All categories

9 days ago

In what processes do proteins in the electron transport chain participate?

Physics

1 answer:

kicyunya [3.2K]9 days ago

8 0

Answer:

1. Reactions involving oxidation and reduction along with proton pumping

2. Reactions involving phosphorylation and proton pumping

Explanation:

During oxidative phosphorylation, there is a transfer of electrons from donors to acceptors, which constitutes a redox reaction.

These redox reactions liberate energy that is utilized to produce ATP. In eukaryotic cells, these reactions are performed by protein complexes found in the mitochondria, while in prokaryotic cells, the proteins are positioned in the intermembrane space of the cells. These interconnected protein complexes are referred to as electron transport chains.

You might be interested in

Several charges in the neighborhood of point P produce an electric potential of 6.0 kV (relative to zero at infinity) and an ele

ValentinkaMS [3465]

Answer:

0.018 J

Explanation:

The work required to bring the charge from infinity to the point P is equal to the change in its electric potential energy. This can be expressed as

$W = q \Delta V$

where

$q=3.0 \mu C = 3.0 \cdot 10^{-6} C$ represents the charge's magnitude

and $\Delta V = 6.0 kV = 6000 V$ signifies the potential difference between point P and infinity.

After substituting into the formula, we arrive at

$W=(3.0\cdot 10^{-6}C)(6000 V)=0.018 J$

4 0

1 month ago

A person is rowing across the river with a velocity of 4.5 km/hr northward. The river is flowing eastward at 3.5 km/hr (Figure 4

Yuliya22 [3333]

Answer: Her velocity magnitude (v) relative to the shore is 5.70 km/h.

Explanation:

Let Q be the speed of the boat, and P be the speed of the river flow.

R represents the resultant velocity combining boat velocity and river current.

According to vector addition using the law of triangles:

$R=\sqrt{P^2+Q^2+2PQCos\theta}$

From the diagram:

P = 3.5 km/h, Q = 4.5 km/h

$\theta= 90^o$

$R=\sqrt{P^2+Q^2+2PQCos\theta}=\sqrt{(3.5)^2+(4.5)^2+3.5\times 4.5\times cos90^o}=5.70$

$(Cos90^o=0),(sin 90^o=1)$

$\alpha =tan^{-1}\frac{Qsin\theta}{P+Qcos\theta}=tan^{-1}\frac{4.5 sin 90^o}{3.5+4.5 cos90^o}=tan^{-1}\frac{4.5}{3.5}=52.12^o$

Therefore, her velocity magnitude relative to the shore is 5.70 km/h.

8 0

2 months ago

Which model is used to describe the interaction of external forces that affect an organization's strategy and ability to compete

ValentinkaMS [3465]

Answer:

Competitive forces model

Explanation:

The Competitive forces model is a crucial instrument in strategic analysis aiming to assess an organization’s competitiveness. Commonly referred to as the "Five Force Model of Porter", this framework includes five key factors: the intensity of rivalry among existing competitors, the negotiating power of buyers, the threat posed by potential new entrants, the bargaining strength of suppliers, and the risk of substitute products or services.

These elements significantly influence an organization's competitive strategy and its likelihood of success.

5 0

1 month ago

During the construction of an office building, a hammer is accidentally dropped from a height of 784 ft. the distance (in feet)

Softa [3030]

T= 24.5 feet per second. That’s the speed it attains just before hitting the ground.

7 0

11 days ago

Read 2 more answers

A 55-kg pilot flies a jet trainer in a half vertical loop of 1200-m radius so that the speed of the trainer decreases at a const

Keith_Richards [3271]

a) $-1.54 m/s^2$

b) 803.4 N

Explanation:

a) At point C (top of the loop), the pilot experiences weightlessness, leading to the normal force from the seat being zero:

N = 0

Consequently, the force balance equation at position C becomes:

$mg=m\frac{v^2}{r}$ where the left term signifies the pilot's weight and the right term represents the centripetal force, with:

= acceleration due to gravity

= jet's velocity at the top $g=9.8 m/s^2$

= loop radius

$v$ By solving for v,

$r=1200 m$

Thus, this is the jet's speed at C.

The speed at position A (bottom) can be derived from $v_C=\sqrt{gr}=\sqrt{(9.8)(1200)}=108.4 m/s$

The distance traveled by the jet corresponds to half the circumference of the circle with radius r, therefore

$v_A=550 km/h =152.8 m/s$

Given the plane's deceleration is consistent, we can obtain it using the following equation:

$s=\pi r=\pi(1200)=3770 m$

b) The pilot experiences a force equal to the normal force from the seat. At point B (half-way through the loop), we find:

$v_C^2-v_A^2=2as\\a=\frac{v_C^2-v_A^2}{2s}=\frac{108.4^2-152.8^2}{2(3770)}=-1.54 m/s^2$ - The normal force from the seat, N, directed towards the center of the loop

- As there are no further forces acting toward the central axis, N must equal the centripetal force:

(1)

where

represents the speed at position B.

To deduce the velocity at B, we note that the distance covered by the jet between positions A and B is a quarter of a circle:

With knowledge of the deceleration, we can implement the equation of motion to find the velocity at the midway point B: $N=m\frac{v_B^2}{r}$

$v_B$

Thus, we then apply eq.(1) to determine the normal force acting on the pilot at B:

$s=\frac{\pi r}{2}=\frac{\pi(1200)}{2}=1885 m$

6 0

28 days ago