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3 years ago

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Which data set has the largest range? A. 55, 57, 59, 60, 61, 49, 48 B. 21, 25, 14, 16, 29, 22, 20 C. 12, 15, 16, 19, 18, 15, 27

D. 31, 36, 38, 41, 42, 54, 37

1 answer:

Simora [160]3 years ago

3 0

Data D has the largest range.

Data A: 61-48=13
Data B: 29-14=15
Data C:27-12=15
Data D:54-31=23

Therefore, Data D has the largest range.

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What are the products of linear electron flow during the light reactions of photosynthesis?

Katena32 [7]

Answer:

NADPH and ATP

Explanation:

In the clear stage the light that "hits" chlorophyll excites an electron to a higher energy level. In a series of reactions, energy is converted (throughout an electron transport process) into ATP and NADPH. Water breaks down in the process releasing oxygen as a secondary product of the reaction. ATP and NADPH are used to make the C-C bonds in the dark stage.

Photophosphorylation is the process of converting the energy of the electron excited by light into a pyrophosphate bond of an ADP molecule. This occurs when water electrons are excited by light in the presence of P680. The energy transfer is similar to the chemosmotic electron transport that occurs in the mitochondria.

Light energy causes the removal of an electron from a P680 molecule that is part of Photosystem II, the electron is transferred to an acceptor molecule (primary acceptor), and then passes downhill to Photosystem I through a conveyor chain of electrons The P680 requires an electron that is taken from the water by breaking it into H + ions and O-2 ions. These O-2 ions combine to form O2 that is released into the atmosphere.

The light acts on the P700 molecule of Photosystem I, causing an electron to be raised to a higher potential. This electron is accepted by a primary acceptor (different from the one associated with Photosystem II).

The electron goes through a series of redox reactions again, and finally combines with NADP + and H + to form NADPH, a carrier of H needed in the independent phase of light.

Electron of photosystem II replaces the excited electron of the P700 molecule.

There is therefore a continuous flow of electrons (non-cyclic) from water to NADPH, which is used for carbon fixation.

Cyclic electron flow occurs in some eukaryotes and in photosynthetic bacteria. NADPH does not occur, only ATP. This also occurs when the cell requires additional ATP, or when there is no NADP + to reduce it to NADPH.

In Photosystem II, the "pumping" of H ions into the thylakoids (from the stroma of the chloroplast) and the conversion of ADP + P to ATP is motorized by an electron gradient established in the thylakoid membrane.

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2 years ago

oee [108]

Answer: a switch can do A, B and E

Explanation:

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2 years ago

What happens when the moon faces one side of the earth?

algol [13]

When the moon faces earth a solar eclipse happens :-)

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2 years ago

Sound waves produced by a source pass a point five times every second. Which of the following choices correctly describes the pe

Ugo [173]

The period is 1/5 second, and the frequency is 5 Hz.

Explanation:

Let's start by reviewing some definitions about waves:

- The period of a wave is the time taken for the wave to make one complete cycle. It is indicated with T and it is measured in seconds (s)

- The frequency of a wave is the number of complete cycles made by the wave in one second. It is indicated with f and it is measured in Hertz (Hz). It is equal to the reciprocal of the period:

$f=\frac{1}{T}$

where f is the frequency and T is the period.

In this problem, we have a wave that passes a given point five time per second. This means that the number of oscillations per second is five, and so its frequency is:

$f=\frac{5 cycles}{1 sec}=5 Hz$

It follows than the period of the wave is:

$T=\frac{1}{f}=\frac{1}{5}s$

Therefore, the correct statement is

The period is 1/5 second, and the frequency is 5 Hz.

Learn more about waves, frequency and period:

brainly.com/question/5354733

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3 0

3 years ago

To visit your favorite ice cream shop, you must travel 490 m west on Main Street and then 920 m south on Division Street. Suppos

topjm [15]

Answer:

a) The magnitude of your average velocity during the 121 s is 8.61 m/s.

b) The direction of the average velocity is 61.9° south of west.

c) Your average speed during the trip is 11.7 m/s

Explanation:

Hi there!

a) The average velocity (a.v) is calculated as the displacement divided by the time it took to do such a displacement.

The displacement is calculated as the distance between the initial position and the final position:

Displacement = Δ(x,y) = final position - initial position

Let's consider that your initial position is the origin of our frame of reference and let's also consider that west and south are positive directions (+x and +y respectively). Then the displacement vector will be:

Δ(x,y) = final positon - initial position

Δ(x,y) = (490, 920) m - (0, 0) m = (490, 920) m

The average velocity will be:

a.v = Δ(x,y) / t

a.v = (490, 920) m / 121 s

a.v = (4.05, 7.60) m/s

The magnitude of the average velocity is calculated as follows:

The magnitude of your average velocity during the 121 s is 8.61 m/s.

b) To find the direction of the average velocity, we have to use trigonometric rules of right triangles. Notice that the x and y-components of the average velocity (vx and vy) together with the average velocity vector (v), with magnitude 8.61 m/s, form a triangle (see figure).

Also, notice that v is the hypotenuse of the triangle and that vx is the side adjacent to the angle θ while vy is the side opposite to θ.

Using trigonometry, we can calculate the value of the angle θ:

cos θ = adjacent side / hypotenuse

cos θ = vx / v

cos θ = 4.05 m/s / 8.61 m/s

θ = 61.9°

The direction of the average velocity is 61.9° south of west.

c) The average speed (a.s) is calculated as the traveled distance (d) divided by the time it took to cover that distance (t). In total, you traveled (490 m + 920 m) 1410 m in 121 s, then the average speed will be:

a.s = d/t

a.s = 1410 m / 121 s

a.s = 11.7 m/s

Your average speed during the trip is 11.7 m/s

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3 years ago