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ohaa [14]
3 years ago
10

Which of the following refers to a full-scale version of a product used to validate performance?

Engineering
2 answers:
kompoz [17]3 years ago
4 0
I’m thinking it would be c sorry if it’s wrong .
Serjik [45]3 years ago
3 0
It’s b i’m pretty sure i’m right
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In order to defend against side channel power analysis, we should: ______________
wariber [46]

Answer:

Some examples of predator and prey are lion and zebra, bear and fish, and fox and rabbit. ... The words "predator" and "prey" are almost always used to mean only animals that eat animals, but the same concept also applies to plants: Bear and berry, rabbit and lettuce, grasshopper and leaf

Explanation:

8 0
3 years ago
MITM can present in two forms: eavesdropping and manipulation. Discuss the process involved when an attacker is eavesdropping an
Nikitich [7]

Answer / Explanation:

Eavesdropping attack is also sometimes refereed to as sniffing attack. It is simply the process by which an attacker or hacker tries to penetrate very suddenly into an unaware individuals network or server with the intention to steal information transmitted over the network or server through that computer.  

To prevent such attack, there are several mean which include installing network monitoring software to check who else is connected to the network but the most common method of preventing such attack is to encrypt the Hypertext Transfer Protocol (http) and the way to do this is by securing it with a sort of security key.

On installing the security key, the network becomes encrypted and secured such that whatever network transmitted over the network becomes encrypted and unable to read. The protocol then converts to (https).

5 0
3 years ago
¿Cuál de los siguientes factores reduciría el riesgo de roturas por un choque térmico a. Alto coeficiente de dilatación térmico
hoa [83]

Answer:

c. Alto módulo de elasticidad

Explanation:

The correct answer to the given question is c. Alto modulo de elasticidad

A Youngs modulus measures the resistance of any material to elastic deformation. It is basically the ratio of the stress applied to a body to the results of the stress which is the response of the body over the pressure applied. This is to test the stiffness of any material and most of time the material stays constant over stressing.

5 0
3 years ago
A small hot surface at temperature Ti-430K having an emissivity 0.8 dissipates heat by radiation into a surrounding area at T2-4
Nat2105 [25]

Answer:

389.6 W/m²

Explanation:

The power radiated to the surroundings by the small hot surface, P = σεA(T₁⁴ - T₂⁴) where σ = Stefan-Boltzmann constant = 5.67 × 10⁻⁸ W/m²-K⁴, ε = emissivity = 0.8. T₁ = temperature of small hot surface = 430 K and T₂ = temperature of surroundings = 400 K

So, P = σεA(T₁⁴ - T₂⁴)

h = P/A = σε(T₁⁴ - T₂⁴)  

Substituting the values of the variables into the equation, we have

h = 5.67 × 10⁻⁸ W/m²-K⁴ × 0.8 ((430 K )⁴ - (400 K)⁴)  

h = 5.67 × 10⁻⁸ W/m²-K⁴ × 0.8 (34188010000 K⁴ - 25600000000 K⁴)  

h = 5.67 × 10⁻⁸ W/m²-K⁴ × 0.8 × 8588010000K⁴

h = 38955213360 × 10⁻⁸ W/m²

h = 389.55213360 W/m²

h ≅ 389.6 W/m²

5 0
2 years ago
Assuming 800 W / m 2 solar irradiance and a 35 % efficient solar panel, how much roof area should be covered to supply 10 A at 1
Korvikt [17]

Answer:

a) 4.286 m² b) $ 55.44/mo

Explanation:

If we assume that the sun is behaving as an isotropic radiator, the power density that is arriving to the house, is constant and equal to the quoted solar irradiance.

If the energy conversion capability of the solar panels were 100%, the roof area needed to supply the power required, would be simply the quotient between the power required and the solar irradiance, as follows:

A = P / SI = 10 A* 120 V / 800 W/m² = 1200 W / 800 W/m²= 1,5 m²

As the solar panels are only 35% efficient in converting the solar energy to useful electrical energy, we will need more roof area, according to this expression:

Ae = At / 0.35 = 1,5 / 0.35 = 4.286 m²

b) If we can get 1200 W during 7 hs/day, the energy supplied by the solar panels will be the product of the power times the time, as follows:

E= 1200 W* 7 hs = 8.4 Kwh

If the cost per Kwh, is $0.22, assuming 7 hs. of use in average during a month (assumed to be of 30 days), we can have savings as follows:

Cost = 0.22($/Kwh)* 8.4 (Kwh/day)*30 (days/mo) = $ 55.44

8 0
3 years ago
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