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4 years ago

Which of the following sequences describes the path by which electrons travel downhill energetically in aerobic respiration

Physics

1 answer:

Vlad1618 [11]4 years ago

5 0

Answer:

glucose → NADH → electron transport chain → oxygen.

Explanation:

Aerobic respiration is also called as cellular respiration, in which electrons are picked up by the FADH, and NADH from the food. Then through a proton pump electrons are transferred to the electron transport chain, in this, the activity of the proton pump generates an electrochemical gradient, then this gradient used by the ATP synthase enzyme to produce ATP.

In aerobic respiration, the last acceptor of electron is oxygen. An electron is donated to oxygen for the formation of the water.

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A car from the beginning was traveling at 27.8m/s when he stepped on the brakes and stopped. According to the DMV, this would le

Yuliya22 [10]

Answer:

-6.44 m/s²

Explanation:

Given:

Δx = 60 m

v₀ = 27.8 m/s

v = 0 m/s

Find: a

v² = v₀² + 2aΔx

(0 m/s)² = (27.8 m/s)² + 2a (60 m)

a = -6.44 m/s²

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3 years ago

Wax, like all matter, comes in many phases. What are the three possible

Levart [38]

A b and c its so simple bro send it 23-89

7 0

3 years ago

2 Points

juin [17]

Answer: c

Explanation:

Analytical methods

3 0

3 years ago

Q 24, 25, 26 i dont get them and need answers for it

Alexandra [31]

Answer:

24) W = 75 [J]; 25) W = 1794[J]; 26) n = 8.8 (times) or 9 (times)

Explanation:

24) This problem can be solved by means of the following equation.

$DU = Q-W\\$

where:

DU = internal energy difference [J]

Q = Heat transfer [J]

W = work [J]

Since there are no temperature changes the internal energy change is equal to zero

DU = 0

therefore:

$Q = W\\$

The work is equal to the heat transfered, W = 75 [J].

25) The heat transfer can be calculated by means of the following equation.

$Q = m*c_{p}*DT\\where:\\m = mass = 0.4[kg]\\c_{p} = specific heat = 897[J/kg*K]\\DT= 5 [C]$

Q = 0.4*897*5 = 1794[J]

Work is equal to heat transfer, W = 1794[J]

26) Each time the bag falls the potential energy is transformed into heat energy, which is released into the environment. In this way the potential energy is equal to the developed heat.

$E_{p}=Q\\\\E_{p}=m*g*h$

where:

m = mass = 0.5[kg]

g = gravity = 9.81[m/s^2]

h = 1.5 [m]

$E_{p}=0.5*9.81*1.5\\E_{p}=7.36[J]$

The heat developed can be calculated by means of the following equation.

$Q=m*c_{p}*DT\\Q=0.5*130*1\\Q=65[J]$

The number of times will be calculated as follows

n = 65/7.36

n = 8.8 (times) or 9 (times)

7 0

4 years ago

There are 5510 lines per centimeter in a grating that is used with light whose wavelegth is 467 nm. A flat observation screen is

Mademuasel [1]

Answer:

1.696 nm

Explanation:

For a diffraction grating, dsinθ = mλ where d = number of lines per metre of grating = 5510 lines per cm = 551000 lines per metre and λ = wavelength of light = 467 nm = 467 × 10⁻⁹ m. For a principal maximum, m = 1. So,

dsinθ = mλ = (1)λ = λ

dsinθ = λ

sinθ = λ/d.

Also tanθ = w/D where w = distance of center of screen to principal maximum and D = distance of grating to screen = 1.03 m

From trig ratios 1 + cot²θ = cosec²θ

1 + (1/tan²θ) = 1/(sin²θ)

substituting the values of sinθ and tanθ we have

1 + (D/w)² = (d/λ)²

(D/w)² = (d/λ)² - 1

(w/D)² = 1/[(d/λ)² - 1]

(w/D) = 1/√[(d/λ)² - 1]

w = D/√[(d/λ)² - 1] = 1.03 m/√[(551000/467 × 10⁻⁹ )² - 1] = 1.03 m/√[(1179.87 × 10⁹ )² - 1] = 1.03 m/1179.87 × 10⁹ = 0.000848 × 10⁻⁹ = 0.848 × 10⁻¹² m = 0.848 nm.

w is also the distance from the center to the other principal maximum on the other side.

So for both principal maxima to be on the screen, its minimum width must be 2w = 2 × 0.848 nm = 1.696 nm

So, the minimum width of the screen must be 1.696 nm

4 0

3 years ago