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True [87]
1 year ago
10

The light reactions could be viewed as analogous to a hydro-electric dam. In that case, the wall of the dam that holds back the

water would be analogous to?.
Physics
1 answer:
Viktor [21]1 year ago
3 0

The light reactions could be viewed as analogous to a hydroelectric dam. In that case, the wall of the dam that holds back the water would be analogous to the thylakoid membrane.

Thylakoid membrane is present in cyanobacteria and chloroplasts of plants. It plays a crucial role in photosynthesis and photosystem II reactions.

In general, these are the regions where light-dependent reactions take place. The thylakoid membrane is a lipid-bound membrane that maintains potential difference and also controls the flow of liquids across the membrane during light reactions.

In the provided case, we can see that the wall of the dam holds back the water, similarly, in light-dependent reactions, thylakoid membranes control the liquid flow and also regulate the potential gradient across the membrane and also allow the selective proteins to pass through.

If you need to learn more about light reactions click here:

brainly.com/question/26623807

#SPJ4

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A rotating paddle wheel is inserted in a closed pot of water. The stirring action of the paddle wheel heats the water. During th
fgiga [73]

Answer:

the final energy of the system is 35.5 kJ.

Explanation:

Given;

initial energy of the system, E₁ = 10 kJ

heat transferred to the system, q₁  30 kJ

Heat lost to the surrounding, q₂ = 5kJ

heat gained by the system, Q = q₁ - q₂ = 30 kJ - 5kJ = 25 kJ

work done on the system, W = 500 J = 0.5 kJ

Apply first law of thermodynamic,

ΔU = Q + W

where;

ΔU  is change in internal energy

Q is the heat gained by the system

W is work done on the system

ΔU = 25kJ + 0.5 kJ

ΔU = 25.5 kJ

The final energy of the system is calculated as;

E₂ = E₁ + ΔU

E₂ = 10 kJ + 25.5 kJ

E₂ =  35.5 kJ.

Therefore, the final energy of the system is 35.5 kJ.

3 0
2 years ago
A piece of aluminum has a volume of 1.50 10-3 m3. the coefficient of volume expansion for aluminum is β = 69 ✕ 10-6 (°c)-1. the
Alex17521 [72]

Answer:

W = 3.12 J

Explanation:

Given the volume is 1.50*10^-3  m^3 and the coefficient of volume for aluminum is β = 69*10^-6 (°C)^-1. The temperature rises from 22°C to 320°C. The difference in temperature is 320 - 22 = 298°C, so ΔT = 298°C. To reiterate our known values we have:

β = 69*10^-6 (°C)^-1       V = 1.50*10^-3  m^3       ΔT = 298°C

So we can plug into the thermal expansion equation to find ΔV which is how much the volume expanded (I'll use d instead of Δ because of format):

dV = \beta V_{0} dT\\dV = (69*10^{-6})( C)^{-1} * (1.50*10^{-3})m^{3} * (298)C\\dV = 3.0843*10^-5

So ΔV = 3.0843*10^-5 m^3

Now we have ΔV, next we have to solve for the work done by thermal expansion. The air pressure is 1.01 * 10^5 Pa

To get work, multiply the air pressure and the volume change.

W = P * dV = (1.01 * 10^5)Pa * (3.0843*10^{-5})m^3\\W = 3.115143J

W = 3.12 J

Hope this helps!

4 0
3 years ago
In a double slit experiment, 450 nm light passes through two slits producing an interference pattern where the first bright frin
musickatia [10]

Answer:

1.3 x 10⁻⁴ m

Explanation:

\lambda = wavelength of the light = 450 nm = 450 x 10⁻⁹ m

n = order of the bright fringe = 1

θ = angle = 0.2°

d = separation between the slits

For bright fringe, Using the equation

d Sinθ = n \lambda

Inserting the values

d Sin0.2° = (1) (450 x 10⁻⁹)

d (0.003491) = (450 x 10⁻⁹)

d = 1.3 x 10⁻⁴ m

6 0
3 years ago
A car starts from rest and goes down a slope with a constant
GREYUIT [131]

Answer:

25m/s

Explanation:

here we use 1st equation of motion

then we find this ans

3 0
3 years ago
you travel 4.0km east, 4.0km north, then 5.0km at 53.1 degrees north of west in a total of 5 hours. What is the magnitude and di
Sergio039 [100]

Answer:

Explanation:

We shall convert all the displacement in vector form .

i and j represents east and north respectively .

D₁ = 4 i

D₂ = 4 j

D₃ = - 5 cos 53.1 i + 5 sin 53.1 j

= -3i + 4 j

Total displacement = D₁ + D₂ + D₃

= 4i + 4 j - 3i + 4 j

= i + 8j

magnitude of displacement = √( 1² + 8² )

= 8.06 km

velocity = 8.06 / 5

= 1.61 km / h

Direction from x axis in anticlockwise direction .

Tanθ =  8 / 1 = 8

θ = 83° north of east .

7 0
3 years ago
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