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3 years ago

C'est quoi le lien entre l'atmosphere et les plantes?

Physics

1 answer:

Rus_ich [418]3 years ago

3 0

Answer:

Les plantes produisent de l'oxygène et ont contribué à faire de la Terre une planète habitable. Grâce au processus de photosynthèse pendant la journée, les plantes absorbent le dioxyde de carbone de l'air, le convertissent en sucre et libèrent de l'oxygène dans l'atmosphère.

Les plantes consomment du dioxyde de carbone - un gaz à effet de serre important - au cours du processus de photosynthèse. La réduction du dioxyde de carbone dans l'atmosphère a un effet de refroidissement indirect. Les plantes refroidissent également l'atmosphère car elles libèrent de la vapeur d'eau lorsqu'elles deviennent chaudes, un processus similaire à la transpiration.

la température, l'humidité et l'intensité lumineuse autour de la plante; la concentration de dioxyde de carbone dans l'air autour des feuilles. La relation est inverse; autrement dit, à mesure que la concentration de CO2 augmente, le nombre de stomates produits diminue, et vice versa.

Explanation:

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Q5:

kap26 [50]

S = 10 + (-4) = 6

Displacement is vector quantity

5 0

3 years ago

Linear Velocity What is the linear velocity in cm>min for any edge point of a 12-cm-diameter CD (compact disc) spinning at 10

daser333 [38]

3,89,988 cm/min is the linear velocity

Given,

Diameter of CD = 12 cm

So, Radius of CD = 6 cm

CD is spinning at 10350 rev/min

Firstly , convert rev/min into rad/min

1 rev = 2π radians

10350 rev/min = 10350 × 2π

= 64998 rad/min

Formula used,

$v=rw$ where,

$v$ is the Linear velocity

$r$ is the radius

$w$ is the angular velocity

$v$ = 6 cm × 64998rad/min

= 3,89,988 cm/min

Thus, linear velocity for any edge point of a 12-cm-diameter CD (compact disc) spinning at 10,350 rev/min is 389988 cm/min.

Learn more about Angular speed here brainly.com/question/540174

#SPJ4

8 0

2 years ago

Use the worked example above to help you solve this problem. An airboat with mass 4.30 102 kg, including the passenger, has an e

umka21 [38]

Answer:

a) a = 1,865 m / s² and b) t = 8.1 s

Explanation:

a) Let's use Newton's second law to find acceleration, we can work the equation in scalar form because displacement and force have the same direction

F = m .a

a = F / m

a = 8.02 10² /4.3 10²

a = 1,865 m / s²

b) We use kinematic relationships in one dimension

vf = vo + at

vf = 0 + a t

t = vf / a

t = 15.1 / 1.865

t = 8.1 s

3 0

3 years ago

Identify the wavelength of this wave.

Zolol [24]

Answer:

Explanation:

The line(A) goes throughout the entire picture. So therefore choice A would be it's length.

3 0

3 years ago

A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o

blondinia [14]

Answer:

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

Explanation:

At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:

$\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}}$ (1)

Where:

$T$ - Period of rotation of the satellite, measured in seconds.

$r$ - Distance of the satellite with respect to the center of the planet, measured in meters.

$G$ - Gravitational constant, measured in newton-square meters per square kilogram.

$M$ - Mass of the Earth, measured in kilograms.

Now we clear the distance of the satellite with respect to the center of the planet:

$r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}$

$r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} }$ (2)

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ and $T = 25800\,s$ , then the distance of the satellite is:

$r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }$

$r \approx 18.897\times 10^{6}\,m$

The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:

$F = \frac{G\cdot m\cdot M}{r^{2}}$ (3)

Where:

$m$ - Mass of the satellite, measured in kilograms.

$F$ - Force exerted on the satellite by the Earth, measured in newtons.

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ , $m = 6105\,kg$ and $r \approx 18.897\times 10^{6}\,m$ , then the gravitational force is:

$F = \frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6105\,kg)\cdot (6\times 10^{24}\,kg)}{(18.897\times 10^{6}\,m)^{2}}$

$F = 6841.905\,N$

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

4 0

3 years ago