The application of electro bioengineering uses principles of nick and computer science to design products is application of electrical engineering principles to biology, medicine, conduct, or health.
<h3>What is Bioelectronics?</h3>
- Bioelectronics is the application of electrical engineering principles to biology, medicine, conduct, or health.
- It advances the fundamental concepts, creates knowledge for the molecular to the organ techniques levels, and develops creative devices or methods for the deterrence, diagnosis, and treatment of disease, for patient rehabilitation, and for improving health.
- Bio electromagnetics, instrumentation, neural networks, robotics, and detector technologies are some of the disciplines necessary to develop new knowledge and creations in this area.
- A keystone of this research area is the building of and real-world devices and systems.
- Onsite facilities for prototyping and testing instrumentation systems, fabricating and measuring the performance of implantable devices, and making robotic prostheses, are readily available.
- New detectors and sensor arrays are microfabricated in a 2,000 sq ft cleanroom.
To learn more about Bioelectronics, refer to:
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Answer:
Percentage change 5.75 %.
Explanation:Given ;
Given
Pressure of condenser =0.0738 bar
Surface temperature=20°C
Now from steam table
Properties of steam at 0.0738 bar
Saturation temperature corresponding to saturation pressure =40°C
So Δh=2573.5-167.5=2406 KJ/kg
Enthalpy of condensation=2406 KJ/kg
So total heat=Sensible heat of liquid+Enthalpy of condensation

Total heat =4.2(40-20)+2406
Total heat=2,544 KJ/kg
Now film coefficient before inclusion of sensible heat



Now film coefficient after inclusion of sensible heat



=5.75 %
So Percentage change 5.75 %.
Answer:
Disaggregation
Explanation:
In a company it is a way to create operational plans that are focused, either by time or by section.
Answer:
Departure rate = 7.65 vehicle/min
Explanation:
See the attached file for the calculation.
An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone. A police car observed the automobile. At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile at a constant acceleration of 1.96 m/s2 . The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s2
Find the total time required for the police car to over take the automobile.
Answer:
15.02 sec
Explanation:
The total time required for the police car to overtake the automobile is related to the distance covered by both cars which is equal from instant point of abreast.
So; we can say :

By using the second equation of motion to find the distance S;





where ;
u = 0





Recall that:



= 46.68 - 7.85 t -2.505 t² = 0
Solving by using quadratic equation;
t = -6.16 OR t = 3.02
Since we can only take consideration of the value with a positive integer only; then t = 3.02 secs
From the question; The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit;
Therefore ; the total time required for the police car to over take the automobile = 12 s + 3.02 s
Total time required for the police car to over take the automobile = 15.02 sec