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bonufazy [111]
3 years ago
7

A particular Table in a relational database contains 100,000 Data Records/rows, each of which Data Record/row requires 200 bytes

. A select statement returns all Data Records/rows in the Table that satisfy an equality search on an attribute. Estimate the time in milliseconds to complete the query when each of the following Indexes on that attribute is used.
A. No Index (Heap File of Data Records)
B. A Static Hash Index (with no overflow buckets/Pages). Assume the cost of applying the hash function is H, negligible.
Engineering
1 answer:
MrRa [10]3 years ago
3 0

The correct question is;

A particular table in a relational database contains 100,000 rows, each of which requires

200 bytes of memory. Estimate the time in milliseconds to to insert a new row into the

table when each of the following indices on the related attribute is used. Assume a page

size of 4K bytes and a page access time of 20 ms.

a. No index (heap file)

b. A clustered, non-integrated B+ tree index, with no node splitting

required. Assume that each index entry occupies 100 bytes. Assume that the

index is 75% occupied and the actual data pages are 100% occupied. Assume

that all matching entries are in a single page.

Answer:

A) 20 ms

B) 120 ms

Explanation:

A) Append (at the end of file). Just one IO, i.e., 20 ms

B) Now, when we assume that each entry in the index occupies 100 bytes, then an index page can thus hold 40 entries. Due to the fact that the data file occupies 5000 pages, the leaf level of the tree must contain at least 5000/40 pages which is 125 pages.

So, the number of levels in the tree (assuming page 75% occupancy in the

index) is (log_30 (125)) + 1 = 3. Now, if we assume that the index is clustered and not integrated with the data file and all matching entries are in a single

page, then 4 I/O operations and 80ms are required to retrieve all matching

records. Two additional I/O operations are required to update the leaf page

of the index and the data page. Hence, the time to do the insertion is

120ms.

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the ____ method is the safest turning technique to use as it does not expose your hands to the airbags deployment area
Slav-nsk [51]

Answer:

Push/pull Method

Explanation:

Push/pull method is seen as the best method to use when a driver what to steer a wheel to point the car in a particular direction reason been that PUSH/PULL method enables the driver to sit perfectly well in order to firmly hold the wheel which is why this method is often recommended as the safest and most effective way of steering than other steering method.

Therefore the PUSH/PULL method is the safest and the best turning technique to use because it does not expose the driver hands to the airbags area reason been that air bag are sometimes place in the centre of the steering wheel which means that when a driver cross his/her arms when steering will block the air bag in a situation where the air bag needed to explode out because airbag can explodes out with a lot of force which in turn could force the driver arms towards his/ her face which is dangerous.

6 0
3 years ago
The temperature of a gas stream is to be measured by a thermocouple whose junction can be approximated as a 1.2-mm-diameter sphe
myrzilka [38]

Answer:

First compute the characteristic length and the Biot number to see if the lumped analysis is applicable

Lc = V/A = (pie*D3/6) / (pie * D2)= 1.2/6 = 0.0012/6= 0.0002m

Bi = hLc/K = (110W/m2.oC)(0.0002)(moC/35W)= 110*0.0002/35 = 0.0006 less than 0.1

Since the Biot number is less than 0.1, we can use the lumped parameter analysis.

In such an analysis, the time to reach a certain temperature is given by

t = -1/bIn(T-Tinfinite/T - Tinfinite)

From the data in the problem we can compute the parameter, b, and then compute the time for the ratio (T – Tinfinite/(Ti – Tinfinite) to reach the desired value.

b = hA/pCpV = h/pCpLc = 110/8500*0.0002 *320*s

b = 110/544s = 0.2022/s

The problem statement is interpreted to read that the measured temperature difference T – Tinfinite has eliminated 98.5% of the transient error in the initial temperature reading Ti – Tinfinite so the value of value of (T – Tinfinite)/(Ti – Tinfinite) to be used in this equation is 0.015

t = -1/bIn(T-Tinfinite/T - Tinfinite)

t = -s/0.1654 (In0.015)

t = (-s*-4.1997)/0.2022

t = 20.77s

It will take the thermocouple 20.77s to reach 98.5% of the initial temperature

3 0
3 years ago
. An ideal vapor compression refrigeration cycle operates with a condenser pressure of 900 kPa. The temperature at the inlet to
tatuchka [14]

Answer:

Actual COP = 5.368

Maximum theoretical COP = 6.368

Explanation:

Given - An ideal vapor compression refrigeration cycle operates with a condenser pressure of 900 kPa. The temperature at the inlet to the compressor is -5oC.

