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3 years ago

11

A colorful design for a product meant to appeal to a preschool audience reflects which of the following points in industrial des

ign?
usability
appeal to emotion
maintenance
effective use of resources

1 answer:

lord [1]3 years ago

6 0

Appeal to emotion

the product doesn’t need to be colorful to function but it would be more appealing to the intended audience if it was

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A BOD test is to be run on a sample of wastewater that has a five-day BOD of 230 mg/L. If the initial DO of a mix of distilled w

Alexxandr [17]

Answer:

Distribution factor P = =38.33

V = 7.826 ml

Explanation:

given details:

BOD =230 mg/l

DO inital = 8.0mg/l

DO final = 2.0mg/l

we know

BOD = [DO inital -DO final] * distribution factor

230 = [8 - 2] D.F

Distribution factor P $= \frac{230}{6}$

Distribution factor P = =38.33

THE RANGE OF WASTE WATER VOLUME IN 300 ml bottle is

distribution factor $= \frac{300}{V}$

$V = \frac{300}{38.33}$

V = 7.826 ml

6 0

3 years ago

The E7018 Electrode produces a/an

julia-pushkina [17]

Answer:

Explanation:

These include the 6010, 6011, 6012, 6013, 7014, 7024 and 7018 electrodes. 6010 electrodes deliver deep penetration and have the ability to “dig” through rust, oil, paint or dirt, making them popular among pipe welders.

7 0

4 years ago

A flywheel made of Grade 30 cast iron (UTS = 217 MPa, UCS = 763 MPa, E = 100 GPa, density = 7100 Kg/m, Poisson's ratio = 0.26) h

hram777 [196]

Answer:

N = 38546.82 rpm

Explanation:

$D_{1}$ = 150 mm

$A_{1}= \frac{\pi }{4}\times 150^{2}$

= 17671.45 $mm^{2}$

$D_{2}$ = 250 mm

$A_{2}= \frac{\pi }{4}\times 250^{2}$

= 49087.78 $mm^{2}$

The centrifugal force acting on the flywheel is fiven by

F = M ( $R_{2}$ - $R_{1}$ ) x $w^{2}$ ------------(1)

Here F = ( -UTS x $A_{1}$ + UCS x $A_{2}$ )

Since density, $\rho = \frac{M}{V}$

$\rho = \frac{M}{A\times t}$

$M = \rho \times A\times t$ $M = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t$

$M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37$

$M = 8252963901$

∴ $R_{2}$ - $R_{1}$ = 50 mm

∴ F = $763\times \frac{\pi }{4}\times 250^{2}-217\times \frac{\pi }{4}\times 150^{2}$

F = 33618968.38 N --------(2)

Now comparing (1) and (2)

$33618968.38 = 8252963901\times 50\times \omega ^{2}$

∴ ω = 4036.61

We know

$\omega = \frac{2\pi N}{60}$

$4036.61 = \frac{2\pi N}{60}$

∴ N = 38546.82 rpm

7 0

3 years ago

- Viscoelastic stress relaxation

My name is Ann [436]

Explanation:

The correct answers to the fill in the blanks would be;

1. Viscoelastic stress relaxation refers to scenarios for which the stress applied to a polymer must decay over time in order to maintain a constant strain. Otherwise, over time, the polymer chains will slip and slide past one another in response to a constant applied load and the strain will increase (in magnitude).

2. Viscoelastic creep refers to scenarios for which a polymer will permanently flow over time in response a constant applied stress.

The polymer whose properties have been mentioned above is commonly known as Kevlar.

It is mostly used in high-strength fabrics and its properties are because of several hydrogen bonds between polymer molecules.

5 0

3 years ago

A furnace wall composed of 200 mm, of fire brick. 120 mm common brick 50mm 80% magnesia and 3mm of steel plate on the outside. I

Liula [17]

Answer:

fire brick / common brick : 1218 °C
common brick / magnesia : 1019 °C
magnesia / steel : 90.06 °C
heat loss: 4644 kJ/m^2/h

Explanation:

The thermal resistance (R) of a layer of thickness d given in °C·m²·h/kJ is ...

R = d/k

so the thermal resistances of the layers of furnace wall are ...

R₁ = 0.200/4 = 0.05 °C·m²·h/kJ

R₂ = 0.120 2.8 = 3/70 °C·m²·h/kJ

R₃ = 0.05/0.25 = 0.2 °C·m²·h/kJ

R₄ = 0.003/240 = 1.25×10⁻⁵ °C·m²·h/kJ

So, the total thermal resistance is ...

R₁ +R₂ +R₃ +R₄ = R ≈ 0.29286 °C·m²·h/kJ

__

The rate of heat loss is ΔT/R = (1450 -90)/0.29286 = 4643.70 kJ/(m²·h)

__

The temperature drops across the various layers will be found by multiplying this heat rate by the thermal resistance for the layer:

fire brick: (4543.79 kJ/(m²·h))(0.05 °C·m²·h/kJ) = 232 °C

so, the fire brick interface temperature at the common brick is ...

1450 -232 = 1218 °C

For the next layers, the interface temperatures are ...

common brick to magnesia = 1218 °C - (3/70)(4643.7) = 1019 °C

magnesia to steel = 1019 °C -0.2(4643.7) = 90.06 °C

_____

<em>Comment on temperatures</em>

Most temperatures are rounded to the nearest degree. We wanted to show the small temperature drop across the steel plate, so we showed the inside boundary temperature to enough digits to give the idea of the magnitude of that.

5 0

3 years ago