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3 years ago

Why should engineers avoid obvious patterns?

Engineering

2 answers:

lozanna [386]3 years ago

5 0

Answer:

Patterns may cloud the vision of the design process and send the engineer off track.

Explanation:

Engineers use a cyclical process and must stay on track during this process. A key point to an engineer is being able to have the ability to use creative and critical thinking.

seraphim [82]3 years ago

5 0

Answer:

Patterns tend to impose a prescribed way of thinking that will limit possible solutions.

Explanation:

While patterns help scientists and mathematicians solve problems, engineers should avoid obvious patterns. These patterns tend to lock in a prescribed way of thinking that will limit possible solutions. While making observations is helpful, don't force them into categories. Watch for observations that are unexpected, and compare and contrast all information before deciding on a final solution.

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A plot of land is an irregular trangle with a base of 122 feet and a height of 47 feet what is the area of the plot?

Reika [66]

Answer:

150 is the area

Explanation:

3 0

3 years ago

A PLL is set up so that its VCO free-runs at 8.9 MHz. The VCO does not change frequency unless its input is within plus or minus

pickupchik [31]

It’s A 75khz because it’s plus or minus so if u add it would be too much

5 0

3 years ago

The time to half-maximum voltage is how long it takes the capacitor to charge halfway. Based on your experimental results, how l

satela [25.4K]

Answer:

Time taken for the capacitor to charge to 0.75 of its maximum capacity = 2 × (Time take for the capacitor to charge to half of its capacity)

Explanation:

The charging of a capacitor/the build up of its voltage follows an exponential progression and is given by

V(t) = V₀ [1 - e⁻ᵏᵗ]

where k = (1/time constant)

when V(t) = V₀/2

(1/2) = 1 - e⁻ᵏᵗ

e⁻ᵏᵗ = 0.5

In e⁻ᵏᵗ = In 0.5 = - 0.693

-kt = - 0.693

kt = 0.693

t = (0.693/k)

Recall that k = (1/time constant)

Time to charge to half of max voltage = T(1/2)

T(1/2) = 0.693 (Time constant)

when V(t) = 0.75

0.75 = 1 - e⁻ᵏᵗ

e⁻ᵏᵗ = 0.25

In e⁻ᵏᵗ = In 0.25 = -1.386

-kt = - 1.386

kt = 1.386

t = 1.386(time constant) = 2 × 0.693(time constant)

Recall, T(1/2) = 0.693 (Time constant)

t = 2 × T(1/2)

Hope this Helps!!!

3 0

3 years ago

Read 2 more answers

An aluminum block weighing 28 kg initially at 140°C is brought into contact with a block of iron weighing 36 kg at 60°C in an in

Anika [276]

Answer:

Equilibrium Temperature is 382.71 K

Total entropy is 0.228 kJ/K

Solution:

As per the question:

Mass of the Aluminium block, M = 28 kg

Initial temperature of aluminium, $T_{a} = 140^{\circ}C$ = 273 + 140 = 413 K

Mass of Iron block, m = 36 kg

Temperature for iron block, $T_{i} = 60^{\circ}C$ = 273 + 60 = 333 K

At 400 k

Specific heat of Aluminium, $C_{p} = 0.949\ kJ/kgK$

At room temperature

Specific heat of iron, $C_{p} = 0.45\ kJ/kgK$

Now,

To calculate the final equilibrium temperature:

Amount of heat loss by Aluminium = Amount of heat gain by Iron

$MC_{p}\Delta T = mC_{p}\Delta T$

$28\times 0.949(140 - T_{e}) = 36\times 0.45(T_{e} - 60)$

Thus

$T_{e} = 109.71^{\circ}C$ = 273 + 109.71 = 382.71 K

where

$T_{e}$ = Equilibrium temperature

Now,

To calculate the changer in entropy:

$\Delta s = \Delta s_{a} + \Delta s_{i}$

Now,

For Aluminium:

$\Delta s_{a} = MC_{p}ln\frac{T_{e}}{T_{i}}$

$\Delta s_{a} = 28\times 0.949\times ln\frac{382.71}{413} = - 2.025\ kJ/K$

For Iron:

$\Delta s_{i} = mC_{i}ln\frac{T_{e}}{T_{i}}$

$\Delta s_{a} = 36\times 0.45\times ln\frac{382.71}{333} = 2.253\ kJ/K$

Thus

$\Delta s =-2.025 + 2.253 = 0.228\ kJ/K$

6 0

3 years ago

Consider a 400 mm × 400 mm window in an aircraft. For a temperature difference of 90°C from the inner to the outer surface of th

Sergeeva-Olga [200]

Answer:

HEAT LOST

polycarbonate = 252 W

soda lime glass = 1680 W

aerogel = 16.8 W

COST associated with heat loss

polycarbonate = $ 262.08

soda lime glass = $ 1,747.2

aerogel = $ 17.472

The cost associated with heat loss is maximum in Soda Lime and minimum in Aerogel

Explanation:

Given that;

surface area for each window = 0.4m * 0.4m = 0.16m^2

DeltaT = 90°C, L = 12mm = 0.012m

thermal conductivity of soda line can be gotten from tables in FUNDAMENTALS OF HEAT AND MASS TRANSFER

so at 300K

KsL = 1.4 W/mK

Kag = 0.014 W/mK

Kpc = 0.21 W/mK

Now HEAT LOSS

for polycarbonate;

Qpc = -KA dt/dx

NOTE ( heat flows from high temperature region to low temperature regions. so the second temperature would be smaller compared to the initial causing a negative in the change in temperature)

so Qag = (0.21 * 0.16 * 90) / 0.012

= 252 W

for soda lime glass;

Qsl = (1.4 * 0.16 * 90) / 0.012

= 1680 W

for aerogel

Qaq = (0.014 * 0.16 * 90) / 0.012

= 16.8 W

Now for COST associated with heat lost

for polycarbonate;

cost = Qpc * 130 * 8 * 1/1000

= 252 * 130 * 8 * 1/1000

= $ 262.08

for soda lime glass;

cost = 1680 * 130 * 8 * 1/1000

= $ 1,747.2

for aerogel

cost = 16.8 * 130 * 8 * 1/1000

= $ 17.472

Therefore the cost associated with heat loss is maximum in Soda Lime and minimum in Aerogel

6 0

3 years ago