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3 years ago

Although the reactions of the calvin cycle do not depend directly on light. True or False

Physics

1 answer:

Harlamova29_29 [7]3 years ago

4 0

Answer:

True

Explanation:

The complete question is:

"Although the reactions of the Calvin cycle do not depend directly on light, they do not usually occur at night. True o False"

The Calvin cycle is also known as the Calvin-Benson cycle or as the CO₂ fixation phase in the photosynthesis process.

The Calvin cycle generates the reactions necessary to fix the carbon in a solid structure for the formation of glucose and, in turn, regenerates the molecules for the continuation of the cycle.

The Calvin cycle is known as the dark phase of photosynthesis, or the carbon fixation phase. It is called the dark phase because this cycle is not dependent on light like other parts that make up the photosynthesis process. But it uses the energy that is produced in the light phase of photosynthesis to fix carbon.

It can be said that it consists of or forms the second stage of photosynthesis, in which the carbon of the carbon dioxide that is absorbed is fixed.

So, the statement is true because the Calvin cycle uses the energy that is produced in the light phase of photosynthesis to fix carbon.

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Identify which one is chemical change, chemical property,physical change and physical property

Ulleksa [173]

Answer:

Heat conductivity: Physical property

Silver tarnishing: chemical change

Sublimation: physical change

Magnetizing steel: physical property

Length of a metal object: physical property

Shortening melting: physical property

exploding dynamite: chemical change

combustible: chemical property

water freezing: physical change

Acid resistance: chemical property

brittleness: physical property

milk souring: chemical change

baking bread: chemical change

Explanation:

First you need to understand the differences between physical and chemical change; physical and chemical property.

Physical properties vs chemical properties:

Physical properties are properties that you can observe without changing the substance's composition. This means you can measure these properties without changing them chemically.

Chemical properties, on the other hand, are properties are not as direct. These properties are generally determined by the way the react with other substances changing their composition.

Physical changes vs chemical changes:

A substance that undergoes physical change does not change in chemical composition. They may look physically different in terms of size and shape, but overall, their chemical composition remains constant. The best example would be water. Water can change phases, from solid to liquid when they melt. Essentially, they look like different substances, but the change is only physical and not chemical.

Chemical change, is different by the fact that they change in chemical composition. Bonds are broken and/or made through the reaction, which changes them not only physically but chemically as well. Some of the most indicative signs of a chemical change occurring are: change in color, odor, production of gas, production of light/heat.

4 0

3 years ago

When a voltage difference is applied to a piece of metal wire, a 5.0 mA current flows through it. If this metal wire is now repl

zheka24 [161]

Answer:

I = 21.13 mA ≈ 21 mA

Explanation:

I₁ = 5 mA

L₁ = L₂ = L

V₁ = V₂ = V

ρ₁ = 1.68*10⁻⁸ Ohm-m

ρ₂ = 1.59*10⁻⁸ Ohm-m

D₁ = D

D₂ = 2D

S₁ = 0.25*π*D²

S₂ = 0.25*π*(2*D)² = π*D²

If we apply the equation

R = ρ*L / S

where (using Ohm's Law):

R = V / I

we have

V / I = ρ*L / S

If V and L are the same

V / L = ρ*I / S

then

(V / L)₁ = (V / L)₂ ⇒ ρ₁*I₁ / S₁ = ρ₂*I₂ / S₂

S₁ = 0.25*π*D² and

S₂ = 0.25*π*(2*D)² = π*D²

we have

ρ₁*I₁ / (0.25*π*D²) = ρ₂*I₂ / (π*D²)

⇒ I₂ = 4*ρ₁*I₁ / ρ₂

⇒ I₂ = 4*1.68*10⁻⁸ Ohm-m*5 mA / 1.59*10⁻⁸ Ohm-m

⇒ I₂ = 21.13 mA

5 0

3 years ago

According to Newton’s First Law of Motion, what will happen to a baseball thrown in outer space

Illusion [34]

Answer:

The baseball will stay in motion until another force act upon it.

