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Vsevolod [243]
2 years ago
9

An engineer is working with archeologists to create a realistic Roman village in a museum. The plan for a balance in a marketpla

ce calls for granite stones, each weighing denarium. (The denarium was a Roman unit of mass: ). The manufacturing process for making the stones will remove of the material. If the granite to be used has a density of , what is the minimum volume of granite (in ) that the engineer should order?
Engineering
1 answer:
NeTakaya2 years ago
3 0

Answer:

The minimum volume requirement for the granite stones is 1543.64 cm³

Explanation:

1 granite stone weighs 10 denarium

100 granted stones will weigh 1000 denarium

1 denarium = 3.396g

1000 denarium = 3396g.

But we're told that 20% of material is lost during the making of these stones.

This means the mass calculated represents 80% of the original mass requirement, m.

80% of m = 3396

m = 3396/0.8 = 4425 g

This mass represents the minimum mass requirement for making the stones.

To now obtain the corresponding minimum volume requirement

Density = mass/volume

Volume = mass/density = 4425/2.75 = 1543.64 cm³

Hope this helps!!!

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A 260 ft (79.25m) length of size 4 AWG uncoated copper wire operating at a temperature of 75°c has a resistance of 0.0792 ohm.

Explanation:

From the given data the area of size 4 AWG of the code is 21.2 mm², then K is the Resistivity of the material at 75°c is taken as ( 0.0214 ohm mm²/m ).

To find the resistance of 260 ft (79.25 m) of size 4 AWG,

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K = 0.0214 ohm mm²/m

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A = 21.2 mm²

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  = 0.0214 * 3.738

  = 0.0792 ohm.

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Tensile Strength (MPa) Number-Average Molecular Weight (g/mol)
IceJOKER [234]

Answer:

\mathbf{T_{S \infty } \ \approx 215.481 \ MPa}

\mathbf{M_n = 49163.56431  \ g/mol }

Explanation:

The question can be well structured in a table format as illustrated below:

Tensile Strength (MPa)            Number- Average Molecular Weight  (g/mol)

82                                                  12,700

156                                                 28,500

The tensile strength and number-average molecular weight for two polyethylene materials given above.

Estimate the number-average molecular weight that is required to give a tensile strength required above. Using the data given find TS (infinity) in MPa.

<u>SOLUTION:</u>

We know that :

T_S = T_{S \infty} - \dfrac{A}{M_n}

where;

T_S = Tensile Strength

T_{S \infty} = Tensile Strength (Infinity)

M_n = Number- Average Molecular Weight  (g/mol)

SO;

82= T_{S \infty} - \dfrac{A}{12700} ---- (1)

156= T_{S \infty} - \dfrac{A}{28500} ---- (2)

From equation (1) ; collecting the like terms; we have :

T_{S \infty} =82+ \dfrac{A}{12700}

From equation (2) ; we have:

T_{S \infty} =156+ \dfrac{A}{28500}

So; T_{S \infty} = T_{S \infty}

Then;

T_{S \infty} =82+ \dfrac{A}{12700} =156+ \dfrac{A}{28500}

Solving by L.C.M

\dfrac{82(12700) + A}{12700} =\dfrac{156(28500) + A}{28500}

\dfrac{1041400 + A}{12700} =\dfrac{4446000 + A}{28500}

By cross multiplying ; we have:

({4446000 + A})*  {12700} ={28500} *({1041400 + A})

(5.64642*10^{10} + 12700A) =(2.96799*10^{10}+ 28500A)

Collecting like terms ; we have

(5.64642*10^{10} - 2.96799*10^{10} ) =( 28500A- 12700A)

2.67843*10^{10}  = 15800 \ A

Dividing both sides by 15800:

\dfrac{ 2.67843*10^{10} }{15800} =\dfrac{15800 \ A}{15800}

A = 1695208.861

From equation (1);

82= T_{S \infty} - \dfrac{A}{12700} ---- (1)

Replacing A = 1695208.861 in the above equation; we have:

82= T_{S \infty} - \dfrac{1695208.861}{12700}

T_{S \infty}= 82 + \dfrac{1695208.861}{12700}

T_{S \infty}= \dfrac{82(12700) +1695208.861 }{12700}

T_{S \infty}= \dfrac{1041400 +1695208.861 }{12700}

T_{S \infty}= \dfrac{2736608.861 }{12700}

\mathbf{T_{S \infty } \ \approx 215.481 \ MPa}

From equation(2);

156= T_{S \infty} - \dfrac{A}{28500} ---- (2)

Replacing A = 1695208.861 in the above equation; we have:

156= T_{S \infty} - \dfrac{1695208.861}{28500}

T_{S \infty}= 156 + \dfrac{1695208.861}{28500}

T_{S \infty}= \dfrac{156(28500) +1695208.861 }{28500}

T_{S \infty}= \dfrac{4446000 +1695208.861 }{28500}

T_{S \infty}= \dfrac{6141208.861}{28500}

\mathbf{T_{S \infty } \ \approx 215.481 \ MPa}

We are to also estimate the number- average molecular weight that is required to give a tensile strength required above.

If the Tensile Strength (MPa) is 82 MPa

Definitely the average molecular weight will be = 12,700 g/mol

If the Tensile Strength (MPa) is 156 MPa

Definitely the average molecular weight will be = 28,500 g/mol

But;

Let us assume that the Tensile Strength (MPa) = 181 MPa for example.

Using the same formula:

T_S = T_{S \infty} - \dfrac{A}{M_n}

Then:

181 = 215.481- \dfrac{1695208.861 }{M_n}

Collecting like terms ; we have:

\dfrac{1695208.861 }{M_n} = 215.481-  181

\dfrac{1695208.861 }{M_n} =34.481

1695208.861= 34.481 M_n

Dividing both sides by 34.481; we have:

M_n = \dfrac{1695208.861}{34.481}

\mathbf{M_n = 49163.56431  \ g/mol }

5 0
3 years ago
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