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lilavasa [31]
3 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]3 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.

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The domain of discourse is the members of a chess club. The predicate B(x, y) means that person x has beaten person y at some po
satela [25.4K]

Answer:A. No one has ever beat Nancy.

Explanation:

The dormain of discourse in a simple language is the set of entities upon which our discussions are based when discussing about something.

The dormain of discourse is also known simply as universe, can also be said to be a set of entities o

upon which certain variables of interest in some formal treatment may range.

The dormain of discourse is generally attributed to Augustus De Morgan, it was also extensively used by George Boole in his Laws of Thought.

THE LOGICAL UNDERSTANDING OF THE THE QUESTION IS THAT NO ONE HAS EVER BEAT NANCY.

8 0
3 years ago
To increase the thermal efficiency of a reversible power cycle operating between thermal reservoirs at TH and Tc, would you incr
alukav5142 [94]

<u></u>\ T_{c} has greater effect.

<u>Explanation</u>:

\eta_{\max }=1-\frac{T_{c}}{T_{A}}

T_{c}\\ = Temperature of cold reservoir

T_{H} = Temperature of hot reservoir

when T_{c} is decreased by 't',

$\eta_{\text {incre }}$ = 1-\frac{\left(\tau_{c}-t\right)}{T_{H}}

=n \ + \frac{t}{T_{n}}      -(i)

when {T_{H}} is increased by 'T'

\eta_{i n c}=\frac{n+\frac{t}{T_{H}}}{\left(1+\frac{k}{T_{H}}\right)}-(ii)

\eta_{\text {incre }} \ T_{c}>\eta_{\text {incre }} T_{\text {H }}

7 0
3 years ago
A gas metal arc welder is also known as a _____ welder.<br> A) TIGB) GTAWC) GMAWD) Resistance spot
nignag [31]

Answer:

GMAW

Explanation:

It's literally the initials of that type of welding

7 0
3 years ago
Exhaust gas from a furnace is used to preheat the combustion air supplied to the furnace burners. The gas, which has a flow rate
Monica [59]

Answer:

The total tube surface area in m² required to achieve an air outlet temperature of 850 K is 192.3 m²

Explanation:

Here we have the heat Q given as follows;

Q = 15 × 1075 × (1100 - t_{A2}) = 10 × 1075 × (850 - 300) = 5912500 J

∴ 1100 - t_{A2} = 1100/3

t_{A2}  = 733.33 K

\Delta \bar{t}_{a} =\frac{t_{A_{1}}+t_{A_{2}}}{2} - \frac{t_{B_{1}}+t_{B_{2}}}{2}

Where

\Delta \bar{t}_{a} = Arithmetic mean temperature difference

t_{A_{1} = Inlet temperature of the gas = 1100 K

t_{A_{2} = Outlet temperature of the gas = 733.33 K

t_{B_{1} =  Inlet temperature of the air = 300 K

t_{B_{2} = Outlet temperature of the air = 850 K

Hence, plugging in the values, we have;

\Delta \bar{t}_{a} =\frac{1100+733.33}{2} - \frac{300+850}{2} = 341\tfrac{2}{3} \, K = 341.67 \, K

Hence, from;

\dot{Q} = UA\Delta \bar{t}_{a}, we have

5912500  = 90 × A × 341.67

A = \frac{5912500  }{90 \times 341.67} = 192.3 \, m^2

Hence, the total tube surface area in m² required to achieve an air outlet temperature of 850 K = 192.3 m².

4 0
3 years ago
A drilling operation is performed on a steel part using a 12.7 mm diameter twist drill with a point angle of 118 degrees. The ho
Masteriza [31]

Answer:

a. Rotational speed of the drill  = 375.96 rev/min

b. Feed rate  = 75 mm/min

c. Approach allowance  = 3.815 mm

d. Cutting time  = 0.67 minutes

e. Metal removal rate after the drill bit reaches full diameter. = 9525 mm³/min

Explanation:

Here we have

a. N = v/(πD) = 15/(0.0127·π) = 375.96 rev/min

b. Feed rate = fr = Nf = 375.96 × 0.2 = 75 mm/min

c. Approach allowance = tan 118/2 = (12.7/2)/tan 118/2 = 3.815 mm

d. Approach allowance T∞ =L/fr = 50/75 = 0.67 minutes

e. R = 0.25πD²fr = 9525 mm³/min.

7 0
3 years ago
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