Find three examples of good websites and three examples of bad websites. List them below, and in a

Which is a method of social engineering that could be used to deliver ransomware?.

bezimeni [28]

The method of social engineering that could be used to deliver ransomware is; email phishing

<h3>What are methods of Social engineering?</h3>

Ransomware is simply a type of social engineering that criminals use to steal data, infect computers, and infiltrate company networks.

Now, the most popular form of social engineering that could be used to deliver ransomware is email phishing. This is because email Phishing has become more pervasive and dangerous these days as it is used when attackers/criminals send fraudulent emails that either contain dangerous links or exploitative requests.

Read more about email phishing at; brainly.com/question/26072214

7 0

3 years ago

19 million Joules of chemical energy are supplied to a boiler in a power plant. 5 x 106 Joules of energy are lost through unburn

Fiesta28 [93]

Answer:

67.89%

Explanation:

Energy Efficiency= $\frac {U_{energy}}{T_{energy}}\times 100$

Where $U_{energy}$ is useful energy and $T_{energy}$ is the total energy

The useful energy= (19-5-1.1) million= 12.9 million

The total energy= 19 million

$Efficiency=\frac {12.9}{19}\times 100=67.89473684\approx 67.89$

Therefore, the efficiency of the boiler is 67.89%

8 0

3 years ago

The Web and Digital Communications pathway is broken down into four main categories of technology. Which category BEST describes

jasenka [17]

D marketing because he’s advertising ski equipment to buy

6 0

3 years ago

Consider a single crystal oriented such that the slip direction and normal to the slip plane are at angles 42.7° and 48.3°, resp

zlopas [31]

Answer:

Applied Stress > 58.29 MPa

Explanation:

Resolved shear stress should be greater than critically resolved shear stress in order to cause the single crystal to yield

Given angles are

∅ = 42.7 degree

Ф = 48.3 degree

Critically resolved shear stress = 28.5 MPa

If we consider

Critically resolved shear stress = resolved shear stress

Applied stress can be found by

$Z_{R} = applied stress X cos\phi X cos\theta$ (1)

Applied Stress = $\frac{Z_{R} }{Cos\phi XCos\theta}$

Applied Stress = $\frac{28.5}{Cos(48.3)XCos(42.7)}$

Applied Stress = 58.29 MPa

We got reference

By putting applied stress values of greater than 58.29 MPa in equation 1 we get

Resolved Shear Stress = 60 x Cos(48.3) x Cos(42.7)

Resolved Shear Stress = 29.33 MPa

Therefore, by the above calculation we conclude that applied stress should be greater than 58.29 MPa, In order to make resolved shear stress to be greater than critically resolved shear stress that is essential for single crystal to yield.

8 0

3 years ago

Can anyone pls help.Will mark as branliest

poizon [28]

Answer:

Explanation:

Hope this helped you!

5 0

3 years ago