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viva [34]
3 years ago
11

.Which equation represents the most accurate model of cellular respiration? Note: C6H12O6 is sugar.

Physics
1 answer:
Shkiper50 [21]3 years ago
4 0

The equation that most accurately represents the model of cellular respiration is: C6H12O6 (sugar) + 6O2 (oxygen) = 6CO2 (carbon dioxide) + 6H2O (water) + energy.

<h3>CELLULAR RESPIRATION:</h3>

Cellular respiration is the process whereby living organisms obtain energy by breaking down food molecules in their cells.

The process of cellular respiration breaks down sugar molecules (glucose) in the presence of oxygen to produce carbon dioxide and water as products, as well as energy in form of ATP.

Therefore, the equation that most accurately represents the model of cellular respiration is: C6H12O6 (sugar) + 6O2 (oxygen) = 6CO2 (carbon dioxide) + 6H2O (water) + energy.

Learn more about cellular respiration at: brainly.com/question/12671790?referrer=searchResults

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4 years ago
Four fixed point charges are at the corners of a square with sides of length L. Q1 is positive and at (OL) Q2 is positive and at
Ne4ueva [31]

Answer:

A) See Annex

B) Fq₁₂ = K *  Q₁*Q₂ /16 [N] (repulsion force)

C)  Fq₃₂  = K * Q₃*Q₂ /16 [N] (repulsion force)

D) Fq₄₂ = K * Q₄*Q₂ /32 [N] (attraction force)

E) Net force (its components)

Fnx = (2,59/64 )* K*Q²  [N] in direction of original Fq₃₂

Fny =(2,59/64 )* K*Q² [N] in direction of original Fq₁₂

Explanation:

For calculation of d (diagonal of the square, we apply Pythagoras Theorem)

d² = L² + L²    ⇒  d² = 2*L²     ⇒ d = √2*L²   ⇒ d= (√2 )*L

d = 4√2 units of length   (we will assume meters, to work with MKS system of units)

B) Force of Q₁ exerts on charge Q₂

Fq₁₂  = K * Q₁*Q₂ /(L)²     Fq₁₂ = K *  Q₁*Q₂ /16 (repulsion force in the direction indicated in annex)

C) Force of Q₃ exerts on charge Q₂

Fq₃₂  = K * Q₃*Q₂ /(L)²     Fq₃₂  = K * Q₃*Q₂ /16  (repulsion force in the direction indicated in annex)

D) Force of -Q₄ exerts on charge Q₂

Fq₄₂ = K * Q₄*Q₂ / (d)²      Fq₄₂ = K * Q₄*Q₂ /32 (Attraction force in the direction indicated in annex)

E) Net force in the case all charges have the same magnitude Q (keeping the negative sign in Q₄)

Let´s take the force that  Q₄ exerts on Q₂  and Q₂ = Q  ( magnitude) and

Q₄ = -Q

Then the force is:

F₄₂ = K * Q*Q / 32       F₄₂  = K* Q²/32  [N]

We should get its components

F₄₂(x) = [K*Q²/32 ]* √2/2   and so is F₄₂(y)  =  [K*Q²/32 ]* √2/2

Note that this components have opposite direction than forces  Fq₁₂  and

Fq₃₂  respectively, and that Fq₁₂ and Fq₃₂ are bigger than F₄₂(x) and  F₄₂(y) respectively

In new conditions

Fq₁₂ = K *  Q₁*Q₂ /16    becomes  Fq₁₂ = K * Q²/ 16 [N]   and

Fq₃₂ = K* Q₃*Q₂ /16      becomes   Fq₃₂ = K* Q² /16  [N]

Note that Fq₁₂ and Fq₃₂ are bigger than F₄₂(x) and  F₄₂(y) respectively

Then over x-axis we subtract Fq₃₂ - F₄₂(x)  = Fnx

and over y-axis, we subtract   Fq₁₂ - F₄₂(y) = Fny

And we get:

Fnx = K* Q² /16 - [K*Q²/32 ]* √2/2  ⇒  Fnx =  K*Q² [1/16 - √2/64]

Fnx = (2,59/64 )* K*Q²

Fny has the same magnitude  then

Fny =(2,59/64 )* K*Q²

The fact that Fq₁₂ and Fq₃₂ are bigger than F₄₂(x) and  F₄₂(y) respectively, means that Fnx and Fny remains as repulsion forces

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Explanation:

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