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siniylev [52]
3 years ago
5

Which of the following is NOT true about drawings and specifications?

Engineering
1 answer:
hoa [83]3 years ago
3 0

Answer:

sorry but we can't tell the answer because your not showing the drawings and specifications.

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Select the correct answer.
boyakko [2]
I think the answer is b
6 0
3 years ago
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Visual perception is a mental process that is non selective<br> True<br> False
NARA [144]

The statement "Visual perception is a mental process that is non selective" is false, it is a psychic function that allows the organism to capture, elaborate and interpret selective information that comes from the environment.

<h2>What is visual perception?</h2>

Visual perception is that inner sensation of apparent knowledge, resulting from a specific stimulus or light impression recorded by the eyes.

<h3>Characteristics of visual perception</h3>

  • It incorporates the sensory stimuli received from objects, situations or events and converts them into a meaningful interpretation experience.

  • It is an active process of the brain through which an external reality is created by transforming the light information captured by the eye.

Therefore, we can conclude that visual perception is the interpretation made by the brain of the different organisms of the stimuli received through the senses.

Learn more about visual perception here: brainly.com/question/10259599

3 0
2 years ago
Radioactive wastes generating heat at a rate of 3 x 106 W/m3 are contained in a spherical shell of inner radius 0.25 m and outsi
MariettaO [177]

Answer:

Inner surface temperature= 783K.

Outer surface temperature= 873K

Explanation:

Parameters:

Heat,e= 3×10^6 W/m^3

Inner radius = 0.25 m

Outside radius= 0.30 m

Temperature at infinity, T(¶)= 10°c = 273. + 10 = 283K.

Convection coefficient,h = 500 W/m^2 . K

Temperature of the surface= T(s) = ?

Temperature of the inner= T(I) =?

STEP 1: Calculate for heat flux at the outer sphere.

q= r × e/3

This equation satisfy energy balance.

q= 1/3 ×3000000(W/m^3) × 0.30 m

= 3× 10^5 W/m^2.

STEP 2: calculus the temperature for the surface.

T(s) = T(¶) + q/h

T(s) = 283 + 300000( W/m^2)/500(W/m^2.K)

T(s) = 283+600

T(s)= 873K.

TEMPERATURE FOR THE OUTER SURFACE is 873 kelvin.

The same TWO STEPS are use for the calculation of inner temperature, T(I).

STEP 1: calculate for the heat flux.

q= r × e/3

q= 1/3 × 3000000(W/m^3) × 0.25 m

q= 250,000 W/m^2

STEP 2:

calculate the inner temperature

T(I) = T(¶) + q/h

T(I) = 283K + 250,000(W/m^2)/500(W/m^2)

T(I) = 283K + 500

T(I) = 783K

INNER TEMPERATURE IS 783 KELVIN

5 0
3 years ago
50POINTS
maxonik [38]

Answer:

Ensure that all material and energy inputs and outputs are as inherently safe and benign as possible. Minimize the depletion of natural resources. Prevent waste. Develop and apply engineering solutions while being cognizant of local geography, aspirations, and cultures.Green engineering is the design, commercialization, and use of processes and products that minimize pollution, promote sustainability, and protect human health without sacrificing economic viability and efficiency.The goal of environmental engineering is to ensure that societal development and the use of water, land and air resources are sustainable. This goal is achieved by managing these resources so that environmental pollution and degradation is minimized.

Explanation:i helped

7 0
3 years ago
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A storage tank, used in a fermentation process, is to be rotationally molded from polyethylene plastic. This tank will have a co
NNADVOKAT [17]

Answer:

The volume up to cylindrical portion is approx  32355 liters.

Explanation:

The tank is shown in the attached figure below

The volume of the whole tank is is sum of the following volumes

1) Hemisphere top

Volume of hemispherical top of radius 'r' is

V_{hem}=\frac{2}{3}\pi r^3

2) Cylindrical Middle section

Volume of cylindrical middle portion of radius 'r' and height 'h'

V_{cyl}=\pi r^2\cdot h

3) Conical bottom

Volume of conical bottom of radius'r' and angle \theta is

V_{cone}=\frac{1}{3}\pi r^3\times \frac{1}{tan(\frac{\theta }{2})}

Applying the given values we obtain the volume of the container up to cylinder is

V=\pi 1.5^2\times 4.0+\frac{1}{3}\times \frac{\pi 1.5^{3}}{tan30}=32.355m^{3}

Hence the capacity in liters is V=32.355\times 1000=32355Liters

3 0
3 years ago
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