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lidiya [134]
3 years ago
9

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?
Physics
1 answer:
Studentka2010 [4]3 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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If you took a white piece of plastic to a depth of 180 m, what color would it appear?.
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Answer:

It would appear blue

Explanation:

3 0
2 years ago
Please help answer question​
nika2105 [10]

Answer:

C = 1.01

Explanation:

Given that,

Mass, m = 75 kg

The terminal velocity of the mass, v_t=60\ m/s

Area of cross section, A=0.33\ m^2

We need to find the drag coefficient. At terminal velocity, the weight is balanced by the drag on the object. So,

R = W

or

\dfrac{1}{2}\rho CAv_t^2=mg

Where

\rho is the density of air = 1.225 kg/m³

C is drag coefficient

So,

C=\dfrac{2mg}{\rho Av_t^2}\\\\C=\dfrac{2\times 75\times 9.8}{1.225\times 0.33\times (60)^2}\\\\C=1.01

So, the drag coefficient is 1.01.

4 0
2 years ago
Do quasars reside within or without side of galaxies?
sveticcg [70]

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6 0
3 years ago
A student environmental group is creating a campaign for locally sourced energy resources.What would be best for them to feature
Ivenika [448]

Answer:

<em>A farmer with a field of solar panels.</em>

Explanation:

The closest to a locally sources energy would have been

A coal mine located in their county.

But coal as an energy source is not environmentally friendly due to carbon emission, and should not be what the group should advocate for.

<em>The best bet for them is </em>

<em>A farmer with a field of solar panels.</em>

As solar panels are a source of green energy and green energy is what the environmental group should often and always advocate for

3 0
2 years ago
If the range of a projectile's trajectory is six times larger than the height of the trajectory, then what was the angle of laun
zvonat [6]

Answer:

H = 1/2 g t^2    where t is time to fall a height H

H = 1/8 g T^2   where T is total time in air  (2 t  = T)

R = V T cos θ       horizontal range

3/4 g T^2 = V T cos θ       6 H = R    given in problem

cos θ = 3 g T / (4 V)           (I)

Now t = V sin θ / g     time for projectile to fall from max height

T = 2 V sin θ / g

T / V = 2 sin θ / g

cos θ = 3 g / 4 (T / V)     from (I)

cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ

tan θ = 2/3      

θ = 33.7 deg

As a check- let V = 100 m/s

Vx = 100 cos 33.7 = 83,2

Vy = 100 sin 33,7 = 55.5

T = 2 * 55.5 / 9.8 = 11.3 sec

H = 1/2 * 9.8 * (11.3 / 2)^2 = 156

R = 83.2 * 11.3 = 932

R / H = 932 / 156 = 5.97        6 within rounding

3 0
2 years ago
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