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3 years ago

What is the best countermeasure against social engineering?

Engineering

1 answer:

Mkey [24]3 years ago

3 0

Answer:

Hello Monk7294!

Answer:

Employee education

Explanation:

The most important countermeasure for social engineering is employee education. All the employees should be trained to keep confidential data safe. As a part of security education, organizations have to provide timely orientation about their security policy to new employees. The security policy should address the consequences of the breaches.

- I Hope this helps Have an awesome day!

~ Chloe marcus <3

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Ainat [17]

Answer:

meepus weepus

Explanation:

8 0

3 years ago

Read 2 more answers

The rate of flow through an ideal clarifier is 8000m3 /d, the detention time is 1h and the depth is 3m. If a full-length movable

Fittoniya [83]

Answer:

a) 35%

b) yes it can be improved by moving the tray near the top

Tray should be located ( 1 to 2 meters below surface )

max removal efficiency ≈ 70%

c) The maximum removal will drop as the particle settling velocity = 0.5 m/h

Explanation:

Given data:

flow rate = 8000 m^3/d

Detention time = 1h

depth = 3m

Full length movable horizontal tray : 1m below surface

a) Determine percent removal of particles having a settling velocity of 1m/h

velocity of critical sized particle to be removed = Depth / Detention time

= 3 / 1 = 3m/h

The percent removal of particles having a settling velocity of 1m/h ≈ 35%

b) Determine if the removal efficiency of the clarifier can be improved by moving the tray, the location of the tray and the maximum removal efficiency

The tray should be located near the top of the tray ( i.e. 1 to 2 meters below surface ) because here the removal efficiency above the tray will be 100% but since the tank is quite small hence the

Total Maximum removal efficiency

= percent removal $_{above}$ + percent removal $_{below}$

= ( d $_{a}$ ,v $_{p}$ ) . $\frac{d_{a} }{depth}$ + ( d $_{a}$ ,v $_{p}$ ) . $\frac{depth - d_{a} }{depth}$ = 100

hence max removal efficiency ≈ 70%

c) what is the effect of moving the tray would be if the particle settling velocity were equal to 0.5m/h?

The maximum removal will drop as the particle settling velocity = 0.5 m/h

7 0

3 years ago

Column arrays: Transpose a row array Construct a row array countValues with elements 1 to endValue, using the double colon opera

White raven [17]

Answer:

Matlab code with step by step explanation and output results are given below

Explanation:

We have to construct a Matlab function that creates a row vector "countValues" with elements 1 to endValue. That means it starts from 1 and ends at the value provided by the user (endValue).

function countValues = CreateArray(endValue)

% Here we construct a row vector countValues from 1:endValue

countValues = 1:endValue;

% then we transpose this row vector into column vector

countValues = countValues';

end

Output:

Calling this function with the endValue=11 returns following output

CreateArray(11)

ans =

Hence the function works correctly. It creates a row vector then transposes it and makes it a column vector.

7 0

3 years ago

The shaft is made of A992 steel. It has a diameter of 1 in. and is supported by bearings at A and D, which allows free rotation.

zysi [14]

Answer:

the angle of twist of B with respect to D is -1.15°

the angle of twist of C with respect to D is 1.15°

Explanation:

The missing diagram that is supposed to be added to this image is attached in the file below.

From the given information:

The shaft is made of A992 steel. It has a diameter of 1 in and is supported by bearing at A and D.

For the Modulus of Rigidity G = 11 × 10³ Ksi = 11 × 10⁶ lb/in²

The objective are :

1) To determine the angle of twist of B with respect to D

Considering the Polar moment of Inertia at the shaft $J\tau$

shaft $J\tau$ = $\dfrac{\pi}{2}r^4$

where ;

r = 1 in /2

r = 0.5 in

shaft $J \tau$ = $\dfrac{\pi}{2} \times 0.5^4$

shaft $J\tau$ = 0.098218

Now; the angle of twist at B with respect to D is calculated by using the expression

$\phi_{B/D} = \sum \dfrac{TL}{JG}$

$\phi_{B/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}$

where;

$T_{CD} \ \ and \ \ L_{CD}$ are the torques at segments CD and length at segments CD

${T_{BC} \ \ and \ \ L_{BC}}$ are the torques at segments BC and length at segments BC

Also ; from the diagram; the following values where obtained:

$L_{BC}}$ = 2.5 in

$J\tau$ = 0.098218

G = 11 × 10⁶ lb/in²

$T_{BC$ = -60 lb.ft

$T_{CD$ = 0 lb.ft

$L_{CD$ = 5.5 in

$\phi_{B/D} = 0+ \dfrac{[(-60 \times 12 )] (2.5 \times 12 )}{ (0.9818)(11 \times 10^6)}$

$\phi_{B/D} = \dfrac{[(-720 )] (30 )}{1079980}$

$\phi_{B/D} = \dfrac{-21600}{1079980}$

$\phi_{B/D} =$ − 0.02 rad

To degree; we have

$\phi_{B/D} = -0.02 \times \dfrac{180}{\pi}$

$\mathbf{\phi_{B/D} = -1.15^0}$

Since we have a negative sign; that typically illustrates that the angle of twist is in an anti- clockwise direction

Thus; the angle of twist of B with respect to D is 1.15°

(2) Determine the angle of twist of C with respect to D.Answer unit: degree or radians, two decimal places

For the angle of twist of C with respect to D; we have:

$\phi_{C/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}$

$\phi_{C/D} = 0+\dfrac{T_{BC}L_{BC}}{JG}$

$\phi_{B/D} = 0+ \dfrac{[(60 \times 12 )] (2.5 \times 12 )}{ (0.9818)(11 \times 10^6)}$

$\phi_{C/D} = \dfrac{21600}{1079980}$

$\phi_{C/D} =$ 0.02 rad

To degree; we have

$\phi_{C/D} = 0.02 \times \dfrac{180}{\pi}$

$\mathbf{\phi_{C/D} = 1.15^0}$

3 0

3 years ago

The thrust F of a screw propeller is known to depend upon the diameter d,speed of advance \nu ,fluid density p, revolution per s

musickatia [10]

Answer:

screw thrust = ML $T^{-2}$

Explanation:

thrust of a screw propeller is given by the equation = p $V^{2}$ $D^{2}$ x $\frac{ND}{V}$ Re

where,

D is diameter

V is the fluid velocity

p is the fluid density

N is the angular speed of the screw in revolution per second

Re is the Reynolds number which is equal to puD/μ

where p is the fluid density

u is the fluid velocity, and

μ is the fluid viscosity = kg/m.s = M $L^{-1}$ $T^{-1}$

Reynolds number is dimensionless so it cancels out

The dimensions of the variables are shown below in MLT

diameter is m = L

speed is in m/s = L $T^{-1}$

fluid density is in kg/ $m^{3}$ = M $L^{-3}$

N is in rad/s = L $L^{-1}$ $T^{-1}$ =

If we substitute these dimensions in their respective places in the equation, we get

thrust = M $L^{-3}$ $(LT^{-1}) ^{2}$ $L^{2}$ $\frac{T^{-1} L}{LT^{-1} }$

= M $L^{-3}$ $L^{2}$ $T^{-2}$

screw thrust = ML $T^{-2}$

This is the dimension for a force which indicates that thrust is a type of force

6 0

3 years ago