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3 years ago

5

How many 10" diameter circles can be cut from a semicircular shape that has a 20"

2 answers:

Iteru [2.4K]3 years ago

5 0

Answer:

<h2> 1</h2>

<em>The</em><em> </em><em>diameter</em><em> </em><em>measurement</em><em> </em><em>of</em><em> </em><em>a</em><em> </em><em>semi circle</em><em> </em><em>having</em><em> </em><em>a</em><em> </em><em>measure</em><em> </em><em>of</em><em> </em><em>1</em><em>0</em><em>"</em><em> </em><em>diameter</em><em> </em><em>,</em><em> </em><em>2</em><em>0</em><em>"</em><em> </em><em>diameter</em><em> </em><em>and</em><em> </em><em>a</em><em> </em><em>length</em><em> </em><em>of</em><em> </em><em>2</em><em>5</em><em>.</em>

N76 [4]3 years ago

3 0

9514 1404 393

Answer:

1

Explanation:

Only one such circle can be drawn. The diameter of the 10" circle will be a radius of the semicircle. In order for the 10" circle to be wholly contained, the flat side of the semicircle must be tangent to the 10" circle. There is only one position in the figure where that can happen. (see attached).

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"A fluid at a pressure of 7 atm with a specific volume of 0.11 m3/kg is constrained in a cylinder behind a piston. It is allowed

AlekseyPX

Answer:

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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.

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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

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Answer:

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3 years ago

Situation: Peter is designing a new hybrid car that functions on solar power. He is currently working on sketches of his design

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2 years ago

An aluminum alloy tube with an outside diameter of 3.50 in. and a wall thickness of 0.30 in. is used as a 14 ft long column. Ass

slega [8]

Answer:

slenderness ratio = 147.8

buckling load = 13.62 kips

Explanation:

Given data:

outside diameter is 3.50 inc

wall thickness 0.30 inc

length of column is 14 ft

E = 10,000 ksi

moment of inertia $= \frac{\pi}{64 (D_O^2 -D_i^2)}$

$I = \frac{\pi}{64}(3.5^2 -2.9^2) = 3.894 in^4$

Area $= \frac{\pi}{4} (3.5^2 -2.9^2) = 3.015 in^2$

$radius = \sqrt{\frac{I}{A}}$

$r = \sqrt{\frac{3.894}{3.015}$

r = 1.136 in

slenderness ratio $= \frac{L}{r}$

$= \frac{14 *12}{1.136} = 147.8$

buckling load $= P_cr = \frac{\pi^2 EI}}{l^2}$

$P_{cr} = \frac{\pi^2 *10,000*3.844}{( 14\times 12)^2}$

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3 0

3 years ago