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3 years ago

The overall function of the calvin cycle is __________.

Physics

1 answer:

AlexFokin [52]3 years ago

6 0

Answer:

the primary function of the Calvin cycle is to make organic products plants need, using the products from the light reactions of photosynthesis (ATP and NADPH).

Explanation:

hope this helped

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Two conditions must<br> be met for work to be done, what are they?

Margarita [4]

The two conditions are:

1) Application of-force on the body.

2) Displacement of the body in the direction of force.

Hope this helps!

6 0

3 years ago

3. Which part of the back seat of a car—the left side, the right side, or the

Nina [5.8K]

Answer:

The sides

Explanation:

Because there's a seat in front of the child to avoid him/her from flying

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4 years ago

Why does the sky appears blue?

SashulF [63]

Explanation:

Sky appears blue because of Scattering.Blue light is scattered in all directions by the tiny molecules of air in Earth's atmosphere. Blue is scattered more than other colors because it travels as shorter, smaller waves. This is why we see a blue sky most of the time.

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4 years ago

A woman fires a rifle with barrel length of 0.5400 m. Let (0, 0) be where the 125 g bullet begins to move, and the bullet travel

dybincka [34]

Answer:

Explanation:

Given that,

Length of barrel =0.54m

Mass of bullet=125g=0.125kg

Force extend

F=16,000+10,000x-26,000x²

a. Work done is given as

W= ∫Fdx

W= ∫(16,000+10,000x-26,000x² dx from x=0 to x=0.54m

W=16,000x+10,000x²/2 -26,000x³/3 from x=0 to x=0.54m

W=16,000x+5,000x²- 8666.667x³ from x=0 to x=0.54m

W= 16,000(0.54) + 5000(0.54²) - 8666.667(0.54³) +0+0-0

W=8640+1458-1364.69

W=8733.31J

The workdone by the gas on the bullet is 8733.31J

b. Work done is given as

Work done when the length=1.05m

W= ∫Fdx

W= ∫(16,000+10,000x-26,000x² dx from x=0 to x=1.05m

W=16,000x+10,000x²/2 -26,000x³/3 from x=0 to x=1.05m

W=16,000x+5,000x²- 8666.667x³ from x=0 to x=1.05mm

W= 16,000(1.05) + 5000(1.05²) - 8666.667(1.05³) +0+0-0

W=16800+5512.5-10032.75

W=12,279.75J

The workdone by the gas on the bullet is 12,279.75J

3 0

4 years ago

A car covers 72 kilometers in the first hour of its journey. In the next hour, it covers 90 kilometers. What is the amount of wo

Nina [5.8K]

The work done is $2.8125 \times 10^{5} \mathrm{J}$

Work Done = Change in Kinetic Energy (ΔKE)

<u>Explanation</u>

In first 1 hour it travels 72 km

So, Velocity = $\frac{\text { distance }}{\text { time }}=\frac{72}{1} k m / h=72 \mathrm{km} / \mathrm{h}=\frac{72000}{3600} \mathrm{m} / \mathrm{s}=20 \mathrm{m} / \mathrm{s}$

or, Initial Velocity (u) = 20 m/s

Similarly for the next hour it covers 90 km

So, Velocity = $\frac{\text { distance }}{\text { time }}=\frac{90}{1} k m / h=90 \mathrm{km} / \mathrm{h}=\frac{90000}{3600} \mathrm{m} / \mathrm{s}=25 \mathrm{m} / \mathrm{s}$

or, Final Velocity (v) = 20 m/s

Work done = Change in Kinetic Energy (ΔKE)

Work done = ΔKE = $\frac{1}{2} m v^{2}-\frac{1}{2} m u^{2}$

ΔKE = $\frac{1}{2} m\left(v^{2}-u^{2}\right)=\frac{1}{2} \times\left(2.5 \times 10^{3}\right) \times\left(25^{2}-20^{2}\right)$

= $\frac{2500 \times(625-400)}{2}=\frac{2500 \times 225}{2}=\frac{562500}{2}$ = 281250 joule

= $2.8125 \times 10^{5} \mathrm{J}$

4 0

3 years ago