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Leno4ka [110]
3 years ago
11

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:

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

<em> </em>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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For a satellite to be in a circular orbit 950 km above the surface of the earth, what orbital speed must it be given?
ANTONII [103]

1.) 7382 m/s

The gravitational attraction between the satellite and the Earth provides the centripetal force that keeps the satellite in circular motion, so we can write

\frac{GMm}{(R+h)^2}=m\frac{v^2}{R+h}

where

G is the gravitational constant

M=5.98\cdot 10^{24}kg is the Earth's mass

m is the satellite's mass

R=6370 km = 6.37\cdot 10^6 m is the Earth's radius

h=950 km = 0.95\cdot 10^6 m is the altitude of the satellite above the Earth's surface

v is the orbital speed

Solving the formula for v, we find

v=\sqrt{\frac{GM}{R+h}}=\sqrt{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24} kg)}{(6.37\cdot 10^6 m +0.95\cdot 10^6 m)}}=7382 m/s

2) 1.73 hours

The period of the orbit is the time taken to complete one revolution around the Earth, therefore:

T=\frac{2\pi (R+h)}{v}

where the numerator is the circumference of the orbit and v the orbital speed, therefore we find

T=\frac{2\pi (6.37\cdot 10^6 m+0.95e6 m)}{7382 m/s}=6227 s

Converting into hours,

T=\frac{6227 s}{3600 s/h}=1.73 h

6 0
3 years ago
A biological community is made up of all the
vazorg [7]
It's 1. A biological community is made up of all biotic species, or populations in an area. Hope this helps! Brainliest please?
8 0
3 years ago
The current in one wire is 5.00<br> a. what is the current in the other wire.
Gemiola [76]
The wire of the the other wire is to connect
5 0
3 years ago
Suppose you have three springs with force constants of k1 = k2 = k3 = 3.70 x 10^3 N/m. What is their effective force constant if
disa [49]

Answer:

The effective force constant is 1233.33 N/m.

Explanation:

It is given that,

Force constant 1, k_1=3.7\times 10^3\ N/m

Force constant 2, k_2=3.7\times 10^3\ N/m

Force constant 3, k_3=3.7\times 10^3\ N/m

The effective force constant if one is hung from the other in series is given by :

\dfrac{1}{K_{eff}}=\dfrac{1}{k_1}+\dfrac{1}{k_2}+\dfrac{1}{k_3}

\dfrac{1}{K_{eff}}=\dfrac{1}{3.7\times 10^3}+\dfrac{1}{3.7\times 10^3}+\dfrac{1}{3.7\times 10^3}

K_{eff}=1233.33\ N/m

So, the effective force constant is 1233.33 N/m. Hence, this is the required solution.

6 0
3 years ago
A car travels eastwards at 60km/h for 2h, then travels northwards at 20km/h for 8h. Find,
sveta [45]

Answer:

a) Average\ Speed=28\ km/h\\b) Average\ Velocity= 20\ km/h

Explanation:

We\ are\ given\ that,\\Velocity\ of\ the\ car\ Eastwards=60\ km/h\\Time\ taken\ by\ the\ car\ Eastwards=2\ h\\Velocity\ of\ the\ car\ Northwards=20\ km/h\\Time\ taken\ by\ the\ car\ Northwards=8\ h\\Hence,\\As\ we\ know\ that,\\Speed=\frac{Distance}{Time}\\Distance= Speed* Time\\Now,\ lets\ find\ the\ distance\ covered\ by\ the\ car\ in\ both\ the\ cases.

Hence,\\Distance\ Covered\ During\ its\ Eastward\ Journey=60*2=120\ km\\Distance\ Covered\ During\ its\ Northward\ Journey=20*8=160\ km\\Now,\\As\ we\ know\ that\ Average\ Speed=\frac{Total\ Distance}{Total\ Time} \\Here,\\Total\ distance\ of\ the\ car=Distance\ Covered\ During\ its\ Northward\ Journey+Distance\ Covered\ During\ its\ Eastward\ Journey\\Hence,\\Total\ distance\ of\ the\ car=120+160=280\ km\\Total\ time\ taken\ by\ the\ car=8+2=10\ hours\\Hence,\\Average\ Speed\ Of\ the\ Car\ throughout\ its\ journey=\frac{280}{10}=28\ km/h

Now,\\For\ Average\ Velocity\ we\ need\ to\ consider\ displacement\ as:\\Average\ Velocity\ =\frac{Total\ Displacement}{Total\ Time} \\Now,\\As\ we\ already\ know\ that\ displacement\ is\ the\ shortest\ distance\\ from\ the\ initial\ to\ the\ final\ point.\\We\ observe\ that, \\The\ car\ forms\ a\ right\ triangle\ during\ its\ complete\ journey.\\Hence,\\As\ we\ already\ know\ that,\\Distance\ travelled\ Eastwards= 120\ km\\Distance\ travelled\ Northwards= 160\ km\\Hence,\\We\ may\ apply\ Pythagoras\ Property\ to\ find\ the\ net\ displacement.\\Hence,\\a^2+b^2=c^2\\120^2+160^2=c^2\\14400+25600=c^2\\40000=c^2\\c=\sqrt{40000}\\c=200\\Hence,\\Total\ displacement=200\ km\\Total\ Time\ taken=2+8=10\ hours\\Hence,\\Average\ Velocity\ Of\ the\ Car=\frac{200}{10}=20\ km/h

7 0
3 years ago
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