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lilavasa [31]
2 years ago
14

Often an attacker crafts e-mail attacks containing malware designed to take advantage of the curiosity or even greed of the reci

pients. Imagine that the message body Amy saw on the e-mail from Davey had been "See our managers’ salaries and SSNs" instead of "Funniest joke you’ll see today." 1. Would it be ethical for Amy to open such a file? 2. If such an e-mail came in, what would be the best action to take?
Engineering
1 answer:
Mademuasel [1]2 years ago
5 0

Answer:

1) The ethical act Amy should display is to open the Email since it came in through her personal inbox. However, further action should not be taken since ordinarily opening the mail will not disclose or corrupt her system either locally of her system on the server and web space.

2)  If such email came in, the first action i will take is to open the mail then take a pause to read through the instruction because most scam or malware comes with an array of further instructions.

After carefully reading through, i will call up Davey to confirm if he sent such mail which answer will obviously be a No and in the case where i am unable to get a call through to Davey, i will immediately delete the mail from my inbox to avoid the mistake of clicking any embedded link within the mail.

You might be interested in
Air enters the compressor of a cold air-standard Brayton cycle with regeneration and reheat at 100 kPa, 300 K, with a mass flow
yanalaym [24]

Answer:

a. 47.48%

b. 35.58%

c. 2957.715 KW

Explanation:

T_2 =T_1 + \dfrac{T_{2s} - T_1}{\eta _c}

T₁ = 300 K

\dfrac{T_{2s}}{T_1} = \left( \dfrac{P_{2}}{P_1} \right)^{\dfrac{k-1}{k} }

T_{2s} = 300 \times (10) ^{\dfrac{0.4}{1.4} }

T_{2s} = 579.21 K

T₂ = 300+ (579.21 - 300)/0.8 = 649.01 K

T₃ = T₂ + \epsilon _{regen}(T₅ - T₂)

T₄ = 1400 K

Given that the pressure ratios across each turbine stage are equal, we have;

\dfrac{T_{5s}}{T_4} = \left( \dfrac{P_{5}}{P_4} \right)^{\dfrac{k-1}{k} }

T_{5s} = 1400×\left( 1/\sqrt{10}  \right)^{\dfrac{0.4}{1.4} }  = 1007.6 K

T₅ = T₄ + (T_{5s} - T₄)/\eta _t = 1400 + (1007.6- 1400)/0.8 = 909.5 K

T₃ = T₂ + \epsilon _{regen}(T₅ - T₂)

T₃ = 649.01 + 0.8*(909.5 - 649.01 ) = 857.402 K

T₆ = 1400 K

\dfrac{T_{7s}}{T_6} = \left( \dfrac{P_{7}}{P_6} \right)^{\dfrac{k-1}{k} }

T_{7s} = 1400×\left( 1/\sqrt{10}  \right)^{\dfrac{0.4}{1.4} }   = 1007.6 K

T₇ = T₆ + (T_{7s} - T₆)/\eta _t = 1400 + (1007.6 - 1400)/0.8 = 909.5 K

a. W_{net \ out} = cp(T₆ -T₇) = 1.005 * (1400 - 909.5) = 492.9525 KJ/kg

Heat supplied is given by the relation

cp(T₄ - T₃) + cp(T₆ - T₅) = 1.005*((1400 - 857.402) + (1400 - 909.5)) = 1038.26349 kJ/kg

Thermal efficiency of the cycle = (Net work output)/(Heat supplied)

Thermal efficiency of the cycle = (492.9525 )/(1038.26349 ) =0.4748 = 47.48%

b. bwr = \dfrac{W_{c,in}}{W_{t,out}}

bwr = (T₂ -T₁)/[(T₄ - T₅) +(T₆ -T₇)]  = (649.01 - 300)/((1400 - 909.5) + (1400 - 909.5)) = 35.58%

c. Power = 6 kg *492.9525 KJ/kg  = 2957.715 KW

3 0
3 years ago
1. A cylindrical casting is 0.3 m in diameter and 0.5 m in length. Another casting has the same metal is rectangular in cross-se
Lorico [155]

Based on the Chvorinov's rule, the diference in the <em>solidification</em> times of the two castings is 14.092 times the <em>solidification</em> time of the prism casting.

