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Alchen [17]
4 years ago
7

SSL/TLS is used for ________.

Engineering
1 answer:
Nana76 [90]4 years ago
4 0

Answer:

1.Originally developed by Netscape, SSL is an Internet security protocol used by Internet browsers and Web servers to transmit sensitive information. SSL has become part of an overall security protocol known as Transport Layer Security (TLS). SSL/TLS can be used for Web applications

2.Small malware programs that download larger malware programs are called downloaders

3. Single network addresses in different single networks are neither unique nor unique if the two single networks use the same technology standards

4. Converting application messages into bits is called encoding

5. Guarantees by a provider to meet certain delivery quality specifications are called service level agreements

6. Advanced persistent threats are particularly dangerous for enterprises, as hackers have ongoing access to sensitive company data. They are extremely dangerous for the victim

7. In antivirus filtering, the best ways to filter currently use BEHAVIOURAL DETECTION

8. The general name for malware on a user's PC that collects sensitive information and sends this information to an attacker is SPYWARE

9. Firewall log files should be read EVERYDAY

10. When a firewall identifies an attack packet, it both discards the packet and copies information about the packet into a log file

11. IT security employees are the most difficult employees

12. In authentication, the SUPPLICANT is the party trying to prove his or her identity.

13. In a firewall, VPN traversal reduces firewall filtering effectiveness

14. DIRECTLY-PROPAGATING WORM can spread more rapidly

15 UDP is an unreliable, connectionless transport layer protocol that provides message-based data transmission. TCP is an example of connection-oriented protocols.

16. Host P transmits a SYN segment to Host Q. If host Q is willing to open the connection, it will transmit a SYN/ACK segment.

17. 802.11 data link layer standards govern neither access point operation nor frame organization

18. Credit card number thieves are called CARDERS

19. Using encryption, you make it impossible for attackers to read your messages even if they intercept them. This is CONFIDENTIALITY

20. Iris scanning is attractive because of its PRECISION

21. The source socket is 60.171.18.22:57707. The source is a CLIENT

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An inductor is energized as in the circuit of Fig. 2-4a. The circuit has L 10 mH and VCC 14 V. (a) Determine the required on tim
Savatey [412]

Answer:

A) 11.1 ms

B) 5.62 Ω

Explanation:

L ( inductance ) = 10 mH

Vcc = 14V

<u>A) determine the required on time of the switch such that the peak energy stored in the inductor is 1.2J </u>

first calculate for the current  ( i )  using the equation for energy stored in an inductor hence

i = \sqrt{\frac{2W}{L} }   ----- ( 1 )

where : W = 1.2j ,  L = 10 mH

Input values into equation 1  

i = 15.49 A

Now determine the time required  with expression below

i( t ) = 15.49 A

L = 10 mH, Vcc = 14

hence the time required ( T-on ) = 11.1 ms

attached below is detailed solution

B) <u>select the value of R such that switching cycle can be repeated every 20 ms </u>

using the expression below

τ = \frac{L}{R}  ---- ( 2 )

but first we will determine the value of τ

τ = t-off / 5 time constants

  = (20 - 11.1 ) / 5  = 1.78 ms

Back to equation 2

R = L / τ

  = (10 * 10^-3) / (1.78 * 10^-3)

  = 5.62 Ω

3 0
3 years ago
40 POINTS IF ANSERED WITHIN 10 MINS
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A It can show changes that occur in many different parts of Earth at the same time is the answer (I think)…
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If increased productivity in the workplace is an example of the macroeconomic impact of the green economy and increased income f
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3 years ago
If 4 squares equal 1 foot, how many inches is one square?
xeze [42]

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A cylindrical rod of a metal alloy is stressed elastically in tension. The original diameter was 11 mm and a force of 55 kN was
andrey2020 [161]

Answer:

10.984mm

Explanation:

by elastic modulus

stress=modulus of elasticity*strain

stress=loading/area                          area" cross-section"

11mm=0.011m

area=π(d/2)^2=π(0.011/2)^2=9.503*10^-5 square meter

stress=55000/(9.503*10^-5)=578.745 MPa

convert MPa and GPa to pascal.

strain=stress/modulus=(578.745*10^6)/(125*10^9)=0.00463............axial strain

v=Poisson  ratio

lateral strain=(-v)*axial strain= -0.31*0.00463

lateral strain= -1.4353*10^-3=change in diameter/ original diameter

change in diameter=(-1.4353*10^-3)*0.011= -1.57883*10^-5 m

negative indicates decrease in diameter.

decrease in dia.=0.01578mm

new diameter=11-0.01578= 10.984mm

3 0
3 years ago
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