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2 years ago

What is a positive effect of increased carbon dioxide within the carbon cycle?

Physics

1 answer:

MaRussiya [10]2 years ago

5 0

Answer:

The positive impact in the cycle is that CO2 will make plants to use water more efficiently and also thereby make them to become drought resistant and thereby grow faster.

Explanation:

Carbon cycle is the process by which carbon dioxide travels from the atmosphere to the earth and returns back to the atmosphere.

The negative impact of carbon dioxide in the carbon cycle is that it will cause green house effect. However, when it comes to the positive impact in the cycle, CO2 will make plants to use water more efficiently and also thereby make them to become drought resistant and thereby grow faster.

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The space station is 4.41 x 10^5 kg and orbits the earth 6.78 x 10^6 m from the center of earth. The mass of earth is 5.97 x 10^

allochka39001 [22]

Answer:

3 820 885 N

Explanation:

Gravitational equation

F = G m1 m2 / r^2

G = gravitational constant = 6.6713 x 10^-11 m^3/kg-s^2

F = 6.6713 x 10^-11 * 4.41 x 10^5 * 5.97 x 10^24 / ( 6.78x 10^6)^2

= 3820885 .3 N

6 0

2 years ago

The magnetic north pole changes location but the geographic north pole does not true or false

a_sh-v [17]

The North Magnetic Pole is the point on the surface of Earth's Northern Hemisphere at which the planet's magnetic field points vertically downwards (in other words, if a magnetic compass needle is allowed to rotate about a horizontal axis, it will point straight down). There is only one location where this occurs, near (but distinct from) the Geographic North Pole and the Geomagnetic North Pole. So yes true

3 0

3 years ago

A 1-megabit computer memory chip contains many 27 fF capacitors. Each capacitor has a plate area of 3.09 × 10−11 m 2 . Determine

valentina_108 [34]

Answer:

Plate separation of each capacitor is 101.132 °A

Explanation:

The formula to calculate the capacitance in empty space as a function of distance (square parallel plates) is:

$C=\epsilon_{0}\frac{Area}{distance}$

clearing for distance:

$distance=\epsilon_0 \frac{Area}{Capacitance} \\\\\epsilon_0=8.8542(10)^{-12}C^2/Nm\\\\Area=3.09(10)^{-11}m^2\\Capacitance=27(10)^{-15}F\\\\Replacing\\\\distance=\frac{8.8542(10)^{-12}*3.09(10)^{-11}}{27(10)^{-15}} =1.0133(10)^{-8}m\\\\In A\\distance= 101.132 A$

8 0

3 years ago

Peg P is driven by the forked link OA along the path described by r = eu, where r is in meters. When u = p4 rad, the link has an

8_murik_8 [283]

Answer:

The transverse component of acceleration is 26.32 $m/s^2$ where as radial the component of acceleration is 8.77 $m/s^2$

Explanation:

As per the given data

u=π/4 rad

ω=u'=2 rad/s

α=u''=4 rad/s

$r=e^u$

So the transverse component of acceleration are given as

$a_{\theta}=(ru''+2r'u')\\$

Here

$r=e^u\\r=e^{\pi/4}\\r=2.1932 m$

$r'=e^u.u'\\r'=2.1932 \times 2\\r'=4.3864 m$

$a_{\theta}=(ru''+2r'u')\\a_{\theta}=(2.1932\times 4+2\times 4.3864 \times 2)\\a_{\theta}=26.32 m/s\\$

The transverse component of acceleration is 26.32 $m/s^2$

The radial component is given as

$a_r=r''-r\theta'^2$

Here

$r''=e^u.u'^2+e^u u''\\r''=2.1932 \times (2)^2+2.1932\times 4\\r''=17.5456 m$

$a_r=r''-ru'^2\\a_r=17.5456-2.1932\times (2)^2\\a_r=8.7728 m/s^2$

The radial component of acceleration is 8.77 $m/s^2$

6 0

3 years ago

A 26.0 kg child plays on a swing having support ropes that are 2.40 m long. A friend pulls her back until the ropes are 45.0 ∘ f

Sloan [31]

A)Ep'=mgh=mgl(1-cosa).At the bottom of the swing Ep=0(reference level),so the potential energy as the child is just released is bigger than the potential energy at the bottom of the swing.;B)The speed of the child at the bottom of the swing-->v=√(2gh)=√[2gl(1-cosa)];C)I don't think that the tension does any work.

8 0

3 years ago