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Arturiano [62]
3 years ago
5

a. To measure the water current in an ocean, a marker is dropped onto it. Determine if the trajectory traced by the drifting mar

ker forms a streamline, streakline, or pathline.b. To learn more about the flow, a large number of similar markers are released at the same point in space in quick succession. Determine if the line connecting these markers at a later instant forms a streamline, streakline, or pathline.

Engineering
1 answer:
ValentinkaMS [17]3 years ago
5 0

Answer:

A) The trajectory traced by the drifting marker form a pathline. The pathline is the trajectory than a single particle does in the fluid (like in this case) along with all its travel.

b) The line connecting these markers released at the same point in space in quick succession is a Streakline. The streaklines are the lines formed by the loci of points in the fluid that have continuously pass throw particulars points in space in the past. The main difference between the pathline and the streakline is that the first one is not affected by flow changes over time, while the second one is. The Pathline is a movie that shows the travel of a single particle in time. The streakline is movie that shows how the particles change their path along time.

In the gif attached to this answer, you can see in red the pathline of particle, in blue the streakline of several particles released at the same point in space in quick succession and in grey the streamline.

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Create a Relational Schema for the following scenario. Include all primary and foreign keys and list any assumptions you make. 
NARA [144]

Answer:

See explaination

Explanation:

Some of the Important Observation used in Entity Relational Diagram.

1. A Chemist can be involved one project or many projects

2. A project can hire just one or many chemists.

3. A chemist can lead over one or many projects

4. A project can have only one project leader chemist.

5. A project can be done in 1 or many laboratories.

6. A laboratory can host 0 or 1 projects.

7. A laboratory can have 1 or many equipments.

8. An Equipment can be in 1 or 0 laboratories.

9. A chemist can have 0 or many types of equipments.

10. An Equipment can have 0 or 1 chemist at a time

Notations:

PK=Primary Key:

A primary key, also called a primary keyword, is a key in a relational database that is unique for each record. It is a unique identifier, such as a driver license number, telephone number

FK=Foreign Key:

A foreign key is a column or group of columns in a relational database table that provides a link between data in two tables. It acts as a cross-reference between tables because it references the primary key of another table, thereby establishing a link between them

Composite key:

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4 0
3 years ago
The soil borrow material to be used to construct a highway embankment has a mass unit weight of 107.0 lb/cf and a water content
MrRissso [65]

Answer:

Option D

Explanation:

Given information

Bulk unit weight of 107.0 lb/cf

Water content of 7.3%,=0.073

Specific gravity of the soil solids is 2.62

Specifications

Dry unit weight is 113 lb/cf  

Water content is 6%.

Volume of embankment is 440,000-cy

Borrow material

Dry_{unit,weight}=\frac {bulk_{unit,weight}}{1+water_{content}}=\frac {107}{1+0.073}= 99.72041 lb/cf  

Embankment

Considering that the volume of embankment is inversely proportional to the dry unit weight

\frac {V_{embankment}}{V_{borrow}}=\frac {Dry_{borrow}}{Dry_{embankment}}

Therefore, V_{borrow}=V_{embankment} *\frac {Dry_{embarkement}}{Dry_{borrow}}

V_{borrow}=440,000-cy*\frac {113 lb/cf }{99.72041 lb/cf }= 498594-cy

Therefore, volume of borrow material is 498594-cy

(b)

The weight of water in embankment is found by multiplying the moisture content and dry unit weight.

Assuming that all the specifications are achieved, weight of water in embankment=0.06*113=6.78 lb/cf

Since 1 yd^{3}= 27 ft^{3}

The embankment requires water of  6.78*27*440000= 80546400 lb

Borrow materials’ water will also be 0.073*99.72041=7.27959 lb/cf

Borrow material requires water of 7.27959*27*498594=97998120 lb

Extra water between borrow material and embankment=97998120 lb-80546400 lb=17451720 lb

Unit_{weight}=\frac {17451720}{498594}=35.00186 lb

1 gallon is approximately 8.35 yd^{3} hence

\frac {35.00186 lb/yd^{3}}{8.35}=4.19184 gallons/yd^{3}

That's approximately 4.2 gallons

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Answer:

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Explanation:

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madreJ [45]

Answer:FALSE

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Most of the personal protective equipment (PPE) which are in use in various industries are examples of Negative pressure respirator device,any leak or damage done to the device will allow the inflows of harmful and toxic Air into the person's respiratory system. AIR SUPPLY SYSTEMS ARE KNOWN TO SUPPLY FRESH UNCONTAMINATED AIR THROUGH AIR STORED INSIDE COMPRESSED CYLINDERS OR OTHER SOURCES AVAILABLE.

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Answer:

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Explanation:

Solution in pen paper form in the attachment section

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