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fgiga [73]
3 years ago
6

Sarah is developing a Risk Assessment for her organization. She is asking each department head how long can they be without thei

r primary system. What recovery goal is Sarah trying to determine.
Engineering
1 answer:
Natali [406]3 years ago
8 0

Answer:

Sarah is asking each department head how long they can be without their primary system. Sarah is trying to determine the Recovery Time Objective (RTO) as this is the duration of time within which the primary system must be restored after the disruption.

Recovery Point Objective is basically to determine the age of restoration or recovery point.

Business recovery and technical recovery requirements are to assess the requirements to recover by Business or technically.

Hence, Recovery Time Objective (RTO) is the correct answer.

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Conduction is thermal energy transfer by:_______.
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Answer:

a. molecular interactions.

Explanation:

Conduction is thermal energy transfer by molecular interactions. Therefore, conduction involves the transfer of electric charge or thermal energy due to the movement of particles. When the conduction relates to electric charge, it is known as electrical conduction while when it relates to thermal energy, it is known as heat conduction.

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5 0
3 years ago
An AC power generator produces 50 A (rms) at 3600 V. The voltage is stepped up to 100 000 V by an ideal transformer and the ener
RSB [31]

Given:

I_{rms} = 50 A

voltage, V = 3600V

step-up voltage, V' = 100000 V

Resistance of line, R = 100\ohm

Solution:

To calculate % heat loss in long distance power line:

Power produced by AC generator, P = 50\times 3600 W

P = 180000 W = 180 kW

At step-up voltage, V = 100000V or 100 kV

current, I = \frac{P}{V'}

I = \frac{1800000}{100000}

I = 1.8 A

Power line voltage drop is given by:

V_{drop} = I\times R

V_{drop} = 1.8\times 100

V_{drop} = 180 V

Power dissipated in long transmission line P_{dissipated} = V_{drop}\times I

Power dissipated in long transmission line P_{dissipated} = 180\times 1.8 = 324 W

% Heat loss in power line, P_{loss} = \frac{P_{dissipated}}{P}\times 100

% Heat loss in power line, P_{loss} = \frac{324}{180000}\times 100

P_{loss} = 0.18%

 

5 0
3 years ago
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