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Oksanka [162]
2 years ago
13

Ok bye guys going offline have a great daydefine metabolism​

Engineering
2 answers:
Rus_ich [418]2 years ago
6 0

Answer:

the chemical processes in plants or animals that change food into energy and help them grow

पेड़-पौधों और जंतुओं में होने वाली रासायनिक प्रक्रियाएँ जो भोजन को ऊर्जा में परिवर्तित कर देती हैं जिससे उनकी वृद्धि होती है; चयापचयन, उपाप्चय

Explanation:

plz mark me as brainiest

IgorLugansk [536]2 years ago
6 0

Metabolism (pronounced: meh-TAB-uh-liz-um) is the chemical reactions in the body's cells that change food into energy. Our bodies need this energy to do everything from moving to thinking to growing. Specific proteins in the body control the chemical reactions of metabolism.

Explanation:

<h2><em>hope</em><em> </em><em>it</em><em> </em><em>is</em><em> </em><em>helpful</em><em> </em><em>for</em><em> </em><em>you</em><em> </em></h2><h2><em>keep</em><em> </em><em>always</em><em> </em><em>smiling</em><em> </em></h2>
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In a short essay, discuss the question, "How are you an innovator?"
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Being innovative means doing things differently or doing things that have never been done before. An innovator is someone who has embraced this idea and creates environments in which employees are given the tools and resources to challenge the status quo, push boundaries and achieve growth.

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If you were driving the blue Prius in the situation pictured above, explain why the red Mustang should be given right-of-way at
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3 years ago
A sheet of steel 3-mm thick has nitrogen atomospheres on both sides at 900 C and is permitted to achieve a steady-state di usion
kati45 [8]

Answer:

X_B = 1.8 \times 10^{-3} m = 1.8 mm

Explanation:

Given data:

Diffusion constant for nitrogen is = 1.85\times 10^{-10} m^2/s

Diffusion flux = 1.0\times 10^{-7} kg/m^2-s

concentration of nitrogen at high presuure = 2 kg/m^3

location on which nitrogen  concentration is 0.5 kg/m^3   ......?

from fick's first law

J = D \frac{C_A C_B}{X_A X_B}

Take C_A as point  on which nitrogen concentration is 2 kg/m^3

x_B = X_A + D\frac{C_A -C_B}{J}

Assume X_A is zero at the surface

X_B = 0 + ( 12\times 10^{-11} ) \frac{2-0.5}{1\times 10^{-7}}

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4 0
3 years ago
A cylindrical specimen of steel has an original diameter of 12.8 mm. It is tested in tension its engineering fracture strength i
Mama L [17]

Answer:

a) The ductility = -30.12%

the negative sign means reduction

Therefore, there is 30.12% reduction

b) the true stress at fracture is 658.26 Mpa

Explanation:

Given that;

Original diameter d_{o} = 12.8 mm

Final diameter d_{f} = 10.7

Engineering stress  \alpha _{E} = 460 Mpa

a) determine The ductility in terms of percent reduction in area;

Ai = π/4(d_{o} )²  ; Ag = π/4(d_{f} )²

% = π/4 [ ( (d_{f} )² - (d_{o} )²) / ( π/4  (d_{o} )²) ]

= ( (d_{f} )² - (d_{o} )²) / (d_{o} )² × 100

we substitute

= [( (10.7)² - (12.8)²) / (12.8)² ] × 100

= [(114.49 - 163.84) / 163.84 ] × 100

= - 0.3012 × 100

= -30.12%

the negative sign means reduction

Therefore, there is 30.12% reduction

b) The true stress at fracture;

True stress  \alpha _{T} = \alpha _{E} ( 1 +  E_{E} )

E_{E}  is engineering strain

E_{E}  = dL / Lo

= (do² - df²) / df² = (12.8² - 10.7²) / 10.7² = (163.84 - 114.49) / 114.49

= 49.35 / 114.49  

E_{E} = 0.431

so we substitute the value of E_{E}  into our initial equation;

True stress  \alpha _{T} = 460 ( 1 +  0.431)

True stress  \alpha _{T} = 460 (1.431)

True stress  \alpha _{T} = 658.26 Mpa

Therefore, the true stress at fracture is 658.26 Mpa

6 0
2 years ago
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