To find -  If this device operates using R134a as the working fluid. Calculate the actual COP of this device as well as the maximum theoretical COP.

Proof -

Given that,

An ideal vapor compression refrigeration cycle operates with a condenser pressure of 900 kPa.

From Refrigerant 134-a Table

At T1 = -5°C

h1 = 247.505 KJ/kg

S1 = 0.93434 KJ/kg

At P2 = 900 KPa

S1 = S2

h2 = 274.679 Kj/Kg

h3 = h4 = 101.61 KJ/g

So,

Compressor work (Wc) = h2 - h1

                                       = 274.679 - 247.505

                                       = 27.174

⇒Compressor work (Wc) = 27.174 KJ/kg

Now,

Heat out (Qout) = h2 - h3

                          = 274.679 - 101.61

                          = 173.069

⇒Heat out (Qout) = 173.069 KJ/kg

Now,

Heat input (Qin) = h1 - h4

                          = 274.505 - 101.61

                          = 145.895

⇒Heat input (Qin) = 145.895 KJ/kg

So,

Actual COP at the refrigerator is -

(COP)R = (Qin)/(Wc)

            = (145.895)/ (27.174)

            = 5.368

⇒Actual COP = 5.368

Now,

Maximum theoretical COP is -

(COP) = (Qout)/(Wc)

          = (173.069)/ (27.174)

          = 6.368

⇒Maximum theoretical COP = 6.368

8 0
3 years ago
What is a two stroke engine and what is a four stroke engine, please keep the definitions as simple as can be and please explain
rewona [7]

Answer:

A two-stroke engine is a type of internal combustion engine that completes a power cycle with two strokes of the piston during only one crankshaft revolution.

A four-stroke engine is an internal combustion engine in which the piston completes four separate strokes while turning the crankshaft. A stroke refers to the full travel of the piston along the cylinder, in either direction.

process for two stroke engine

This is in contrast to a "four-stroke engine", which requires four strokes of the piston to complete a power cycle during two crankshaft revolutions.In a two-stroke engine, the end of the combustion stroke and the beginning of the compression stroke happen simultaneously, with the intake and exhaust (or scavenging) functions occurring at the same time.Two-stroke engines often have a high power-to-weight ratio, power being available in a narrow range of rotational speeds called the "power band". Compared to four-stroke engines, two-stroke engines have a greatly reduced number of moving parts.

process for four stroke engine

A stroke refers to the full travel of the piston along the cylinder, in either direction. The four separate strokes are termed:

Intake: Also known as induction or suction. This stroke of the piston begins at top dead center (T.D.C.) and ends at bottom dead center (B.D.C.). In this stroke the intake valve must be in the open position while the piston pulls an air-fuel mixture into the cylinder by producing vacuum pressure into the cylinder through its downward motion. The piston is moving down as air is being sucked in by the downward motion against the piston.

Compression: This stroke begins at B.D.C, or just at the end of the suction stroke, and ends at T.D.C. In this stroke the piston compresses the air-fuel mixture in preparation for ignition during the power stroke (below). Both the intake and exhaust valves are closed during this stage.

Combustion: Also known as power or ignition. This is the start of the second revolution of the four stroke cycle. At this point the crankshaft has completed a full 360 degree revolution. While the piston is at T.D.C. (the end of the compression stroke) the compressed air-fuel mixture is ignited by a spark plug (in a gasoline engine) or by heat generated by high compression (diesel engines), forcefully returning the piston to B.D.C. This stroke produces mechanical work from the engine to turn the crankshaft.

Exhaust: Also known as outlet. During the exhaust stroke, the piston, once again, returns from B.D.C. to T.D.C. while the exhaust valve is open. This action expels the spent air-fuel mixture through the exhaust valve.

characteristics of two stroke and four stroke engines (which one gets hotter)

Since there are twice as many power strokes during the operation of a two-stroke engine as there are during the operation of a four-stroke engine, the engine tends to heat up more, and thus is likely to have a shorter life. Also, in the two-stroke engine lubricating oil must be mixed with the fuel.

6 0
3 years ago
Read 2 more answers
Bulk wind shear is calculated by finding the vector difference between the winds at two different heights. Using the supercell w
ivanzaharov [21]

Answer:

See explaination

Explanation:

2. 0-1 km shear value: taking winds at 1000mb and 850 mb

15 kts south easterly and 50 kts southerly

Vector difference 135/15 and 180/50 will be 170/61 or southerly 61 kts

3. 0-6 km shear value: taking winds at 1000 mb and 500 mb

15 kts south easterly and 40 kts westerly

Vector difference 135/15 and 270/40 will be 281/51 kts

please see attachment

5 0
3 years ago
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