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Andrew [12]
4 years ago
9

When designing solid rockets, thrust and mass flow must be considered time dependent. a) True b) False

Engineering
1 answer:
Novay_Z [31]4 years ago
8 0

Answer:

the answer is true when designing sold rockets thrust and mass flow

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2) Consider schedules S3, S4, and S5 below. Determine whether each schedule is strict, cascadeless, recoverable, or non-recovera
gulaghasi [49]

Answer:

Explanation:

Consider schedules S3, S4, and S5 below. Determine whether each schedule is strict, cascadeless, recoverable, or non-recoverable. You need to explain your reason.

S3: r1(x), r2(z), r1(z), r3(x), r3(y), w1(x), c1, w3(y), c3, r2(y), w2(z),w2(y),c2

S4: r1(x), r2(z), r1(z), r3(x), r3(y),w1(x),w3(y), r2(y),w2(z),w2(y), c1,c2, c3

S5: r1(x), r2(z), r3(x), r1(z), r2(y), r3(y), w1(x), c1, w2(z), w3(y), w2(y), c3, c2

Strict schedule:

A schedule is strict if it satisfies the following conditions:

Tj reads a data item X after Ti has written to X and Ti is terminated means aborted or committed.

Tj writes a data item X after Ti has written to X and Ti is terminated means aborted or committed.

S3 is not strict because In a strict schedule T3 must read X after C1 but here T3 reads X (r3(X)) before Then T1 has written to X (w1(X)) and T3 commits after T1.

S4 is not strict because In a strict schedule T3 must read X after C1, but here T3 reads X (r3(X)) before T1 has written to X (w1(X)) and T3 commits after T1.

S5 is not strict because T3 reads X (r3(X)) before T1 has written to X (w1(X))

but T3 commits after T1. In a strict schedule T3 must read X after C1.

Cascadeless schedule:

Cascadeless schedule follows the below condition:

Tj reads X only? after Ti has written to X and terminated means aborted or committed.

S3 is not cascadeless schedule because T3 reads X (r3(X)) before T1 commits.

S4 is not cascadeless schedule because T3 reads X (r3(X)) before T1 commits.

S5 is not cascadeless schedule because T3 reads X (r3(X)) before T1 commits or T2 reads Y (r2(Y)) before T3 commits.

But while come to the definition of cascadeless schedules S3, S4, and S4 are not cascadeless, and T3 is not affected if T1 is rolled back in any of the schedules, that is,

T3 does not have to roll back if T1 is rolled back. The problem occurs because these

schedules are not serializable.

Recoverable schedule:

Schedule that follows the below condition:

-----Tj commits after Ti if Tj has?read any data item written by Ti.

Ci > Cj means that Ci happens before Cj. Ai denotes abort Ti. To test if a schedule is

recoverable one has to include abort operations. Thus in testing the recoverability abort

operations will have to used in place of commit one at a time. Also the strictest condition is

------where a transaction neither reads nor writes to a data item, which was written to by a transaction that has not committed yet.

If A1?>C3>C2, then schedule S3 is recoverable because rolling back of T1 does not affect T2 and

T3. If C1>A3>C2. schedule S3 is not recoverable because T2 read the value of Y (r2(Y)) after T3 wrote X (w3(Y)) and T2 committed but T3 rolled back. Thus, T2 used non- existent value of Y. If C1>C3>A3, then S3 is recoverable because roll back of T2 does not affect T1 and T3.

Strictest condition of schedule S3 is C3>C2.

If A1?>C2>C3, then schedule S4 is recoverable because roll back of T1 does not affect T2 and T3. If C1>A2>C3, then schedule S4 is recoverable because the roll back of T2 will restore the value of Y that was read and written to by T3 (w3(Y)). It will not affect T1. If C1>C2>A3, then schedule S4 is not recoverable because T3 will restore the value of Y which was not read by T2.

7 0
3 years ago
Engineering controls are the physical changes that employers make to the work environment or to equipment that make it safer to
____ [38]

Answer:

Engineering Controls. The best engineering controls to prevent heat-related illness is to make the work environment cooler and to reduce manual workload with mechanization. A variety of engineering controls can reduce workers' exposure to heat: Air conditioning, Increased general ventilation , Cooling fans , Local exhaust ventilation at points of high heat production or moisture, Reflective shields to redirect radiant heat , Insulation of hot surfaces Elimination of steam leaks , Cooled seats or benches for rest breaks , Use of mechanical equipment to reduce manual work, Misting fans that produce a spray of fine water droplets.

Hope this helped you!

Explanation:

5 0
4 years ago
Write a program that prompts the user to enter two characters and display the corresponding major and year status. The first cha
lara31 [8.8K]

Answer:

Following is attached the images for source code and the result according to given question.

I hope it will help you a lot!

Explanation:

7 0
4 years ago
Refrigerant 134a enters an air conditioner compressor at 4 bar, 208C, and is compressed at steady state to 12 bar, 808C. The vol
SCORPION-xisa [38]

Answer:

heat transfer rate is -15.71 kW

Explanation:

given data

Initial pressure  = 4 bar

Final pressure  = 12 bar

volumetric flow rate = 4 m³ / min

work input to the compressor = 60 kJ per kg

solution

we use here super hated table for 4 bar and 20 degree temperature and 12 bar and 80 degree is

h1 = 262.96 kJ/kg

v1 = 0.05397 m³/kg

h2 = 310.24 kJ/kg

and here mass balance equation will be

m1  = m2

and mass flow equation is express as

m1 = \frac{A1\times V1}{v1}       .......................1

m1 = \frac{4\times \frac{1}{60}}{0.05397}  

m1 = 1.2353 kg/s

and here energy balance equation is express as

0 = Qcv - Wcv + m × [ ( h1-h2) + \frac{v1^2-v2^2}{2} + g (z1-z2) ]      ....................2

so here Qcv will be

Qcv =  m × [  \frac{Wcv}{m} + (h2-h1)  ]    ......................3

put here value and we get

Qcv =  1.2353 × [ {-60}+ (310.24-262.96) ]

Qcv =  -15.7130 kW

so here heat transfer rate is -15.71 kW

6 0
3 years ago
A diesel engine with an engine volume of 4.0 L and an engine speed of 2500 rpm operates on an air–fuel ratio of 18 kg air/kg fue
Mars2501 [29]

Answer:

Explanation:

i) (18 X 336) air/h = 6048 air/h

Engine speed = 2500 rpm; Capacity of diesel = 800 ppm

∴ 2500/800 = 3.125 kg/pm

Mass flow rate = (3.125 X 6048) Kg/h/pm =18900 Kg/h/pm

ii) Mass flow rate = 32 X 4/1000 X 2500 = 320 Kg/h/pm

3 0
4 years ago
Read 2 more answers
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