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Sedbober [7]
3 years ago
15

Assuming that a query has a buffer holding up to 3 blocks, and each block can hold two records. Use merge-sort to sort the follo

wing records in ascending order: 10, 11, 1, 5, 90, 1, 2, 101
How many runs will be produced in the whole algorithm and what are the contents of the runs?
Engineering
1 answer:
kipiarov [429]3 years ago
5 0

Answer:

a) 5 runs will be generated.

b) Since the buffer can hold 3 records, the first 3 runs are (1, 10, 11), (1, 5, 90) and (2, 10). Then we need to reserve one block as output buffer, the algorithm can merge two runs at most at the same time. As a result, we can choose to merge the first two runs into a larger one: (1,1,5,10,11,90), which is merged with the run (2, 10) to generate the final output.

Explanation:

5 runs will be generated.

Since the buffer can hold 3 records, the first 3 runs are (1, 10, 11), (1, 5, 90) and (2, 10). Then we need to reserve one block as output buffer, the algorithm can merge two runs at most at the same time. As a result, we can choose to merge the first two runs into a larger one: (1,1,5,10,11,90), which is merged with the run (2, 10) to generate the final output.

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For a bronze alloy, the stress at which plastic deformation begins is 266 MPa and the modulus of elasticity is105 GPa.
pentagon [3]

Answer:

88750 N

Explanation:

given data:

plastic deformation σy=266 MPa=266*10^6 N/m^2

cross-sectional area Ao=333 mm^2=333*10^-6 m^2

solution:

To determine the maximum load that can be applied without

plastic deformation (Fy).

Fy=σy*Ao

   =88750 N

7 0
3 years ago
What are the dimensions of the base of the pyramid?
Cloud [144]

Answer:

b

Explanation:

bcus it is

7 0
3 years ago
An air standard cycle with constant specific heats is executed in a closed system with 0.003 kg of air and consists of the follo
Vsevolod [243]

Answer:

a) Please see attached copy below

b) 0.39KJ

c)  20.9‰

Explanation:

The three process of an air-standard cycle are described.

Assumptions

1. The air-standard assumptions are applicable.

2. Kinetic and potential energy negligible.

3. Air in an ideal gas with a constant specific heats.

Properties:

The properties of air are gotten from the steam table.

b) T₁=290K ⇒ u₁=206.91 kj/kg, h₁=290.16 kj/kg.

P₂V₂/T₂=P₁V₁/T₁⇒ T₂=P₂T₁/P₁ = 380/95(290K)= 1160K

T₃=T₂(P₃/P₂)⁽k₋1⁾/k =(1160K)(95/380)⁽⁰°⁴/₁.₄⁾ =780.6K

Qin=m(u₂₋u₁)=mCv(T₂-T₁)

=0.003kg×(0.718kj/kg.k)(1160-290)K= 1.87KJ

Qout=m(h₃₋h₁)=mCp(T₃₋T₁)

=0.003KG×(1.005kj/kg.k(780.6-290)K= 1.48KJ

Wnet, out= Qin-Qout = (1.87-1.48)KJ =0.39KJ

c)ηth= Wnet/W₍in₎ =0.39KJ/1.87KJ = 20.9‰

7 0
3 years ago
Calculate the pressure drop in a duct (measured by a differential oil manometer) if the differential height between the two flui
Burka [1]

Answer:

The pressure drop is 269.7N/m^2

Explanation:

∆P = ∆h × rho × g

∆h = 3.2cm = 3.2/100 = 0.032m, rho = 860kg/m^3, g = 9.8m/s^2

∆P = 0.032×860×9.8 = 269.7N/m^2

6 0
3 years ago
Determine the general lighting load for a two-story office building that measures 125 feet by 150 feet.
Irina-Kira [14]

The general lighting load for a two-story office building that measures 125 feet by 150 feet is 112, 500 sq ft.

<h3>What is lighting load?</h3>

Lighting loads are the energy used to power electric lights and they make up nearly a third of US commercial building energy use.

Lighting load = n(LW)

where;

  • L is length of the building
  • W is width of the building
  • n is number of story building

For one story building, = 3

For two story building, n = 6

Lighting load = 6 x 125 x 150 = 112, 500 sq ft.

Learn more about lighting load here: brainly.com/question/14070748

#SPJ12

7 0
2 years ago
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