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2 years ago

Which stage of cellular respiration involves a gradient of hydrogen ions filtering through atp synthase?

1 answer:

3 0

Electron transport is a stage of cellular respiration that involves a gradient of hydrogen ions filtering through ATP synthase.

<h3>Electron transport:</h3>

Aerobic respiration ends with electron transport. Energy from NADH and FADH2 is transferred to ATP during this phase. Hydrogen ions are pumped through the mitochondrial inner membrane and into the intermembrane gap using energy during the electron transport process.

The oxidative phosphorylation process, also known as the electron transport chain, is a collection of four protein complexes that combine redox events to produce an electrochemical gradient that results in the production of ATP. Both photosynthesis and cellular respiration take place in mitochondria. Hydrogen ions are pumped through the mitochondrial inner membrane and into the intermembrane gap using energy during the electron transport process.

Learn more about electron transport here:

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2. What is the percent composition of sulfur in H2SO4?

allochka39001 [22]

Answer:

C. 32.7%

Explanation:

% composition = ( mass S / mass H2SO4 ) × 100 = 32.08/ 98.10 × 100 = 32.7 % pls mark brainliest

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3 years ago

Evelynn is measuring the pitch of a piano note. What unit of measurement is she most likely recording her value

sweet [91]

Answer:

hertz

Explanation: play piano

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2 years ago

Two hockey pucks, labeled A and B, are initially at rest on a smooth ice surface and are separated by a distance of 18.0 m . Sim

Nonamiya [84]

Answer:

The distance covered by puck A before collision is $z = 8.56 \ m$

Explanation:

From the question we are told that

The label on the two hockey pucks is A and B

The distance between the two hockey pucks is D 18.0 m

The speed of puck A is $v_A = 3.90 \ m/s$

The speed of puck B is $v_B = 4.30 \ m/s$

The distance covered by puck A is mathematically represented as

$z = v_A * t$

=> $t = \frac{z}{v_A}$

The distance covered by puck B is mathematically represented as

$18 - z = v_B * t$

=> $t = \frac{18 - z}{v_B}$

Since the time take before collision is the same

$\frac{18 - z}{V_B} = \frac{z}{v_A}$

substituting values

$\frac{18 -z }{4.3} = \frac{z}{3.90}$

=> $70.2 - 3.90 z = 4.3 z$

=> $z = 8.56 \ m$

8 0

3 years ago

Watching your fish swim in their tank, you notice that when one fish repeatedly jumps, it causes a standing wave given by the ti

ella [17]

Answer:

Check the attached image below.

Explanation:

Kindly check the attached image below to get the step by step explanation to the above question.

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3 years ago

A group of students are provided with three objects all of the same mass and radius. The objects include a solid cylinder, a thi

SOVA2 [1]

Answer:

Sphere, cylinder hoop

Explanation:

To analyze Which student is right it is best to propose the solution of the problem. Let's look for the speed of the center of mass. Let's use the concept of mechanical energy

In the highest part of the ramp

Em₀ = U = mg y

In the lowest part

Here the energy has part of translation and part of rotation

$E_{mf}$ = $K_{T}$ + $K_{R}$

$E_{mf}$ = ½ m $v_{cm}$ ² + ½ I w²

Where I is the moment of inertia of the body and w the angular velocity that relates to the velocity of the center of mass

$v_{cm}$ = w r

w = $v_{cm}$ / r

Let's replace

$E_{mf}$ = ½ I ( $v_{cm}$ / r)²

Energy is conserved

mg y = ½ m $v_{cm}$ ² + ½ I $v_{cm}$ ² / r2

½ (m + I / r²) $v_{cm}$ ² = m g y

½ (1 + I / m r²) $v_{cm}$ ² = g y

$v_{cm}$ = √ [2gy / (1 + I / mr²)]

This is the velocity of the center of mass of the bodies, as they all have the same radius with comparing this point is sufficient. Now let's use the speed definition

v = d / t

t = d / v

t = d / (√ [2gy / (1 + I / mr²)])

t = (d / √ 2gy) √(1 + I / m r²)

Therefore we see that time is proportional to the square root. All quantities are constant and the one that varies is the moment of inertia.

The moments of inertia of

Sphere is Is = 2/5 M r²

Cylinder Ic = ½ M r²

Hoop Ih = M r²

Let's replace each one and calculate the time

Sphere

ts = (d / √2gy) √ (1 + 2/5 Mr² / mr²)

ts = (d / √ 2gy) √ (1 +2/5) = (d / √ 2gy) √(1.4)

ts = (d / √ 2gy) 1.1

Cylinder

tc = (d / √2gy) √ (1 + 1/2 Mr² / Mr²)

tc = (d / √2gy) √ (1 + ½) = (d / √ 2gy) √ 1.5

tc = (d / √ 2gy) 1.2

Hoop

th = (d / √2gy) √ (1 + mr² / mr²)

th = (d / √2gy) √(1 + 1) = (d / √ 2gy) √ 2

th = (d / √ 2gy) 1.41

We have the results for the time the body that arrives the fastest is the sphere and the one that is the most hoop. Therefore the correct answer is

ts < tc < th

Sphere, cylinder hoop

5 0

3 years ago