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balu736 [363]
2 years ago
13

Why is it important that ironworkers wait to begin work until after cement masons are finished with their job?

Engineering
1 answer:
ziro4ka [17]2 years ago
3 0

Answer:

Constructing a building's framework requires a solid foundation

is the answer

Explanation:

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The purification of hydrogen gas is possible by diffusion through a thin palladium sheet. Calculate the number of kilograms of h
diamong [38]

Answer:

M=0.0411 kg/h or 4.1*10^{-2} kg/h

Explanation:

We have to combine the following formula to find the mass yield:

M=JAt

M=-DAt(ΔC/Δx)

The diffusion coefficient : D=6.0*10^{-8} m/s^{2}

The area : A=0.25 m^{2}

Time : t=3600 s/h

ΔC: (0.64-3.0)kg/m^{3}

Δx: 3.1*10^{-3}m

Now substitute the  values

M=-DAt(ΔC/Δx)

M=-(6.0*10^{-8} m/s^{2})(0.25 m^{2})(3600 s/h)[(0.64-3.0kg/m^{3})(3.1*10^{-3}m)]

M=0.0411 kg/h or 4.1*10^{-2} kg/h

8 0
2 years ago
Evan notices a small fire in his workplace. Since the fire is small and the atmosphere is not smoky he decides to fight the fire
Norma-Jean [14]

Answer:

not calling the firemean

Explanation:

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2 years ago
An annealed copper strip of 228 mm wide and 25 mm thick being rolled to a thickness of 20 mm, in one pass. The roll radius is 30
sdas [7]

Answer:

The roll force is 1.59 MN

The power required in this operation is 644.96 kW

Explanation:

Given;

width of the annealed copper, w = 228 m

thickness of the copper, h₀ = 25 mm

final thickness, hf = 20 mm

roll radius, R = 300 mm

The roll force is given by;

F = LwY_{avg}

where;

w is the width of the annealed copper

Y_{avg} is average true stress of the strip in the roll gap

L is length of arc in contact, and for frictionless situation it is given as;

L = \sqrt{R(h_o-h_f)} \\\\L = \sqrt{300(25-20)}\\\\L = 38.73 \ mm

Now, determine the average true stress, Y_{avg}, for the annealed copper;

The absolute value of the true strain, ε = ln(25/20)

ε = 0.223

from true stress vs true strain graph; at true strain of 0.223, the true stress is 280 MPa.

Then, the average true stress = ¹/₂(280 MPa.) = 180 MPa

Finally determine the roll force;

F = LwY_{avg}

F = (\frac{38.73 }{1000})(\frac{228}{1000})*180 \ MPa\\\\F =   1.59 \ MN

The power required in this operation is given by;

P = \frac{2\pi FLN}{60}\\\\P =  \frac{2\pi (1.59*10^6)(0.03873)(100)}{60}\\\\P = 644955.2 \ W\\\\P = 644.96 \ kW

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2 years ago
Who want to play "1v1 lol unblocked games 76"
kotegsom [21]

Answer:I want to play what game?

Explanation:

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2 years ago
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A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 3.5 mm (0.14 in
RideAnS [48]

To resolve this problem we have,

R=3.5mm\\F_f1=950N\\L_1=50mm\\b=12mm\\L_2=40mm

F_{f2} is unknown.

With these dates we can calculate the Flexural strenght of the specimen,

\sigma{fs}=\frac{F_{f1}L}{\pi R^3}\\\sigma{fs}=\frac{(950)(50*10^{-3})}{\pi 3.5*10^{-3}}\\\sigma{fs}=352.65Mpa

After that, we can calculate the flexural strenght for the square cross section using the previously value.

\sigma{fs}=\frac{F_{f2}L}{\pi R^3}\\(352.65*10^6)=\frac{3Ff(40*10^{-3})}{2(12*10^{-10})}\\F_{f2}=\frac{352.65*10^6}{34722.22}\\F_{f2}=10156.32N\\F_{f2}=10.2kN

6 0
2 years ago
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