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Digiron [165]
3 years ago
14

Miguel’s presentation on the Iron Age moving into the Middle Ages was detailed. He outlined the process of smelting and clarifie

d how metals such as copper, tin, and bronze could be combined with intense heat to create strong iron. He added some images of tools that came into wide use, such as adzes and axes and even armor. Miguel left out one tool, however, that was highly important to the civilization’s progress. What might have been that tool, and why is it important?
A.
hammer, invaluable for building homes

B.
drinking vessel, essential for carrying water

C.
fencing, to create separation between properties

D.
heavy plough, which advanced food production
Engineering
1 answer:
tatuchka [14]3 years ago
3 0
Hello!! The correct answer would be C!! I took the same test!
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An insulated piston-cylinder device contains 0.15 of saturated refrigerant-134a vapor at 0.8 MPa pressure. The refrigerant is no
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Answer:

Assumption:

1. The kinetic and potential energy changes are negligible

2. The cylinder is well insulated and thus heat transfer is negligible.

3. The thermal energy stored in the cylinder itself is negligible.

4. The process is stated to be reversible

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a. This is reversible adiabatic(i.e isentropic) process and thus s_{1} =s_{2}

From the refrigerant table A11-A13

P_{1} =0.8MPa   \left \{ {{ {{v_{1}=v_{g}  @0.8MPa =0.025645 m^{3/}/kg } } \atop { {{u_{1}=u_{g}  @0.8MPa =246.82 kJ/kg } -   also  {{s_{1}=s_{g}  @0.8MPa =0.91853 kJ/kgK } } \right.

sat vapor

m=\frac{V}{v_{1} } =\frac{0.15}{0.025645} =5.8491 kg\\and \\\\P_{2} =0.2MPa  \left \{ {{x_{2} =\frac{s_{2} -s_{f} }{s_{fg }}=\frac{0.91853-0.15449}{0.78339}   = 0.9753 \atop {u_{2} =u_{f} +x_{2} }(u_{fg}) =  38.26+0.9753(186.25)= 38.26+181.65 =219.9kJ/kg \right. \\s_{1} = s_{2}

T_{2} =T_{sat @ 0.2MPa} = -10.09^{o}  C

b.) We take the content of the cylinder as the sysytem.

This is a closed system since no mass leaves or enters.

Hence, the energy balance for adiabatic closed system can be expressed as:

E_{in} - E_{out}  =ΔE

w_{b, out}  =ΔU

w_{b, out} =m([tex]u_{1} -u_{2)

w_{b, out}  = workdone during the isentropic process

=5.8491(246.82-219.9)

=5.8491(26.91)

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3 years ago
An 18-in.-long titanium alloy rod is subjected to a tensile load of 24,000 lb. If the allowable tensile stress is 60 ksi and the
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Answer:

Required Diameter = 302.65 inches

Explanation:

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Allowable tensile stress = 60 ksi

Weight of tensile load = 24,000 lb

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Now, we know that the formula for stress is;

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Thus,

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So for this stress, area required is;

A_req = 24000/60 = 4000 in²

So let's find the required diameter here.

Area = πd²/4

So, 4000 = πd²/4

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Required diameter here is;

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Strain = stress/E

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stress = P/A = 24,000/A

strain = elongation/original length = 0.05/18 = 0.00278

Thus;

0.00278 = P/(A•E)

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Making A the subject to obtain;

A = 24000/(120 x 0.00278)

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Required diameter here is;

d = √91599.4

d = 302.65 in

The larger diameter is 302.65 inchesand it's therefore the required one.

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