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vivado [14]
4 years ago
11

To increase fault-tolerance, the security administrator for Corp has installed an active/passive firewall cluster where the seco

nd firewall is held in reserve in case of primary firewall failure. Stateful firewall inspection is being used in the firewall implementation. There have been numerous reports of dropped connections with external clients. Which of the following is MOST likely the cause of this problem?
(A) All packets are traversing the passive firewall causing the connections to be dropped.
(B) The heartbeat between the firewalls is not enabled.
(C) Inbound packets are traversing the active firewall and return traffic is being sent through the passive firewall.
(D) All packets are traversing the active firewall causing the connections to be dropped.
Engineering
1 answer:
zheka24 [161]4 years ago
5 0

Answer:

Inbound packets are traversing the active firewall and return traffic is being sent through the passive firewall

Explanation:

You might be interested in
A 993-kg car starts from rest at the bottom of a drive way and has a speed of 3.00 m/s at a point where the drive way has risen
tino4ka555 [31]

Answer:

13177.34 J

Explanation:

Work done = force × distance

work done by the engine = kinetic energy + potential energy + work done friction

kinetic energy due to the car's speed = 1/2mv² = 4468.5 J

potential energy due to the height = mgh = 993 kg × 9.8 m/s² × 0.6 m = 5838.84 J

work done by friction = 2870 J

work done by engine = 5838.84 J + 2870 J + 4468.5 J = 13177.34 J

7 0
3 years ago
Consider two different types of motors. Motor A has a characteristic life of 4100 hours (based on a MTTF of 4650 hours) and a sh
Daniel [21]

Answer:B

Explanation:

Given

For motor A

Characteristic life(r)=4100 hr

MTTF=4650 hrs

shape factor(B )=0.8

For motor B

Characteristic life(r)=336 hr

MTTF=300 hr

Shape Factor (B)=3

Reliability for 100 hours

R_a=e^{-\left ( \frac{T-r}{n}\right )B}

R_a=e^{-\left ( \frac{4650-4100}{100}\right )0.8}

R_a=e^{-4.4}=0.01227

For B

R_b=e^{-\left ( \frac{300-336}{100}\right )3}

R_b=e^{1.08}=2.944

B is better for 100 hours

(b)For 750 hours

R_a=e^{-0.5866}=0.55621

R_b=e^{0.144}=1.154

So here B is more Reliable.

3 0
3 years ago
Container need to be inspected
Sedaia [141]

Answer:

than look inside it

Explanation:

well if you need to inspect something, looking is very important

7 0
3 years ago
What separates the work of technology transfer research from implementation of the products of such research?
tatuchka [14]

Answer:

The thing that separates the work of technology transfer research from implementation of the products of such research is:

Technology transfer research looks at how technology may be transferred but does not actually make the transfer.

Explanation:

This suggests that technology transfer research is different from the technology transfer (implementation) itself.  The first stops at making scientific investigations into technology transfer activities while the next step performs the actual transfer or implementation.  In other words, technology transfer conveys the results of scientific and technological research to the marketplace and to the wider society.  It is the bridge-builder between the research and the implementation.

4 0
3 years ago
1.00-L insulated bottle is full of tea at 90.08°C. You pour out one cup of tea and immediately screw the stopper back on the bot
liberstina [14]

Answer:

T_{f} = 90.07998 ° C

Explanation:

This is a calorimetry process where the heat given by the Te is absorbed by the air at room temperature (T₀ = 25ºC) with a specific heat of 1,009 J / kg ºC, we assume that the amount of Tea in the cup is V₀ = 100 ml. The bottle being thermally insulated does not intervene in the process

                 Qc = -Qb

                M c_{e_Te} (T₁ -T_{f}) = m c_{e_air} (T_{f}-T₀)

Where M is the mass of Tea that remains after taking out the cup, the density of Te is the density of water plus the solids dissolved in them, the approximate values are from 1020 to 1200 kg / m³, for this calculation we use 1100 kg / m³

   ρ = m / V  

   V = 1000 -100 = 900 ml  

   V = 0.900 l (1 m3 / 1000 l) = 0.900 10⁻³ m³  

   V_air = 0.100 l = 0.1 10⁻³ m³  

Tea Mass  

     M = ρ V_te  

     M = 1100 0.9 10⁻³  

     M = 0.990 kg  

Air mass  

     m = ρ _air V_air  

     m = 1.225 0.1 10⁻³  

     m = 0.1225 10⁻³ kg  

(m c_{e_air} + M c_{e_Te}) T_{f}. = M c_{e_Te} T1 - m c_{e_air} T₀  

T_{f} = (M c_{e_Te} T₁ - m c_{e_air} T₀) / (m c_{e_air} + M c_{e_Te})  

Let's calculate  

T_{f} = (0.990 1100 90.08– 0.1225 10⁻³ 1.225 25) / (0.1225 10⁻³ 1.225 + 0.990 1100)  

T_{f} = (98097.12 -3.75 10⁻³) / (0.15 10⁻³ +1089)  

T_{f} = 98097.11 / 1089.0002  

T_{f} = 90.07998 ° C  

This temperature decrease is very small and cannot be measured

3 0
3 years ago
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