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3 years ago

11

What happens to the free energy released as electrons are passed from photosystem II to photosystem I through a series of electr

on carriers?

1 answer:

lana [24]3 years ago

6 0

<span>It is used to establish and maintain a proton gradient.</span>

You might be interested in

How high is the highest waterfall

LenKa [72]

Angel Falls is 3,230 feet, which is the tallest waterfall in the world.
Hope this helps!
Happy studying,
~Mistermistyeyed.

5 0

3 years ago

A cubical box measuring 1.29 m on each side contains a monatomic ideal gas at a pressure of 2.0 atm How much thermal energy do t

Marrrta [24]

Answer:

a) $U = 652.545\,kJ$ , b) $v \approx 659.568\,\frac{m}{s}$

Explanation:

a) According to the First Law of Thermodinamics, the system is not reporting any work, mass or heat interactions. Besides, let consider that such box is rigid and, therefore, heat contained inside is the consequence of internal energy.

$Q = U$

The internal energy for a monoatomic ideal gas is:

$U = \frac{3}{2} \cdot n \cdot R_{u} \cdot T$

Let assume that cubical box contains just one kilomole of monoatomic gas. Then, the temperature is determined from the Equation of State for Ideal Gases:

$T = \frac{P\cdot V}{n\cdot R_{u}}$

$T = \frac{(202.65\,kPa)\cdot(1.29\,m)^{3}}{(1\,kmole)\cdot(8.314\,\frac{kPa\cdot m^{3}}{kmole\cdot K} )}$

$T = 52.325\,K$

The thermal energy contained by the gas is:

$U = \frac{3}{2}\cdot (1\,kmole)\cdot (8.314\,\frac{kPa\cdot m^{3}}{kmole\cdot K})\cdot (52.325\,K)$

$U = 652.545\,kJ$

b) The physical model for the cat is constructed from Work-Energy Theorem:

$U = \frac{1}{2}\cdot m_{cat} \cdot v^{2}$

The speed of the cat is obtained by isolating the respective variable and the replacement of every known variable by numerical values:

$v = \sqrt{\frac{2 \cdot U}{m_{cat}}}$

$v = \sqrt{\frac{2\cdot (652.545 \times 10^{3}\,J)}{3\,kg} }$

$v \approx 659.568\,\frac{m}{s}$

3 0

3 years ago

A large storage tank, open to the atmosphere at the top and filled with water, develops a small hole in its side at a point 15.6

VladimirAG [237]

To solve this problem it is necessary to take into account the kinematic equations of motion and the change that exists in the volume flow.

By definition the change in speed is given by

$v_f^2-v_i^2 = 2ax$

Where,

x= distance

$v_f =$ final velocity

$v_i =$ initial velocity

a = acceleration

On the other hand we know that the flow of a fluid is given by

$\dot{V} = Av$

Where,

A = Area

v = Velocity

PART A )

Applying this equation to the previously given values we have to

$v_f^2-v_i^2 = 2ax$

$v_f^2-0 = 2*(9.8)(15.6)$

$v_f^2=305.76$

$v_f = 17.48$

Therefore the velocity of the water leaving the hole is 17.48m/s

PART B )

In the case of the hole we take the area of a circle, therefore replacing in the flow equation we have to,

$\dot{V} = \pi r^2 v$

$r = \sqrt{\frac{\dot{V}}{\pi v}}$

$r = \sqrt{\frac{3*10^{-3}*\frac{1}{60}}{\pi (17.48)}$

$r = \sqrt{9.10*10^{-7}}$

$r = 0.54*10^{-4}$

The diameter is 2 times the radius, then is $1.91*10^{-3}$ m or 1.91mm

<em>Note: The rate flow was converted from minutes to seconds.</em>

8 0

3 years ago

A 6.0 kg block initially at rest is pulled to the right along a horizontal, frictionless surface by a constant, horizontal force

weeeeeb [17]

Answer:

F= MASS *ACCELERATION

12=6*a

a=2 m/sec^2

v^2-u^2=2*a*s

v^2=2*2*3

v=sqrt(12)

v=3.5 m/sec

work done by force= change in kinetic energy

F*S=0.5*M*U^2-0.5*M*V^2

12*3=0 0.5*6*V^2

V=3.46 M/SEC

Explanation:

8 0

3 years ago

the video identifies the force pair produced when an apple falls through the air. which force belongs in a free-body diagram of

trapecia [35]

The free-body diagram of an apple falling through the air has weight of the apple pointing downwards and the air-resistance on the apple acting upwards.

When an object falls from up to the ground, the object falls under in the influence of acceleration due to gravity.

The vertical component of the force on the apple as it falls trough the air is given as;

∑Fy = 0

Fₙ - W = 0

Fₙ = W

where;

<em>Fₙ is the frictional force on the apple acting upwards</em>
<em>W is the weight of the apple acting downwards</em>

The free-body diagram of the apple is represented as follows;

↑ Fₙ

Ο

↓ W

Thus, the free-body diagram of an apple falling through the air has weight of the apple pointing downwards and the air-resistance on the apple acting upwards.

Learn more here:brainly.com/question/18770265

6 0

2 years ago