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Ksenya-84 [330]
3 years ago
7

Determine whether or not each of the following four transaction execution histories is serializable. If a history is serializabl

e, specify a serial order of transaction execution to which it is equivalent
A. r1[x] r2[y] w2[x] r1[z] r3[z] w3[z] w1[z]
B. w1[x] w1[y] r2[u] w2[x] r2[y] w2[y] w1[z]
C. w1[x] w2[y] r2[u] w1[z] w2[x] r2[y] w1[u]
D. w1[x] w2[u] w2[y] w1[y] w3[x] w3[u] w1[z]
Engineering
1 answer:
ludmilkaskok [199]3 years ago
8 0

Answer:

Option D. w1[x] w2[u] w2[y] w1[y] w3[x] w3[u] w1[z]

Explanation:

The execution in the option D is correct. This is because there is more than one reasonable criterion.

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Show the ERD with relational notation with crowfoot. Your ERD must show PK, FKs, min and max cardinality, and correct line types
zhenek [66]

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

3 0
3 years ago
Help I need to know if it’s true or false
e-lub [12.9K]

Answer:  False

explanation: for a bloodborne pathogen to spread you would have to have an open wound as well as the blood would have to get in it.

3 0
3 years ago
A water pump delivers 3 hp of shaft power when operating. If the pressure differential between the outlet and the inlet of the p
Natali [406]

Answer:

Mechanical Efficiency =  83.51%

Explanation:

Given Data:

Pressure difference = ΔP=1.2 Psi

Flow rate = V=8ft^3/s\\

Power of Pump = 3 hp

Required:

Mechanical Efficiency

Solution:

We will first bring the change the units of given data into SI units.

P=1.2*6.895 = 8.274KPa\\V=8*0.00283=0.226 m^3/s\\P=3*0.746=2.238KW

Now we will find the change in energy.

Since it is mentioned in the statement that change in elevation (potential energy) and change in velocity (Kinetic Energy) are negligible.

Thus change in energy is

=(Mass * change in P)/density\\= \frac{M*P}{p}\\\\

As we know that Mass = Volume x density

substituting the value

Energy = Volume * density x ΔP / density

Change in energy = Volumetric flow x ΔP

Change in energy = 0.226 x 8.274 = 1.869 KW

Now mechanical efficiency = change in energy / work done by shaft

Efficiency = 1.869 / 2.238

Efficiency = 0.8351 = 83.51%

5 0
3 years ago
An AM radio transmitter operating at 3.9 MHz is modulated by frequencies up to 4 kHz. What are the maximum upper and lower side
Aliun [14]

Answer:

total width bandwidth = 8kHz

Explanation:

given data

transmitter operating = 3.9 MHz

frequencies up to =  4 kHz

solution

we get here upper side frequencies that is

upper side frequencies = 3.9 × 10^{6} + 4  × 10³

upper side frequencies = 3.904 MHz

and

now we get lower side frequencies that is

lower side frequencies = 3.9 × 10^{6} - 4  × 10³

lower side frequencies =  3.896 MHz

and now we get total width bandwidth

total width bandwidth = upper side frequencies  - lower side frequencies

total width bandwidth = 8kHz

6 0
3 years ago
Write using about 10-15 lines for each of the six materials (metals, ceramics, glasses, polymers, composites, and semiconductors
Svetradugi [14.3K]

Answer:

See Answer below- Explanation is the entire answer

Explanation:

Metals:

Properties: Ductile, good heat conductivity, good electrical conductivity, high strength;

Drawbacks: Relatively high weight, reactive with oxygen to create oxides- corrosion is presented;

Examples: steel, aluminum alloys, brass, copper, titanium

Applications: Body of the vehicles, structures in the skyscrapers, cooking pots.

Ceramics:

Properties: Brittle, poor heat conductors, poor electrical conductors, high wear resistance, corrosion resistance;

Drawbacks: Deforms by fracturing, shock resistance is low, no conductivity of electricity;

Examples: concrete, tungsten carbide, diamond

Applications: bricks for constructions, clay pots to keep heat, cutting tools for metals;

Glasses:

Properties: amorphous, transparent, high weight

Drawbacks: poor conductors of heat and electricity; brittle; low shock resistance;

Examples: Silica, lead glass, glaze;

Applications: windows, protection screens;

Polymers:

Properties: low density, recyclable, poor heat and electrical conductors, plastic deformation;

Drawbacks: low strength, low operating temperatures;

Examples: polyethylene, nylon, ABS-plastic, rubber;

Applications: toys, tires, insulation covers for the wires.

Composites:

Properties: high strength to weight ratio, can get combination of properties from the used materials, rarely conductive, good shock resistance;

Drawbacks: high cost, hard to recycle, expensive;

Examples: steel-reinforced concrete, carbon fiber, fiber glass, Nomex, sandwich roof panels;

Applications: buildings, bullet proof vests, body of the Formula 1 cars, rockets, roof panels.

Semiconductors:

Properties: brittle, change conductive behavior under certain scenario, poor heat conductors;

Drawbacks: hard to manufacture, expensive;

Examples: Silicon-based semiconductors, Germanium-based semiconductors, Ga-based semiconductors;

Applications: chips, LED, diodes, transistors, op-amps, microprocessors.

8 0
3 years ago
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