Answer:
enables the representation, analysis and communication of various aspects of an information system. These aspects correspond to varying and incomplete views of information systems and the processes therein.
Answer: c. The VW engineers involved were ethically obligated to hold paramount the health, welfare and safety of the public even if their supervisors directed them to implement software and hardware that enabled cheating on the emissions testing software.
Explanation: The National Society of professional Engineers, NSPE define the code of ethics which must guide engineers in their duty. These codes act as principles of personal conduct, towards the public and their employers.
One of the areas covered by these codes is overriding importance of the safety and health of the public to any other factor. In addition, engineers are to avoid deception and maintain the reputation of their profession. These cannot be sacrificed for the financial gain of their employers or explained away by saying they are following the direction of their employers. While they have certain responsibilities to their employers, the health welfare and safety of the public is more important.
R01= 14.1 Ω
R02= 0.03525Ω
<h3>Calculations and Parameters</h3>
Given:
K= E2/E1 = 120/2400
= 0.5
R1= 0.1 Ω, X1= 0.22Ω
R2= 0.035Ω, X2= 0.012Ω
The equivalence resistance as referred to both primary and secondary,
R01= R1 + R2
= R1 + R2/K2
= 0.1 + (0.035/9(0.05)^2)
= 14.1 Ω
R02= R2 + R1
=R2 + K^2.R1
= 0.035 + (0.05)^2 * 0.1
= 0.03525Ω
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Answer:
battery life in year = 9 years and 48 days
Explanation:
given data
Battery Ampere-hours = 1.5
Pulse voltage = 2 V
Pulse width = 1.5 m sec
Pulse time period = 1 sec
Electrode heart resistance = 150 Ω
Current drain on the battery = 1.25 µA
to find out
battery life in years
solution
we get first here duty cycle that is express as
duty cycle = 
...............1
duty cycle = 1.5 × 
and applied voltage will be
applied voltage = duty energy × voltage ...........2
applied voltage = 1.5 × × 2
applied voltage = 3 mV
so current will be
current = 
................3
current = 
current = 20 µA
so net current will be
net current = 20 - 1.25
net current = 18.75 µA
so battery life will be
battery life = 
battery life = 80000 hours
battery life in year = 
battery life in year = 9.13 years
battery life in year = 9 years and 48 days