Explanation:

3 0

3 years ago

Read 2 more answers

A player at first base catches a throw traveling 38 m/s. The baseball, which has a mass of 0.145 kg, comes to a complete stop in

mixas84 [53]

Answer:

F = 39.36 N

Explanation:

given,

initial speed, u = 38 m/s

final speed, v = 0 m/s

mass of ball = 0.145 Kg

time, t = 0.14 s

Force = ?

using impulse formula

J = change in momentum

J = F x t

m(v - u) = F x t

0.145 x (0 - (-38)) = F x 0.14

F x 0.14 = 5.51

F = 39.36 N

force exerted by the ball is equal to 39.36 N.

5 0

3 years ago

Two 22.7 kg ice sleds initially at rest, are placed a short distance apart, one directly behind the other, as shown in Fig. 1. A

boyakko [2]

Newton's third law of motion sates that force is directly proportional to the rate of change of momentum produced

(a) The final speeds of the ice sleds is approximately 0.49 m/s each

(b) The impulse on the cat is 11.0715 kg·m/s

The reason for arriving at the above values is as follows:

The given parameters are;

The masses of the two ice sleds, m₁ = m₂ = 22.7 kg

The initial speed of the ice, v₁ = v₂ = 0

The mass of the cat, m₃ = 3.63 kg

The initial speed of the cat, v₃ = 0

The horizontal speed of the cat, v₃ = 3.05 m/s

(a) The required parameter:

The final speed of the two sleds

For the first jump to the right, we have;

By the law of conservation of momentum

Initial momentum = Final momentum

∴ m₁ × v₁ + m₃ × v₃ = m₁ × v₁' + m₃ × v₃'

Where;

v₁' = The final velocity of the ice sled on the left

v₃' = The final velocity of the cat

Plugging in the values gives;

22.7 kg × 0 + 3.63 × 0 = 22.7 × v₁' + 3.63 × 3.05

∴ 22.7 × v₁' = -3.63 × 3.05

v₁' = -3.63 × 3.05/22.7 ≈ -0.49

The final velocity of the ice sled on the left, v₁' ≈ -0.49 m/s (opposite to the direction to the motion of the cat)

The final speed ≈ 0.49 m/s

For the second jump to the left, we have;

By conservation of momentum law, m₂ × v₂ + m₃ × v₃ = m₂ × v₂' + m₃ × v₃'

Where;

v₂' = The final velocity of the ice sled on the right

v₃' = The final velocity of the cat

Plugging in the values gives;

22.7 kg × 0 + 3.63 × 0 = 22.7 × v₂' + 3.63 × 3.05

∴ 22.7 × v₂' = -3.63 × 3.05

v₂' = -3.63 × 3.05/22.7 ≈ -0.49

The final velocity of the ice sled on the right = -0.49 m/s (opposite to the direction to the motion of the cat)

The final speed ≈ 0.49 m/s

(b) The required parameter;

The impulse of the force

The impulse on the cat = Mass of the cat × Change in velocity

The change in velocity, Δv = Initial velocity - Final velocity

Where;

The initial velocity = The velocity of the cat before it lands = 3.05 m/s

The final velocity = The velocity of the cat after coming to rest =

∴ Δv = 3.05 m/s - 0 = 3.05 m/s

The impulse on the cat = 3.63 kg × 3.05 m/s = 11.0715 kg·m/s

(c) The required information

The average velocity

Impulse = $F_{average}$ × Δt

Where;

Δt = The time of collision = The time it takes the cat to finish landing = 12 ms

12 ms = 12/1000 s = 0.012 s

We get;

$F_{average} = \mathbf{\dfrac{Impulse}{\Delta \ t}}$

∴ $F_{average} = \dfrac{11.0715 \ kg \cdot m/s}{0.012 \ s} = 922.625 \ kg\cdot m/s^2 = 922.625 \ N$

The average force on the right sled applied by the cat while landing, $\mathbf{F_{average}}$ = 922.625 N

Learn more about conservation of momentum here:

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brainly.com/question/22257327

8 0

2 years ago