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Firdavs [7]
3 years ago
7

Identify the ways in which carbon is added to the atmosphere?

Physics
2 answers:
ZanzabumX [31]3 years ago
8 0

Answer:

Decay organisms, dead organisms and waste products, root respiration, animal respiration, plant respiration and factory emission.

Explanation:

Carbon cycle is one of the biogeochemical cycle in which the different forms of carbon circulates in the biosphere, atmosphere, lithosphere, hydrosphere and pedosphere. In the given situation, the the process of photosynthesis will take up the carbon dioxide from the atmosphere. But other processes like decay organisms, dead organisms and waste products, root respiration, animal respiration, plant respiration and factor emission will add carbon to the atmosphere.

The decay and dead organisms and their waste products: These dead organisms will be decayed by the soil microbes which will produce organic carbon matter in the soil. This organic carbon will be liberated into the atmosphere as carbon dioxide.

Root respiration: The roots are the underground parts of the plant which does not receive oxygen directly from air instead they receive oxygen which is present in the soil pores. The respiration process produces carbon dioxide as product which is released into the atmosphere.

Plant respiration: Plants have small pores called as stomata present on the surface of their leaves. The plants receives oxygen through these pores and expel out carbon dioxide into the atmosphere.

Animal respiration: The animals have prominent and complex respiratory system. The animals take oxygen from the air and expel out carbon dioxide.

Factory emission: Factories use fossil fuels for the generation of electricity. The burning of the fossil fuels liberate carbon gases such as carbon dioxide and carbon monoxide into the atmosphere.

sashaice [31]3 years ago
6 0
Animal respiration, dead organisms, decay organisms and factory emmissions<span />
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Multiple choice: Sunspots appear dark because
zepelin [54]

Answer: C. They are regions in which strong magnetic fields make it difficult for fresh supplies of hot, ionized gas to reach the photosphere

Explanation:

Sunspots appear dark because they are regions in which strong magnetic fields make it difficult for fresh supplies of hot, ionized gas to reach the photosphere.

Sunspots are typically cooler than their surroundings. For example, the temperature of a large sunspot can be about 4,000 Kelvin which is lower than the temperature of the photosphere around it which is about 5,800 Kelvin.

3 0
2 years ago
A weather emergency siren is mounted on a tower, 105 m above the ground. On one hand, it would be a good idea to make the siren
Svetradugi [14.3K]

Answer:

Explanation:

101 dB = 10.1 B.

Maximum intensity of sound allowed = 10.1 B

Intensity of sound in terms of W/m² can be found as follows

log (I / I₀) = 10.1

I / I₀ = 10¹⁰°¹

I = I₀ X 10¹⁰°¹

= 10⁻¹² X  10¹⁰°¹

= 10⁻¹°⁹ W/m²

105 m above the ground the this intensity will be 105² times

intensity at source point = 10⁻¹°⁹ x 105²

= 138.79 W/m²

energy of sound from source

= 4π times

= 4 x 3.14 x 138.79

= 1743.28W/m²

To calculate in terms of decibel :

log 1743.28 / 10⁻¹²

= log 1743.28 +12

= 15.24 B

= 152.4 dB .

152.4 dB .

4 0
3 years ago
A massless spring with force constant ????=200N/m hangs from the ceiling. A 2.0-kg block is attached to the free end of the spri
Makovka662 [10]

Answer:

-0.4454 Joules

Explanation:

m = Mass of block = 2 kg

h = Height of extension = 17 cm = x

g = Acceleration due to gravity = 9.81 m/s²

Potential energy of the spring

P=mgh\\\Rightarrow P=2\times 9.81\times 0.17\\\Rightarrow P=3.3354\ J

The kinetic energy of the spring

K=\frac{1}{2}mx^2\\\Rightarrow K=\frac{1}{2}\times 200\times 0.17^2\\\Rightarrow K=2.89\ J

In this system as the potential and kinetic energy is conserved from work energy equivalence we get

W=P-K\\\Rightarrow W=2.89-3.3354\\\Rightarrow W=-0.4454\ J

The work done by friction is -0.4454 Joules

8 0
3 years ago
Two long, parallel transmission lines, 40.0cm apart, carry 25.0-A and 73.0-A currents.A). Find all locations where the net magne
In-s [12.5K]

Answer:

a) If the currents are in the same direction, the magnetic field is zero at x = 0.298 m = 29.8 cm

That is, in between the wires, 29.8 cm from the 73.0 A wire and 10.2 cm from the 25.0 A wire.

b) If the currents are in opposite directions, the magnetic field is zero at x = 0.608 m = 60.8 cm

That is, along the positive x-axis, 60.8 cm from the 73.0 A wire and 20.8 cm from the 25.0 A wire.

Explanation:

The origin is at the 73.0 A wire and the 25.0 A wire is at x = 0.40 m

The magnetic field in a current carrying wire at a distance r from the wire is given by

B = (μ₀I/2πr)

μ₀ = magnetic constant = (4π × 10⁻⁷) H/m

a) If the currents are in the same direction, at what positions is the magnetic field equal to 0.

According to laws describing the direction.of magnetic fields, this position will be at some point between the two wires.

The magnetic field due to the 73.0 A wire points out of the book, at points along the positive x-axis while the magnetic field due to the 25.0 A wire points into the plane of the book, moving in the negative x-direction.

Hence,

For the 73.0 A wire, I₁ = 73.0 A, r₁ = x

For the 25.0 A wire, I₂ = 25.0 A, r₂ = (0.4 - x)

B = B₁ - B₂ = 0

(μ₀/2π) [(I₁/r₁) - (I₂/r₂)] = 0

(I₁/r₁) = (I₂/r₂)

(I₁/x) = [I₂/(0.4-x)]

(73/x) = [25/(0.4-x)]

73(0.4-x) = 25x

29.2 - 73x = 25x

73x + 25x = 29.2

98x = 29.2

x = (29.2/98) = 0.298 m

b) If the currents are in the opposite directions, at what positions is the magnetic field equal to 0?

According to laws describing the direction.of magnetic fields, this position will be at some point beyond the second wire (since we're initially concerned about the positive x-direction).

The magnetic field due to the 73.0 A wire points out of the book, at points along the positive x-axis while the magnetic field due to the 25.0 A wire (whose direction is now in the opposite direction to the current in the first wire) is also along the positive x-direction.

Hence,

For the 73.0 A wire, I₁ = 73.0 A, r₁ = x

For the 25.0 A wire, I₂ = 25.0 A, r₂ = (x - 0.4)

B = B₁ - B₂ = 0

(μ₀/2π) [(I₁/r₁) - (I₂/r₂)] = 0

(I₁/r₁) = (I₂/r₂)

(I₁/x) = [I₂/(x-0.4)]

(73/x) = [25/(x-0.4)]

73(x-0.4) = 25x

73x - 29.2 = 25x

73x - 25x = 29.2

48x = 29.2

x = (29.2/48) = 0.608 m

Hope this Helps!!!

5 0
3 years ago
(please help ASAP!)
makkiz [27]

Answer:

It's C. Length

Explanation:

Just think of a ruler. Hope I helped! :)

8 0
3 years ago
Read 2 more answers
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