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2 years ago

Which technical practice incorporates build-time identification of security vulnerabilities in the code?

Engineering

1 answer:

jonny [76]2 years ago

3 0

Answer:

Penetration testing

<h3>What is Penetrating Testing?</h3>

A penetration test, colloquially known as a pen test or ethical hacking, is an authorized simulated cyberattack on a computer system, performed to evaluate the security of the system; this is not to be confused with a vulnerability assessment.

To learn more about it, refer

to brainly.com/question/22654163

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A 15 ft high vertical wall retains an overconsolidated soil where OCR-1.5, c'-: O, ф , --33°, and 1 1 5.0 lb/ft3.

Tamiku [17]

Answer:

magnitude of thrust uis 11061.65 lb/ft

location is 5 ft from bottom

Explanation:

Given data:

Height of vertical wall is 15 ft

OCR is 1.5

$\phi = 33^o$

saturated uit weight $\gamma_{sat} = 115.0 lb/ft^3$

coeeficent of earth pressure $K_o$

$K_o = 1 -sin \phi$

= 1 - sin 33 = 0.455

for over consolidate

$K_{con} = K_o \times OCR$

$= 0.455 \times 1.5 = 0.683$

Pressure at bottom of wall is

$P =K_{con} \times (\gamma_{sat} - \gamma_{w}) + \gamma_w \times H$

$= 0.683 \times (115 - 62.4) \times 15 + 62.4 \times 15$

P = 1474.88 lb/ft^3

Magnitude pf thrust is

$F= \frac{1}{2} PH$

$=\frac{1}{2} 1474.88\times 15 = 11061.65 lb/ft$

the location must H/3 from bottom so

$x = \frac{15}{3} = 5 ft$

5 0

4 years ago

Compressed Air In a piston-cylinder device, 10 gr of air is compressed isentropically. The air is initially at 27 °C and 110 kPa

Helen [10]

Answer:

(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ

Explanation:

Solution

Recall that:

A 10 gr of air is compressed isentropically

The initial air is at = 27 °C, 110 kPa

After compression air is at = a450 °C

For air, R=287 J/kg.K

cv = 716.5 J/kg.K

y = 1.4

Now,

(a) W efind the pressure on [MPa]

Thus,

T₂/T₁ = (p₂/p₁)^r-1/r

=(450 + 273)/27 + 273) =

=(p₂/110) ^0.4/1.4

p₂ becomes 2390.3 kPa

So, p₂ = 2.39 MPa

(b) For the increase in total internal energy, is given below:

ΔU = mCv (T₂ - T₁)

=(10/100) (716.5) (450 -27)

ΔU =3030 J

ΔU =3.03 kJ

ΔW = mR ( (T₂ - T₁) / k- 1

(10/100) (287) (450 -27)/1.4 -1

ΔW = 3035 J

Hence, the total work required is = 3.035 kJ

4 0

4 years ago

After the 2015 AFC Championship football game between the New England Patriots and the Indianapolis Colts, it was alleged that t

mars1129 [50]

Answer:

the final temperature would have been 2.81 °C if the pressure drop was 2 psi

Explanation:

if we assume that is no change in volume, there are not leaks present and also that the gas inside the football behaves as an ideal gas, we have:

initial state) P1 V = n R T1

final state) P2 V = n R T2

where P = absolute pressure , V = volume occupied by the gas, n = number of moles of gas, R = ideal gas constant T= absolute temperature

therefore since V= constant (constant volume) and n= constant ( no leaks), if we divide both equations

P2/P1 = T2/T1

therefore

T2 = T1 *(P2/P1)

since P1 absolute = P1 relative + P atmospheric (14.7 psi) = 12.5 psi + 14.7 psi = 27.2 psia

also P2 = P1 - 2 psi = 25.2 psia

T1 = 24.7°C + 273 °C = 297.7 K

therefore

T2 = T1 * (P2/P1) = 297.7 K ( 25.2 psia/27.2 psia) = 275.81 K

thus T2 = 275.81 K = 2.81 °C

8 0

3 years ago

I can’t find the code

Ymorist [56]

Answer:

what code for

Explanation:

8 0

4 years ago

Draw a circuit diagram connected in a reverse bias showing a p-n material and a battery

juin [17]

The Circuit Diagram is connected in a reverse bias attached below.

<h3>What is Reverse Bias? </h3>

Reverse biased p-n junction diode refers to the method by which a p-n junction diode stops electric current in the presence of applied voltage.

The positive terminal of the battery is connected to the n-type semiconductor material in a reverse biased p-n junction diode, while the negative terminal is connected to the p-type semiconductor material.

Learn more about Reverse biased here:

brainly.com/question/14703617

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8 0

2 years ago