HELP PLEASE

Two parallel Rivers (A and B) are separated by confined and unconfined aquifer estimate the RATE of seepage of river A to River

gizmo_the_mogwai [7]

Answer:

Explanation:

A confined aquifer is one that's impermeable and does not allow seepage's of water through it.

An unconfined aquifer is a body of water that has permeable membranes which allows passage of water through it.

There will be little or no transmission of water trough these bodies of water.

3 0

2 years ago

A manufacturer provides a warranty against failure of a carbon steel product within the first 30 days after sale. Out of 1000 so

irina1246 [14]

Answer: idk sorry we both on the same thing

7 0

3 years ago

Which website suffixes are usually the least credible? Check all that apply.

zmey [24]

.com hope this helps

7 0

3 years ago

Read 2 more answers

A wire of diameter d is stretched along the centerline of a pipe of diameter D. For a given pressure drop per unit length of pip

JulsSmile [24]

Answer:

Part A: (d/D=0.1)

DeltaV percent=42.6%

Part B:(d/D=0.01)

DeltaV percent=21.7%

Explanation:

We are going to use the following volume flow rate equation:

$DeltaV=\frac{\pi * DeltaP}{8*u*l}(R^{4}-r^{4} -\frac{(R^{2}-r^{2})}{ln\frac{R}{r}}^{2})$

Above equation can be written as:

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(\frac{r}{R} )^{4}+\frac{(1-(\frac{r}{R} )^{2})}{ln\frac{r}{R}}^{2})$

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(\frac{d}{D} )^{4}+\frac{(1-(\frac{d}{D})^{2})}{ln\frac{d}{D}}^{2})$

First Consider no wire i.e d/D=0

Above expression will become:

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(0)^{4}+\frac{(1-(0)^{2})}{ln0}^{2})$

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}$

Part A: (d/D=0.1)

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(0.1)^{4}+\frac{(1-(0.1)^{2})}{ln0.1}^{2})$

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}*0.574$

$DeltaV percent=\frac{(\frac{\pi*R^{4}*DeltaP}{8*u*l})-\frac{\pi *R^{4}*DeltaP}{8*u*l}*0.574}{\frac{\pi*R^{4}*DeltaP}{8*u*l} }*100$

$DeltaV percent=\frac{1-0.574}{1}*100$

DeltaV percent=42.6%

Part B:(d/D=0.01)

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(0.01)^{4}+\frac{(1-(0.01 )^{2})}{ln0.01}^{2})$

$DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}*0.783$

$DeltaV percent=\frac{(\frac{\pi *R^{4}*DeltaP}{8*u*l})-\frac{\pi *R^{4}*DeltaP}{8*u*l}*0.783}{\frac{\pi *R^{4}*DeltaP}{8*u*l} }*100$

$DeltaV percent=\frac{1-0.783}{1}*100$

DeltaV percent=21.7%

5 0

2 years ago

What application of bioengineering uses principles of electronics and computer science to design products?

Flauer [41]

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:

brainly.com/question/21819443

#SPJ4

4 0

1 year ago