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Vsevolod [243]
3 years ago
8

n the Spring of 2015, three utility companies in the Ukraine received email purporting to come from Ukraine's parliament, the Ra

da. It was addressed to employees that were used to receiving communications from the Rada dealing with utility regulations. The email included an attached Microsoft Excel spreadsheet with malicious Macros. This was an example of what kind of attack
Engineering
1 answer:
Tresset [83]3 years ago
5 0

Answer:

Trojan horse

Explanation:

A trojan horse attack is a type of malware that misleads users, as it appears unsuspicious at first, but actually presents a threat to the user. A common example is that of an email that contains a malicious attachment. Another common example is that of a fake advertisement. The name comes from the Greek story of the Trojan horse that led to the fall of the city of Troy.

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A stainless steel ball (rho = 8055 kg/m3, cp = 480 J/kg·K) of diameter D = 0.21 m is removed from the oven at a uniform temperat
Nataliya [291]

Answer:

Explanation:

The complete detailed  explanation which answer the question efficiently is shown in the attached files below.

I hope it helps a lot !

5 0
3 years ago
Read 2 more answers
Which option identifies why Ethan’s skills are valuable to his team in the following scenario?
larisa [96]

Answer:

Explanation:

The options are:

- In an isometric drawing, multiple angles and axes can be shown in one sketch.

- There is no room for detail in an isometric drawing, so the detail is shown in the orthographic projection.

- Only one sketch will be needed since all other previous designs will no longer be necessary.

- Computer programs will not be necessary to create the exact dimensions of the design.

Orthographic projections are in either the First or Third Angles but the angles are fixed and do not provide perspective view.  Isometric drawings are perspective views from different angles.

So Ethan's skill is valuable because "In an isometric drawing, multiple angles and axes can be shown in one sketch."

4 0
3 years ago
One kg of an idea gas is contained in one side of a well-insulated vessel at 800 kPa. The other side of the vessel is under vacu
laiz [17]

Answer:

Option C = internal energy stays the same.

Explanation:

The internal energy will remain the same or unchanged because this question has to do with a concept in physics or classical chemistry (in thermodynamics) known as Free expansion.

So, the internal energy will be equals to the multiplication of the change in temperature, the heat capacity (keeping volume constant) and the number of moles. And in free expansion the internal energy is ZERO/UNCHANGED.

Where, the internal energy, ∆U = 0 =quantity of heat, q - work,w.

The amount of heat,q = Work,w.

In the concept of free expansion the only thing that changes is the volume.

7 0
4 years ago
A steel bar 100 mm long and having a square cross section 20 mm x 20 mm is pulled in
Ierofanga [76]

Answer:

222.5 Gpa

Explanation:

From definition of engineering stress, \sigma=\frac {F}{A}

where F is applied force and A is original area

Also, engineering strain, \epsilon=\frac {\triangle l}{l} where l is original area and \triangle l is elongation

We also know that Hooke's law states that E=\frac {\sigma}{\epsilon}=\frac {\frac {F}{A}}{\frac {\triangle l}{l}}=\frac {Fl}{A\triangle l}

Since A=20 mm* 20 mm= 0.02 m*0.02 m

F= 89000 N

l= 100 mm= 0.1 m

\triangle l= 0.1 mm= 0.1\times 10^{-3} m

By substitution we obtain

E=\frac {89000\times 0.1}{0.02^{2}\times 0.1\times 10^{-3}}=2.225\times 10^{11}= 225.5 Gpa

5 0
3 years ago
Assume a program requires the execution of 50 x 10^6 FP instructions, 110 x 10^6 INT instructions, 80 x 10^6 Load/Store (L/S) in
Svetradugi [14.3K]

Answer:

We can not improve CPI of FP instructions when we run the program two times faster because it would be negative.

Explanation:

Processor clock rate = 2 GHz

Execution Time =   ∑  (\frac{Clock cyles}{Clock rate})

Clock cycles can be determined using following formula

Clock cycles = (CPI_{FP} x  No. FP instructions )+ ( CPI_{INT} x No. INT instructions) + ( CPI_{L/S}  x No. L/S instructions ) + ( CPI_{branch} x No. branch instructions)

Clock cycles = ( 50 x 10^{6} x 1) + (  110 x 10^{6} x 1) + ( 80 x 10^{6} x 4) + ( 16 x 10^{6} x 2)

Clock cycles = 512 x 10⁶

So,Initial Execution time for FP instructions is,

    = \frac{512(10^{6}) }{2(10^{9}) }

 Initial execution Time =  256 x 10⁻³

For 16 processors ,

clock cycle = 512 x 10⁶

Execution Time = 256 x 10⁻³

To run the program two times faster, half the number of clock cycles

(\frac{Clockcycles}{2} )=   (CPI_{FP} x  No. FP instructions )+ ( CPI_{INT} x No. INT instructions) + ( CPI_{L/S}  x No. L/S instructions ) + ( CPI_{branch} x No. branch instructions)

CPI_{FP improved} x No. FP instructions  =  (\frac{Clockcycles}{2} ) -[ ( CPI_{INT} x No. INT instructions) + ( CPI_{L/S}  x No. L/S instructions ) + ( CPI_{branch} x No. branch instructions)]

CPI_{FP improved} x 50 x 10^{6}  = ( \frac{512(10)^{6} }{2} ) - [ (  110 x 10^{6} x 1) + ( 80 x 10^{6} x 4) + ( 16 x 10^{6} x 2)]

CPI_{FP improved} x 50 x 10^{6}  =  - 206 x 10^{6}

CPI_{FP improved}  = - 206 x 10^{6} / 50 x 10^{6}

CPI_{FP improved} = - 4.12 < 0

3 0
4 years ago
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