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Amanda [17]
3 years ago
9

Working at a ski resort in the mountains has its own unique security issues. Kenny is the chief information officer for Sundance

Ski Resort, and he is faced with both physical and information security threats every month. Since the resort implemented a new software system, they have been having larger number of threats and breaches of company information. He suspects that this may be the cause of an internal employee. He needs to clarify and establish what type of plan to help reduce further problems?
A) An information security plan
B) An ethical information policy
C) An anti-virus plan
D) None of these
Physics
2 answers:
Naily [24]3 years ago
7 0

Answer:

An information security plan ( A )

Explanation:

An information security plan is a plan set in place by an organization through its information security expert to protect the vital information contained in the database of the company from unauthorized access by employees or external bodies not allowed to have access to such information.

one way of achieving an Information security plan includes setting up very strong passwords which will be frequently updated/changed to protect vital information. The information security plan is needed by Ski resort because the rate of threats and breaches of company information increased after the implementation of a new software by the company

Liono4ka [1.6K]3 years ago
6 0

Answer:

A) An information security plan

Explanation:

Information Security is the practice of preventing unauthorized access to an organization's information. This helps to protect sensitive business information and data base to be accessed by unauthorized parties.

Since the increase in the number of information threats encountered by Ski Resort is due to the implementation of a new software system, this means their has been a loophole in the course of implementing the software system that allows malignant access to the company's information.

Kenny, the chief information officer, needs to establish a plan that ensures that the company's information and database is secured from parties that are not authorized.

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the length of iron rod at 100 C is 300.36 cm and at 159 C is 300.54 cm.Calculate its length at 0 c and coefficient of linear exp
Ugo [173]

Answer:

The length at 0 °C is 300.05 cm

Coefficient of linear expansion of iron is 1.02×10¯⁵ C¯¹

Explanation:

From the question given above, the following data were obtained:

Length (L₁) at 100 °C = 300.36 cm

Temperature 1 (θ₁) = 100 °C

Length (L₂) at 159 °C = 300.54 cm

Temperature 2 (θ₂) = 159 °C

Length (L₀) at 0 °C =?

Coefficient of linear expansion (α) =?

L₁ = L₀ (1 + θ₁α)

300.36 = L₀ (1 + 100α) ....(1)

L₂ = L₀ (1 + θ₂α)

300.54 = L₀ (1 + 159α) ..... (2)

Divide equation (2) by (1)

300.54 / 300.36 = L₀ (1 + 159α) / L₀ (1 + 100α)

1.0006 = (1 + 159α) / (1 + 100α)

Cross multiply

1.0006 (1 + 100α) = (1 + 159α)

1.0006 + 100.06α = 1 + 159α

Collect like terms

1.0006 – 1 = 159α – 100.06α

0.0006 = 58.94α

Divide both side by 58.94

α = 0.0006 / 58.94

α = 1.02×10¯⁵ C¯¹

Substitute the value of α into anything of the equation to obtain L₀. Here we shall use equation (2).

300.54 = L₀ (1 + 159α)

α = 1.02×10¯⁵ C¯¹

300.54 = L₀ (1 + 159 ×1.02×10¯⁵)

300.54 = L₀ (1 + 0.0016218)

300.54 = L₀ (1.0016218)

Divide both side by 1.0016

L₀ = 300.54 / 1.0016

L₀ = 300.05 cm

Summary:

The length at 0 °C is 300.05 cm

Coefficient of linear expansion of iron is 1.02×10¯⁵ C¯¹

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3 years ago
Does the mass of a pendulum affects the period of oscillation
pshichka [43]
Mass does not affect the pendulum's swing. The longer the length of string, the farther the pendulum falls; and therefore, the longer the period, or back and forth swing of the pendulum. The greater the amplitude, or angle, the farther the pendulum falls; and therefore, the longer the period.
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2 years ago
Label and describe what is happening in this picture
SOVA2 [1]
Something is reproducing.
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2 years ago
A uniformly charged rod (length = 2.0 m, charge per unit length = 3.0 nc/m) is bent to form a semicircle. What is the magnitude
Artist 52 [7]

Answer:

84.82N/C.

Explanation:

The x-components of the electric field cancel; therefore, we only care about the y-components.

The y-component of the differential electric field at the center is

$dE = \frac{kdQ }{R^2} sin(\theta )$.

Now, let us call \lambda the charge per unit length, then we know that

dQ = \lambda Rd\theta;

therefore,

$dE = \frac{k \lambda R d\theta }{R^2} sin(\theta )$

$dE = \frac{k \lambda  d\theta }{R} sin(\theta )$

Integrating

$E = \frac{k \lambda   }{R}\int_0^\pi sin(\theta )d\theta$

$E = \frac{k \lambda   }{R}*[-cos(\pi )+cos(0) ]$

$E = \frac{2k \lambda   }{R}.$

Now, we know that

\lambda = 3.0*10^{-9}C/m,

k = 9*10^9kg\cdot m^3\cdot s^{-4}\cdot A^{-2},

and the radius of the semicircle is

\pi R = 2.0m,\\\\R = \dfrac{2.0m}{\pi };

therefore,

$E = \frac{2(9*10^9) (3.0*10^{-9})   }{\dfrac{2.0}{\pi } }.$

$\boxed{E = 84.82N/C.}$

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Inga [223]

Answer:

area 4

Explanation:

area 4 has low pressure

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