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3 years ago

What is required for both the light-dependent and light-independent reactions to proceed?

1 answer:

3 0

ATP is required for both light-dependent and light-independent reactions.
ATP stands for adenosine triphosphate.
Hope this helps ;)

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Kendall has empty graduated cylinder with markings and an identical graduated cylinder partway filled with water. She also has a

Elis [28]

The answer is B.
She can measure the mass of the water, marble and the graduated cylinder with the balance.
The volume of the water can be shown on the marked graduated cylinder, the volume of the marble can be measured by the volume difference of the water before and after the marble is put in.

4 0

3 years ago

Read 2 more answers

Newtons third lawWhat action-reaction forces are involved when a rocket engine fires? Why doesnt a rocket need air to push on? A

dangina [55]

Answer: action forc roketorce

reaction force is engine fires

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3 years ago

An offshore oil well is 2 kilometers off the coast. The refinery is 4 kilometers down the coast. Laying pipe in the ocean is twi

shusha [124]

Answer:

Rectangular path

Solution:

As per the question:

Length, a = 4 km

Height, h = 2 km

In order to minimize the cost let us denote the side of the square bottom be 'a'

Thus the area of the bottom of the square, A = $a^{2}$

Let the height of the bin be 'h'

Therefore the total area, $A_{t} = 4ah$

The cost is:

C = 2sh

Volume of the box, V = $a^{2}h = 4^{2}\times 2 = 128$ (1)

Total cost, $C_{t} = 2a^{2} + 2ah$ (2)

From eqn (1):

$h = \frac{128}{a^{2}}$

Using the above value in eqn (1):

$C(a) = 2a^{2} + 2a\frac{128}{a^{2}} = 2a^{2} + \frac{256}{a}$

$C(a) = 2a^{2} + \frac{256}{a}$

Differentiating the above eqn w.r.t 'a':

$C'(a) = 4a - \frac{256}{a^{2}} = \frac{4a^{3} - 256}{a^{2}}$

For the required solution equating the above eqn to zero:

$\frac{4a^{3} - 256}{a^{2}} = 0$

a = 4

Also

$h = \frac{128}{4^{2}} = 8$

The path in order to minimize the cost must be a rectangle.

8 0

3 years ago

Your friends sit in a sled in the snow. If you apply a force pf 75 N to them, they have an acceleration of 0.9 m/s ^ 2. What is

Elza [17]

Answer:

Mass of the sled in the snow 83.33 kg.

Explanation:

Given that,

Force applied to move the sled in the snow (F) = 75N

$\text { Acceleration }(a)=0.9 \mathrm{m} / \mathrm{s}^{2}$

We know that

Newton's second law of motion is

$\text { Force }=\text { mass } \times \text { acceleration }$

F = ma (Or "force" is equal to "mass" times "acceleration".)

So if we move this around we can isolate mass and get mass

$\text { Mass }=\frac{\text { force }}{\text { accelearation }}$

$\mathrm{M}=\frac{75}{0.9}$

M = 83.33 kg

Mass of the sled in the snow 83.33 kg.

3 0

3 years ago

5.54 Two kilograms of air within a piston–cylinder assembly WP execute a Carnot power cycle with maximum and minimum temper atur

Step2247 [10]

Answer:

A) 60%

B) p2 = 1237.2 kPa

v2 = 0.348 m^3

C) w1-2 = w3-4 = 1615.5 kJ

Q2-3 = 60 kJ

Explanation:

A) calculate thermal efficiency

Л = 1 - $\frac{Tl}{Th}$

where Tl = 300 k

Th = 750 k

hence thermal efficiency ( Л ) = [1 - ( 300 / 750 )] * 100 = 60%

B) calculate the pressure and volume at the beginning of the isothermal expansion

calculate pressure ( P2 ) :

= P3v3 = mRT3 ----- (1)

v3 = 0.4m , mR = 2* 0.287, T3 = 750

hence P3 = 1076.25

next equation to determine P2

Qex = p3v3 ln( p2/p3 )

60 = 1076.25 * 0.4 ln(p2/p3)

hence ; P2 = 1237.2 kpa

calculate volume ( V2 )

p2v2 = p3v3

v2 = p3v3 / p2

= (1076.25 * 0.4 ) / 1237.2

= 0.348 m^3

C) calculate the work and heat transfer for each four processes

work :

W1-2 = mCv( T2 - T1 )

= 2*0.718 ( 750 - 300 ) = 1615.5 kJ

W3-4 = 1615.5 kJ

heat transfer

Q2-3 = W2-3 = 60KJ

Q3-4 = 0

D ) sketch of the cycle on p-V coordinates

attached below

6 0

3 years ago