOVA2 [1]2 years ago

4 0

The three best practices that can help defend against social engineering attacks are:

Do not provide password resets in a c.ha.t window.
Resist the urge to click on enticing web links.
Educate employees regarding policies.

<h3>What is Social Engineering?</h3>

This refers to the art of manipulation of people through social interactions with the aim of getting access to sensitive information that gives unauthorized access.

Hence, we can see that some of the best practices that can be used to defend against social engineering include not clicking on suspicious links, educating people about the dangers, etc.

Read more about social engineering here:

brainly.com/question/26072214

#SPJ12

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4. In the Hyatt Regency walkway case study, it is reported that Jack Gillum stamps the 42 shop drawings, including the revised S

mario62 [17]

Answer:

Responsibility

Explanation:

By stamping the drawings that he was looking over, Jack Gillum conveys the fact that he is accepting responsibility for this work. The purpose of Gillum's stamp is to explain that such work has been under engineering review, and that it has fulfilled all the requirements that he watches our for. By putting his stamp in this work, Gillum accepts responsibility in case an error or a discrepancy is found in the drawings.

3 0

2 years ago

H2O enters a conical nozzle, operates at a steady state, at 2 MPa, 300 oC, with the inlet velocity 30 m/s and the mass flow rate

Colt1911 [192]

Answer:

The flow velocity at outlet is approximately 37.823 meters per second.

The inlet radius of the nozzle is approximately 0.258 meters.

Explanation:

A conical nozzle is a steady state device used to increase the velocity of a fluid at the expense of pressure. By First Law of Thermodynamics, we have the energy balance of the nozzle:

Energy Balance

$\dot m \cdot \left[\left(h_{in}+\frac{v_{in}^{2}}{2} \right)-\left(h_{out}+\frac{v_{out}^{2}}{2} \right)\right]= 0$ (1)

Where:

$\dot m$ - Mass flow, in kilograms per second.

$h_{in}$ , $h_{out}$ - Specific enthalpies at inlet and outlet, in kilojoules per second.

$v_{in}, v_{out}$ - Flow speed at inlet and outlet, in meters per second.

It is recommended to use water in the form of superheated steam to avoid the appearing of corrosion issues on the nozzle. From Property Charts of water we find the missing specific enthalpies:

Inlet (Superheated steam)

$p = 2000\,kPa$

$T = 300\,^{\circ}C$

$h_{in} = 3024.2\,\frac{kJ}{kg}$

$\nu_{in} = 0.12551\,\frac{m^{3}}{kg}$

Where $\nu_{in}$ is the specific volume of water at inlet, in cubic meters per kilogram.

Outlet (Superheated steam)

$p = 600\,kPa$

$T = 160\,^{\circ}C$

$h_{out} = 2758.9\,\frac{kJ}{kg}$

If we know that $\dot m = 50\,\frac{kJ}{kg}$ , $h_{in} = 3024.2\,\frac{kJ}{kg}$ , $h_{out} = 2758.9\,\frac{kJ}{kg}$ and $v_{in} = 30\,\frac{m}{s}$ , then the flow speed at outlet is:

$35765-25\cdot v_{out}^{2} = 0$ (2)

$v_{out} \approx 37.823\,\frac{m}{s}$

The flow velocity at outlet is approximately 37.823 meters per second.

The mass flow is related to the inlet radius ( $r_{in}$ ), in meters, by this expression:

$\dot m = \frac{\pi \cdot v_{in}\cdot r_{in}^{2} }{\nu_{in}}$ (3)

If we know that $\dot m = 50\,\frac{kJ}{kg}$ , $v_{in} = 30\,\frac{m}{s}$ and $\nu_{in} = 0.12551\,\frac{m^{3}}{kg}$ , then the inlet radius is:

$r_{in} = \sqrt{\frac{\dot m\cdot \nu_{in}}{\pi\cdot v_{in}}}$

$r_{in}\approx 0.258\,m$

The inlet radius of the nozzle is approximately 0.258 meters.

7 0

2 years ago

Basic Question please help

Dvinal [7]

1(A)
2(B)
3(E)
4(C)
5(D)
6(B)
7(C)

7 0

2 years ago

A 20 kg mass is thrown from the ground to a height of 50 m. (a) find the kinetic energy of the mass at this height. (b) find the

horrorfan [7]

Answer:

(a) 0 kJ

(b) 9.81 kJ

Explanation:

(a)

From the law of conservation of energy, energy can only be transformed from one state to another. At a height of 50 m, all the kinetic energy is converted to potential energy hence KE=0

(b)

Potential energy, PE=mgh where m is the mass, g is acceleration due to gravity and h is the height

Substituting 50 m for h and 20 Kg for m, taking g as 9.81 then

PE=20*9.81*50=9810 J=9.81 kJ

(c)

Relating the equation of potential energy to the equation of kinetic energy, which is $0.5mv^{2}$