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slavikrds [6]
3 years ago
14

The two parking space configurations you'll see in garages and parking lots are _____.

Engineering
2 answers:
Brrunno [24]3 years ago
6 0

Answer:

Angled parking spaces and parallel parking and perpendicular parking space

Explanation:

solution

parking vehicle is garages and parking lot in properly is very necessary to avoid conjunction and save space

pattern of parking are Angled parking spaces and parallel parking and perpendicular parking space

Angled parking is that when vehicle is part at some angle like 45° or 60°  

and in parallel parking vehicle is park in one line as each back bumper of adjacent 1 vehicle

and perpendicular parking it is bay parking

so correct answer is Angled parking spaces and parallel parking and perpendicular parking space

den301095 [7]3 years ago
4 0

Answer:

straight-in spaces and angle spaces  Hope this helps!

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Answer:

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Explanation:

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7 0
3 years ago
You have just finished your OST takeoffs for a single-story home and found 175 LF of interior walls with 2x6 studs 14" OC. The h
zimovet [89]

Answer:

Total BF for the interior wall is 7.50BD

Explanation:

Given Data:

· Size of stud = 2” x 6”

· Height of Wall = 8 ft

· Top plates = 2

· Bottom Plate = 1

BF stands for board feet in lumber/wood terminology. It is the unit of volume.

1 BF (Board feet) = 1 ft x 1 ft x 1 inch

Since there are total three plates at top and bottom, we have to deduct their thickness from wall height to calculate height of stud.

Height of stud = 8’ – 3 x 2” = 7’6” = 7.5 ft

Board feet of one stud = 7.50 6/12 x 2 = 7.50 BD

Total BF for the interior wall is 7.50BD

7 0
3 years ago
The in situ moist unit weight of a soil is 17.3 kN/m3 and the moisture content is 16%. The specific gravity of soil solids is 2.
Temka [501]

Answer:

Explanation:

Given that,

Moist content w = 16%

The in situ moist unit weight of the soil : γ(in situ) = 17.3 kN/m³

Specific gravity of the soil

G(s) = 2.72

Minimum dry unit weight of the soil

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Question

how many cubic meters of soil from the excavation site are needed to produce 2000 m³ of compacted fill?

Let determine the in situ dry unit weight γd(in-situ) using the relation

γd(in-situ) = γ(in-situ) / [1 + (w/100)]

γd(in-situ) = 17.3/ [1 + (18/100)]

γd(in-situ) = 17.3 / ( 1 + 0.18)

γd(in-situ) = 17.3 / 1.18

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To determine the Volume of the soil to be excavated (Vex)

Let the Volume to be excavated = V

We can use the relation

V=V(fill) × γd(compacted) / γd(in situ)

Given that, V(fill) = 2000m³

V(fill) is the volume of the compacted fill

Therefore,

V=V(fill) × γd(compacted) / γd(in situ)

Vex = 2000 × 18.1 / 14.66

Vex = 2469.13 m³

So, the excavated volume of the soil is 2469.13 m³

3 0
4 years ago
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dedylja [7]

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6 0
3 years ago
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