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wolverine [178]
3 years ago
12

Suppose we store a relation R (x,y) in a grid file. Both attributes have a range of values from 0 to 1000. The partitions of thi

s grid file happen to be uniformly spaced; for x there are partitions every 20 units, at 20, 40, 60, and so on, while for y the partitions are every 50 units, at 50, 100, 150, and so on.
Required:
a. How many buckets do we have to examine to answer the range query?
b. We wish to perform a nearest-neighbor query for the point (110, 205). We begin by searching the bucket with lower-left corner at (100, 200) and upper-right corner at (120, 250), and we find that the closest point in this bucket is (115, 220). What other buckets must be searched to verify that this point is the closest?
Engineering
1 answer:
leva [86]3 years ago
3 0

Answer:

For (a) The total number of buckets from the given query for the relation is 25 buckets (b) the nearest neighboring query is (80, 200) (80, 150), (100, 150), (120,150) and (120, 200)

Explanation:

From the question stated, we need to define what a Grid file is

Grid File it is a structure of data that are used to divide the total space into a grid non-periodic, where set of point (small) are defined by more than one cells of the grid.

(a)Finding buckets for the query

The relation is divided into two parts which ranges from 0 to 1000, the first part is partitioned in every 20 units, at 20, 40, 60 etc; a second part is partitioned into every 50 units at 50, 100, 150 etc.

The total number of buckets from the given query for the relation is 25 buckets

(b)Finding the closest point or nearest point

The closest point discovered in the distance is little above 15

These points are are the points closer to the point target (110, 205) which can be found in five neighboring rectangles with left corners lower is stated as follows:

(80, 200) (80, 150), (100, 150), (120,150) and (120, 200)

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If the bolt head and the supporting bracket are made of the same material having a failure shear stress of 'Tra;i = 120 MPa, det
Nina [5.8K]

Answer:

P=361.91 KN

Explanation:

given data:

brackets and head of the screw are made of material with T_fail=120 Mpa

safety factor is F.S=2.5

maximum value of force P=??

<em>solution:</em>

to find the shear stress

                            T_allow=T_fail/F.S

                                         =120 Mpa/2.5

                                         =48 Mpa

we know that,

                               V=P

<u>Area for shear head:</u>

                              A(head)=π×d×t

                                           =π×0.04×0.075

                                           =0.003×πm^2

<u>Area for plate:</u>

                               A(plate)=π×d×t  

                                            =π×0.08×0.03

                                            =0.0024×πm^2

now we have to find shear stress for both head and plate

<u>For head:</u>

                                   T_allow=V/A(head)

                                    48 Mpa=P/0.003×π                 ..(V=P)

                                             P =48 Mpa×0.003×π

                                                =452.16 KN

<u>For plate:</u>

                                   T_allow=V/A(plate)

                                    48 Mpa=P/0.0024×π                 ..(V=P)

                                             P =48 Mpa×0.0024×π

                                                =361.91 KN

the boundary load is obtained as the minimum value of force P for all three cases. so the solution is

                                                P=361.91 KN

note:

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7 0
4 years ago
1. Represent each of the following combinations of units in the correct SI form using the appropriate prefix: (a) μMN, (b) N/μm,
Burka [1]

Answer:

Kindly check explanation

Explanation:

(a) μMN, (b) N/μm, (c) MN/ks2, and (d) kN/ms.

(a) μMN = (10^-6 * 10^6) = 10^(-6 + 6) = 10^0 = N

b) N/μm = N / 10^-6 m = 10^6 * N/m = MN/m

(c) MN/ks2 = 10^6N / (10^3 s)^2

10^6 N / 10^6s^2 = 10^6 * 10^-6 N /s^2 = N/s^2

D) kN/ms = 10^3N / 10^-3 s = 10^3 * 10^3 * N/s = 10^6N / s = MN/s

2)

a) 0.000431kg = 431 × 10^6 kg = 431 * 10^9g = 431Gg

b) 35.3 × 10^3 N

10^3 = kilo(K)

35.3 KN

C) 0.00532km = 5.32 * 10^3 km = 5.32 * 10^3 * 10^3 = 5.32 * 10^6 = 5.32Mm

3) Represent each of the following combinations of units in the correct SI form: (a) Mg/ms, (b) N/nm, and (c) mN/(kg⋅μs).

a) Mg/ms = 10^6g / 10^-3s = 10^6 * 10^3 g/s = 10^9 g/s = Gg/s

b) N/nm = N / 10^-9 m = 10^9 N/m = GN/m

c) mN/(kg⋅μs) = 10^6N / kg(10^-6s) = 10^12N/(kg.s)

= TN/(kg.s)

6 0
3 years ago
5. What number filter lens is recommended when welding with SMAW and using 1/8" (3.2 mm)
Inessa05 [86]

Answer:

What filter lens is recommended for SMAW?

Table of Filter Lenses for Protection From Radiant Energy

Type of Operation Electric Size 1/32 in. Arc Current (Amp)

Shield metal arc welding (SMAW) 3 to 5 60 to 160

5 to 8 160 to 250

More than 8 250 to 500

Explanation:

7 0
3 years ago
What is a perpetual motion machine of the second kind?
d1i1m1o1n [39]

Answer:

perpetual motion machine of second type is a machine that generates job from a single source of heat.

Explanation:

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3 years ago
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vichka [17]

Answer:

b. spark plugs

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