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Ksenya-84 [330]
2 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]2 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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Consider the series solution, Equation 5.42, for the plane wall with convection. Calculate midplane (x* = 0) and surface (x* = 1
VashaNatasha [74]

Answer:

We conclude that the approximate series solution (with only one eigein value) provides systematically high results but by less than 1.5%, for the biot number range from 0.11 to 10. See attached image.

Explanation:

8 0
3 years ago
A horizontal curve on a two-lane road is designed with a 2,300-ft radius, 12-ft lanes, and a 65-mph design speed. Determine the
Ierofanga [76]

Answer:

distance = 22.57 ft

superelevation rate = 2%

Explanation:

given data

radius = 2,300-ft

lanes width = 12-ft

no of lane = 2

design speed = 65-mph

solution

we get here sufficient sight distance SSD that is express as

SSD = 1.47 ut + \frac{u^2}{30(\frac{a}{g}\pm G)}     ..............1

here u is speed and t is reaction time i.e 2.5 second and a is here deceleration rate i.e 11.2 ft/s² and g is gravitational force i.e 32.2 ft/s² and G is gradient i.e 0 here

so put here value and we get

SSD = 1.47 × 65 ×2.5  + \frac{65^2}{30(\frac{11.2}{32.2}\pm 0)}

solve it we get

SSD = 644 ft  

so here minimum distance clear from the inside edge of the inside lane is

Ms = Rv ( 1  - cos (\frac{28.65 SSD}{Rv}) )        .....................2

here Rv is = R - one lane width

Rv = 2300 - 6 = 2294 ft

put value in equation 2 we get

Ms = 2294 ( 1  - cos (\frac{28.65 \times 664}{2294})  )  

solve it we get

Ms = 22.57 ft

and

superelevation rate for the curve will be here as

R  = \frac{u^2}{15(e+f)}  ..................3

here f is coefficient of friction that is 0.10

put here value and we get e

2300 = \frac{65^2}{15(e+0.10)}

solve it we get

e = 2%

3 0
2 years ago
Consider a dip-coating process where a very long (assume infinitely long) wire(solid) with radius, ri, is being pulled verticall
Gekata [30.6K]

Answer:

See explaination and attachment.

Explanation:

Navier-Stokes equation is to momentum what the continuity equation is to conservation of mass. It simply enforces F=ma in an Eulerian frame.

The starting point of the Navier-Stokes equations is the equilibrium equation.

The first key step is to partition the stress in the equations into hydrostatic (pressure) and deviatoric constituents.

The second step is to relate the deviatoric stress to viscosity in the fluid.

The final step is to impose any special cases of interest, usually incompressibility.

Please kindly check attachment for step by step solution.

6 0
3 years ago
Consider CO at 500 K and 1000 kPa at an initial state that expands to a final pressure of 200 kPa in an isentropic manner. Repor
REY [17]

Answer:

T_2=315.69k

Explanation:

Initial Temperature T_1=500K

Initial Pressure P_1=1000kPa

Final Pressure P_2=200kPa

Generally the gas equation is mathematically given by

\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{n-1}{n}}

Where

n for CO=1.4

Therefore

\frac{T_2}{500}=\frac{200}{1000}^{\frac{1.4-1}{1.4}}

T_2=315.69k

7 0
3 years ago
1. Sewage-treatment plant, a large concrete tank initially contains 440,000 liters liquid and 10,000 kg fine suspended solids. T
Elenna [48]

Answer:

Concentration = 10.33 kg/m³

Explanation:

We are given;

Mass of solids; 10,000 kg

Volume; V = 440,000 L = 440 m³

Rate at which water is pumped out = 40,000 liter/h

Thus, at the end of 5 hours we amount of water that has been replaced with fresh water is = 40,000 liter/h x 5 hours = 200,000 L = 200 m³

Now, since the tank is perfectly mixed, therefore we can calculate a ratio of fresh water to sewage water as;

200m³/440m³ = 5/11

Thus, the amount left will be calculated by multiplying that ratio by the amount of solids;

Thus,

Amount left; = 10000 x (5/11) = 4545 kg

The concentration would be calculated by:

Concentration = amount left/initial volume

Thus,

Concentration = 4545/440 = 10.3 kg/m³

8 0
3 years ago
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