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Nesterboy [21]
3 years ago
11

How do high-energy electrons from glycolysis and the krebs cycle contribute to the formation of atp from adp in the electron tra

nsport chain?
Physics
1 answer:
ivanzaharov [21]3 years ago
5 0

Answer

Together with glycolysis, The Krebs cycle, and the electron transport chain release about 36 molecules of ATP per molecule of glucose.The Krebs cycle uses the two molecules of pyruvic acid formed in glycolysis and yields high-energy molecules of NADH and flavin adenine dinucleotide (FADH2), as well as some ATP. The electron transport chain forms a proton gradient across the inner mitochondrial membrane, which drives the synthesis of ATP

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Diferencia entre transferencia y transformación de energía. Da ejemplos de cada caso.
Firdavs [7]

Answer:

La transformación de energía es un proceso en el que la energía se intercambia entre un sistema y el medio ambiente en al menos dos formas de energía diferentes entre sí. Por ejemplo, un panel solar convierte la energía lumínica en energía eléctrica.

En cambio, en la transferencia de energía, esta no cambia su forma sino que es transmitida de un cuerpo a otro. El ejemplo más claro es el de la fogata, que transmite calor al medio ambiente a través de radiación.

6 0
2 years ago
The tires on your truck have 0.35 m radius. In a straight line, you drive 2600 m. What is the angular displacement of the tire,
Travka [436]
1) In a circular motion, the angular displacement \theta is given by
\theta =  \frac{S}{r}
where S is the arc length and r is the radius. The problem says that the truck drove for 2600 m, so this corresponds to the total arc length covered by the tire: S=2600 m. Using the information about the radius, r=0.35 m, we find the total angular displacement:
\theta =  \frac{2600 m}{0.35 m} =7428 rad

2) If we put larger tires, with radius r=0.60 m, the angular displacement will be smaller. We can see this by using the same formula. In fact, this time we have:
\theta =  \frac{2600 m}{0.60 m}=4333 rad
8 0
3 years ago
8.) If a car moving at 50km/h skids 15m with locked brakes, how far does the same car moving at 100km/h
pantera1 [17]

(8) A car starting with a speed <em>v</em> skids to a stop over a distance <em>d</em>, which means the brakes apply an acceleration <em>a</em> such that

0² - <em>v</em>² = 2 <em>a</em> <em>d</em> → <em>a</em> = - <em>v</em>² / (2<em>d</em>)

Then the car comes to rest over a distance of

<em>d</em> = - <em>v</em>² / (2<em>a</em>)

Doubling the starting speed gives

- (2<em>v</em>)² / (2<em>a</em>) = - 4<em>v</em>² / (2<em>a</em>) = 4<em>d</em>

so the distance traveled is quadrupled, and it would move a distance of 4 • 15 m = 60 m.

Alternatively, you can explicitly solve for the acceleration, then for the distance:

A car starting at 50 km/h ≈ 13.9 m/s skids to a stop in 15 m, so locked brakes apply an acceleration <em>a</em> such that

0² - (13.9 m/s)² = 2 <em>a</em> (15 m) → <em>a</em> ≈ -6.43 m/s²

So the same car starting at 100 km/h ≈ 27.8 m/s skids to stop over a distance <em>d</em> such that

0² - (27.8 m/s)² = 2 (-6.43 m/s²) <em>d</em> → <em>d</em> ≈ 60 m

(9) Pushing the lever down 1.2 m with a force of 50 N amounts to doing (1.2 m) (50 N) = 60 J of work. So the load on the other end receives 60 J of potential energy. If the acceleration due to gravity is taken to be approximately 10 m/s², then the load has a mass <em>m</em> such that

60 J = <em>m g h</em>

where <em>g</em> = 10 m/s² and <em>h</em> is the height it is lifted, 1.2 m. Solving for <em>m</em> gives

<em>m</em> = (60 J) / ((10 m/s²) (1.2 m)) = 5 kg

(10) Is this also multiple choice? I'm not completely sure, but something about the weight of the tractor seems excessive. It would help to see what the options might be.

4 0
3 years ago
Which of these is an uneven rhythm movement? Question 5 options:
Musya8 [376]

Answer:

Sliding would be an uneven rhythm because Galloping and skipping has a constant flow of the same movement and same noise.

Explanation:

8 0
2 years ago
Read 2 more answers
A car travels due east with a speed of 52.0 km/h. Raindrops are falling at a constant speed vertically with respect to the Earth
serg [7]

Answer:

a) v = 6.43 m/s

b) v = 15.8 m/s

Explanation:

Speed of car = 56 km/h

56 km/h = 14.4 m/s

Angle rain makes on the glass to the vertical = 66°

Thus knowing that the opposite side of the angle is the distance moved by the car, and the adjacent side is the distance traveled by the rain in the same time

both of which are directly proportional to their velocities

Then

tan(66°) = 14.44m/s ÷ x

or x = 14.44/tan(66°)

Which is the vertical raindrop velocity of the relative to earth

v = 6.43 m/s vertically towards earth

For v relative to the car is we have vector sum of both velocities

v = √(14.44^2 + 6.43^2) = 15.8 m/s which is the velocity relative to car

= 15.8 m/s

6 0
3 years ago
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