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4 years ago

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Josh manages security at a power plant. The facility is sensitive, and security is very important. He would like to incorporate

two-factor authentications with physical security. Which of the options below is the best way to meet this requirement?

1 answer:

Studentka2010 [4]4 years ago

8 0

Question: Josh manages security at a power plant. The facility is sensitive, and security is very important. He would like to incorporate two-factor authentications with physical security. Which of the options below is the best way to meet this requirement? A) Smart cards B) A mantrap with a smart card at one door and a pin keypad at the other door C) A mantrap with video surveillance D) A fence with smart card gate access Answer: The correct answer is (B) A mantrap with a smart card at one door and a pin keypad at the other door Explanation: Multi-factor access control systems are access control systems that authorises access only after more than one piece of evidence is presented to authenticate a user. The mechanisms used in multi-factor authentication include 1. Knowledge (something the user and only the user knows), 2. Possession (something the user and only the user has), and 3. Inherence (something the user and only the user is). A two-factor authentication system uses two of these mechanisms to grant or deny access. An authentication system that uses a mantrap with a smart card (possession) on one door and a pin keypad (knowledge ) on the other door is a two factor authentication system.

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3 years ago

Which of the following are NOT two of the best ways to StaySafe around electricity?

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1) Sleeping with an electronic in charge right next to you.

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3 years ago

Emmy is standing on a moving sidewalk that moves at +2 m/s. Suddenly, she realizes she might miss her flight, so begins to speed

likoan [24]

Answer:

Refer to the attachment for the diagram.

3.53 m/s.

Explanation:

Acceleration is the first derivative of velocity relative to time. In other words, the acceleration is the same as the slope (gradient) of the velocity-time graph. Let $t$ represents the time in seconds and $v$ the speed in meters-per-second.

For $0 < x \le 1$ :

Initial value of $v$ : $\rm 2\;m\cdot s^{-1}$ at $t = 0$ ; Hence the point on the segment: $(0, 2)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 2\; m\cdot s^{-2}$ .
Find the equation of this segment in slope-point form: $v - 2 = 2 (t - 0) \implies v = 2t + 2, \quad 0 < t \le 1$ .

Similarly, for $1 < x \le 2$ :

Initial value of $v$ is the same as the final value of $v$ in the previous equation at $t = 1$ : $t = 2t + 2 = 4$ ; Hence the point on the segment: $(1, 4)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 1\; m\cdot s^{-2}$ .

Find the equation of this segment in slope-point form: $v - 4 = (t - 1) \implies v = t + 3 \quad 1 < t \le 2$ .

For $2 < x \le 3$ :

Initial value of $v$ is the same as the final value of $v$ in the previous equation at $t = 2$ : $t = t + 3 = 5$ ; Hence the point on the segment: $(2, 5)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 0\; m\cdot s^{-2}$ . There's no acceleration. In other words, the velocity is constant.
Find the equation of this segment in slope-point form: $v - 5 = 0 (t - 2) \implies v = 5 \quad 2 < t \le 3$ .

For $3 < x \le 4$ :

Initial value of $v$ is the same as the final value of $v$ in the previous equation at $t = 3$ : $t = 5$ ; Hence the point on the segment: $(3, 5)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm -3\; m\cdot s^{-2}$ . In other words, the velocity is decreasing.
Find the equation of this segment in slope-point form: $v - 5 = -3 (t - 3) \implies v = -3t + 14 \quad 3 < t \le 4$ .

For $4 < x \le 5$ :

Initial value of $v$ is the same as the final value of $v$ in the previous equation at $t = 4$ : $t = -3t + 14$ ; Hence the point on the segment: $(4, 2)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 0\; m\cdot s^{-2}$ . In other words, the velocity is once again constant.
Find the equation of this segment in slope-point form: $v - 2 = 0 (t - 4) \implies v = 2\quad 4 < t \le 5$ .

$t = \rm 3.49\;s$ is in the interval $3 < x \le 4$ . Apply the equation for that interval: $v = -3t +14 = \rm 3.53\; m \cdot s^{-1}$ .

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3 years ago

A marble is placed at Location W. When the marble is released, it rolls down the track.

Norma-Jean [14]

Answer:

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Explanation:

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3 years ago

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