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2 years ago

Which of the following does NOT describe product design.

Engineering

1 answer:

Andrew [12]2 years ago

6 0

Question:

Which of the following does NOT describe product design for maximum supply chain flexibility?

A. Create an increased number of modular products with fewer components.

B. Complete product assembly as close to receipt of actual orders as possible.

C. Reengineer products to use the same components.

D. Create many products variations.

Answer:

The correct option is;

D. Create many products variations

Explanation:

The options are analysed as follows

A. Create an increased number of modular products with fewer components.

Modularity increases the utilization of components such that manufacturers can create a wide variety of possible products that increases costumer satisfaction while maintaining utilization convenience

B. Complete product assembly as close to receipt of actual orders as possible.

Here we have requirement for improved supply chain flexibility as such improving raw material utilization rate and overall manufacturing efficiency

C. Reengineer products to use the same components.

Here we have increasing modularization which improves design flexibility and manufacturing comparative advantage

D. Create many products variations.

The above creates many varying components which increases the number of components to be assembled into products hence it does not describe product design for maximum supply chain flexibility.

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A manufacturer makes two types of drinking straws: one with a square cross-sectional shape, and the other type the typical round

Harlamova29_29 [7]

Answer:

$\frac{Q_{square}}{Q_{circle}} = 0.785$

Explanation:

given data

types of drinking straws

square cross-sectional shape
round shape

solution

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so we can say that

perimeter of square = perimeter of circle

4 × S = π × D

here S is length and D is diameter

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and

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$\frac{Q_{square}}{Q_{circle}} = \frac{AV^2}{AV^2}$

$\frac{Q_{square}}{Q_{circle}} = \frac{S^2}{\frac{\pi D^2}{4}}$

$\frac{Q_{square}}{Q_{circle}} = \frac{(\frac{\pi D}{4})^2}{\frac{\pi D^2}{4}}$

$\frac{Q_{square}}{Q_{circle}} = \frac{\pi }{4}$

$\frac{Q_{square}}{Q_{circle}} = 0.785$

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3 years ago

A logic chip used in a computer dissipates 3 W of power in an environment at 120°F, and has a heat transfer surface area of 0.08

UkoKoshka [18]

Answer:

attached below

Explanation:

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3 years ago

Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320

lesya [120]

Answer:

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

Explanation:

The complete statement of this question is "Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P = 450 kN"

Each bolt is subjected to shear forces. In this case, safety factor is the ratio of the ultimate shear strength to maximum allowable shear stress. That is to say:

$n = \frac{S_{uts}}{\tau_{max}}$

Where:

$n$ - Safety factor, dimensionless.

$S_{uts}$ - Ultimate shear strength, measured in pascals.

$\tau_{max}$ - Maximum allowable shear stress, measured in pascals.

The maximum allowable shear stress is consequently cleared and computed: ( $n = 4.2$ , $S_{uts} = 320\times 10^{6}\,Pa$ )

$\tau_{max} = \frac{S_{uts}}{n}$

$\tau_{max} = \frac{320\times 10^{6}\,Pa}{4.2}$

$\tau_{max} = 76.190\times 10^{6}\,Pa$

Since each bolt has a circular cross section area and assuming the shear stress is not distributed uniformly, shear stress is calculated by:

$\tau_{max} = \frac{4}{3} \cdot \frac{V}{A}$

Where:

$\tau_{max}$ - Maximum allowable shear stress, measured in pascals.

$V$ - Shear force, measured in kilonewtons.

$A$ - Cross section area, measured in square meters.

As connection consist on five bolts, shear force is equal to a fifth of the applied load. That is:

$V = \frac{P}{5}$

$V = \frac{450\,kN}{5}$

$V = 90\,kN$

The minimum allowable cross section area is cleared in the shearing stress equation:

$A = \frac{4}{3}\cdot \frac{V}{\tau_{max}}$

If $V = 90\,kN$ and $\tau_{max} = 76.190\times 10^{3}\,kPa$ , the minimum allowable cross section area is:

$A = \frac{4}{3} \cdot \frac{90\,kN}{76.190\times 10^{3}\,kPa}$

$A = 1.640\times 10^{-3}\,m^{2}$

The minimum allowable cross section area can be determined in terms of minimum allowable bolt diameter by means of this expression:

$A = \frac{\pi}{4}\cdot D^{2}$

The diameter is now cleared and computed:

$D = \sqrt{\frac{4}{\pi}\cdot A}$

$D =\sqrt{\frac{4}{\pi}\cdot (1.640\times 10^{-3}\,m^{2})$

$D = 0.0457\,m$

$D = 45.7\,mm$

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

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3 years ago

A homeowner consumes 260 kWh of energy in July when the family is on vacation most of the time. Determine the average cost per k

Llana [10]

Answer:

16.2 cents

Explanation:

Given that a homeowner consumes 260 kWh of energy in July when the family is on vacation most of the time.

Where Base monthly charge of $10.00. First 100 kWh per month at 16 cents/kWh. Next 200 kWh per month at 10 cents/kWh. Over 300 kWh per month at 6 cents/kWh.

For the first 100 kWh:

16 cent × 100 = 1600 cents = 16 dollars

Since 1 dollar = 100 cents

For the remaining energy:

260 - 100 = 160 kwh

10 cents × 160 = 1600 cents = 16 dollars

The total cost = 10 + 16 + 16 = 42 dollars

Note that the base monthly of 10 dollars is added.

The cost of 260 kWh of energy consumption in July is 42 dollars

To determine the average cost per kWh for the month of July, divide the total cost by the total energy consumed.

That is, 42 / 260 = 0.1615 dollars

Convert it to cents by multiplying the result by 100.

0.1615 × 100 = 16.15 cents

Approximately 16.2 cents

7 0

3 years ago

Hot water at an average temperature of 88°C and an average velocity of 0.4 m/s is flowing through a 5-m section of a thin-walled

SIZIF [17.4K]

Answer:

a) The rate of heat transfer will be 19.58 Watts.

b) The temperature drop of the hot water will be 0.024 Degree Celcius.

Explanation:

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3 years ago