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dexar [7]
3 years ago
15

If the outside diameter of a pipe is 2 m, the length of a piece of insulation wrapped around it would be a)- 628 cm b)- 12.56 m.

c)- 12.56 cm d)-62.8 m
Engineering
1 answer:
AlexFokin [52]3 years ago
6 0

The distance a piece of insulation would be, would be the circumference of the pipe.

The circumference is found by multiplying the diameter by PI.

Circumference - 3.14 x 2 meters = 6.28 meters

1 meter is equal to 100 cm.

Multiply the total meters by 100 to convert to cm:

6.28 meters x 100 = 628 cm.

The answer would be a. 628 cm.

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aivan3 [116]

Answer:

7/7

Explanation:

7/56 ÷1

7/56÷ 7/7

7/56 ×7/7

1/8

5 0
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The specific volume of mercury is .00007 m^3/Kg. What is its density in lbm/ft^3?
aalyn [17]

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891.027 lbm/ft³

Explanation:

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3 years ago
Six forces act on a beam that forms part of a building's
IrinaVladis [17]

Answer:

<h2> FA = 13 kN </h2><h2>FG = 15.3 kN</h2>

Explanation:

write each force in terms of magnitude and directions  

Fx = F sin Ф

Fy = F cos Ф

where Ф is to be measured from x axis.

∑F at y = o

FAy + FBy + FCy + FDy + FEy + FGy = 0

∑F at x = o

FAx + FBx + FCx + FDx + FEx + FGx = 0

Let  

FA = FA sin (110)   +   FA cos (110)

FB = 20 sin (270)  +  20 cos (270)

FC = 16 sin (140)    +  16 cos (140)

FD = 9 sin (40)       +  9 cos (40)

FE = 20 sin (270)    +  20 cos (270)

FG = FG sin (50)     +  FG cos (50)

add x and y forces:

FAx + FBx + FCx + FDx + FEx + FGx = 0

FAy + FBy + FCy + FDy + FEy + FGy = 0

FA sin (110)  + 0  + 16 sin (140)  + 9 sin (40)  + 0   + FG sin (50) = 0

FA cos (110) - 20 + 16 cos (140) + 9 cos (40) - 20 + FG cos (50 = 0

FA sin (110)  + 0  + 10.285  + 5.785  + 0   + FG sin (50) = 0

FA cos (110) - 20 - 12.257 + 6.894 - 20 + FG cos (50) = 0

FA sin (110)  + 16.070 + FG sin (50) = 0        

FA cos (110) - 45.363 + FG cos (50) = 0

solving for FA, and FG

FA = 13 kN

FG = 15.3 kN

7 0
3 years ago
A train consists of a 50 Mg engine and three cars, each having a mass of 30 Mg . If it takes 75 s for the train to increase its
ohaa [14]

Answer:

T = 15 kN

F = 23.33 kN

Explanation:

Given the data in the question,

We apply the impulse momentum principle on the total system,

mv₁ + ∑\int\limits^{t2}_{t1} {Fx} \, dt = mv₂

we substitute

[50 + 3(30)]×10³ × 0 + FΔt = [50 + 3(30)]×10³ ×  ( 45 × 1000 / 3600 )  

F( 75 - 0 ) =  1.75 × 10⁶

The resultant frictional tractive force F is will then be;

F =  1.75 × 10⁶ / 75

F = 23333.33 N

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Applying the impulse momentum principle on the three cars;

mv₁ + ∑\int\limits^{t2}_{t1} {Fx} \, dt = mv₂

[3(30)]×10³ × 0 + FΔt = [3(30)]×10³ ×  ( 45 × 1000 / 3600 )  

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The force T developed is then;

T =  1.125 × 10⁶ / 75

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T = 15 kN

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9966 [12]

Answer:

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Manufacturer’s Recommendations means the written instructions, procedures and recommendations which are issued by the original equipment manufacturer of any plant or equipment used at the Utility Plant relating to the operation, maintenance and repair of such plant or equipment and any revisions thereto issued by the manufacturer, which are valid and applicable at the time such operation, maintenance or repair is undertaken. Notwithstanding the above, Manufacturer’s Recommendations shall not include any instructions, procedures or recommendations of a manufacturer of any plant or equipment that the Owner and the Operator have agreed in writing to exclude from this definition or have agreed in writing should not be followed.

Explanation:

4 0
2 years ago
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