<h3>How to apply the Chvorinov's rule for casting processes</h3>

The Chvorinov's rule is an empirical method to estimate the cooling time of a casting in terms of a <em>reference</em> time. This rule states that cooling time (<em>t</em>) is directly proportional to the square of the volume (<em>V</em>), in cubic meters, divided to the surface area (<em>A</em>), in square meters. Now we proceed to model each casting:

<h3>Cylindrical casting</h3>

t = C · [0.25π · D² · L/(0.5π · D² + π · D · L)]²

t = C · [0.25 · D · L/(0.5 · D + L)]²    (1)

<h3>Prism casting</h3>

t' = C · [3 · T² · L/(6 · T · L + 2 · T · L + 6 · T²)]²

t' = C · [3 · T · L/(8 · L + 6 · T)]²     (2)

<h3>Relationship between the cross sections of both castings</h3>

3 · T² = 0.25π · D²     (3)

Where:

  • <em>t</em> - Cooling time of the cylindrical casting, in time unit.
  • <em>t'</em> - Cooling time of the prism casting, in time unit.
  • <em>C</em> - Cooling factor, in time unit per square meter.
  • <em>D</em> - Diameter of the cylinder, in meters.
  • <em>L</em> - Length of the casting, in meters.
  • <em>T</em> - Width of the cross section of the prism casting, in meters.

If we know that <em>D =</em> <em>0.3 m</em>, then the thickness of the prism casting is:

T = \sqrt{\frac{\pi}{12} }\cdot D

<em>T ≈ 0.153 m</em>

<em />

And (1) and (2) simplified into these forms:

<h3>Cylindrical casting</h3>

t = C · {0.25π · (0.3 m) · (0.5 m)/[0.5 · (0.3 m) + 0.5 m]}²

t = 0.0329 · C     (1b)

<h3>Prism casting</h3>

t' = C · {3 · (0.153 m) · (0.5 m)/[8 · (0.5 m) + 6 · (0.153 m)]}²

t' = 0.00218 · C     (2b)

Lastly we find the <em>percentual</em> difference in the solidification times of the two castings by using the following expression:

<em>r = (</em>1 <em>- t'/t) ×</em> 100 %

<em>r = (</em>1 <em>-</em> 0.00218<em>/</em>0.0329<em>) ×</em> 100 %

<em>r =</em> 93.374 %

The <em>cooling</em> time of the <em>prism</em> casting is 6.626 % of the <em>solidification</em> time of the <em>cylindrical</em> casting. The diference in the <em>solidification</em> times of the two castings is 14.092 times the <em>solidification</em> time of the <em>prism</em> casting. \blacksquare

To learn more on solidification times, we kindly invite to check this verified question: brainly.com/question/13536247

3 0
2 years ago
A(n) _____________ is used commonly in open split-phase motors to disconnect the start winding from the electrical circuit when
Alenkasestr [34]

A <u>centrifugal switch</u> is used commonly in open split-phase motors to disconnect the start winding from the electrical circuit when the motor reaches approximately 75% of its rated speed.

Hope that helps!

7 0
2 years ago
If there are 16 signal combinations (states) and a baud rate (number of signals/second) of 8000/second, how many bps could I sen
Mice21 [21]

Answer:

32000 bits/seconds

Explanation:

Given that :

there are 16  signal combinations (states) = 2⁴

bits  n = 4

and a baud rate (number of signals/second) = 8000/second

Therefore; the number of bits per seconds can be calculated as follows:

Number of bits per seconds = bits  n × number of signal per seconds

Number of bits per seconds =  4 × 8000/second

Number of bits per seconds = 32000 bits/seconds

6 0
2 years ago
15 POINTS! Help.
IRISSAK [1]

Answer: it would  overload

Explanation:

4 0
2 years ago
Read 2 more answers
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