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Drupady [299]
3 years ago
5

Manufacturers frequently make choices about their suppliers of raw materials based on their impact on society and the environmen

t.
Examples include the use of recyclable components, fair trade textiles and biodegradable packaging.
Discuss how these choices may improve the ethical image of a company
Engineering
1 answer:
love history [14]3 years ago
4 0
Manufacturers seek attention
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A 2-cm-diameter vertical water jet is injected upward by a nozzle at a speed of 15 m/s. Determine the maximum weight of a flat p
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Answer:58.28 N

Explanation:

Given data

dia. of nozzle \left ( d\right )=2 cm

initial velocity\left ( u\right )=15 m/s

height\left ( h\right )=2m

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What type of intersection is this?
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lord [1]

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5 0
2 years ago
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
4 years ago
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