Geography High School

Why
does grass turn yellow in very hot weather. Especially the color
why is grass green and dry grass yellow

Answers

Answer 1

When grass is dead or exposed to the light, chlorophyll, the pigment that gives plants their green color, breaks down, causing the grass to become yellow or brown.

The ability of plants to absorb light energy and utilise it to make food through photosynthesis is due to the presence of chlorophyll. Chlorophyll degrades when a plant dies or is exposed to intense sunlight for an extended period of time, making other pigments like carotenoids and anthocyanins more apparent. The plant appears yellow or brown as a result of these pigments.

Grasses have developed very good drought tolerance. They stop photosynthesis, store nutrients in their roots, and patiently wait for the rains to come again. When there is sufficient water, they swiftly shoot up green blades and continue.

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Related Questions

Given an annual primary productivity of 300 g C m-2
yr-1 Calculate the annual maximum yield of herring
using an ecological efficiency of 1%
(you do not have any unit conversions in this problem)

Answers

The annual maximum yield of herring can be calculated using the annual primary productivity and an ecological efficiency of 1%.

To calculate this number, we need to take the annual primary productivity of 300 g C m-2yr-1 and multiply it by the ecological efficiency of 1%. This will give the result of 3 g C m-2yr-1, which is the annual maximum yield of herring. In other words, the annual maximum yield of herring is equal to 3 g C m-2yr-1, when using an ecological efficiency of 1% and an annual primary productivity of 300 g C m-2yr-1.

This result demonstrates the importance of using an ecological efficiency when calculating the annual maximum yield of herring based on an annual primary productivity. Without the ecological efficiency, our results would be off by at least a factor of 100. As such, it is important to understand how to use ecological efficiency when calculating annual maximum yields of species such as herring.

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the oldowan tool industry involved removing ________ from ________.

Answers

The Oldowan tool industry involved removing flakes from corn.

The name Oldowan comes from the Tanzanian archaeological site of Olduvai Gorge, where archaeologist Louis Leakey originally found the first Oldowan stone tools in the 1930s.

In prehistoric times, the Oldowan was a significant stone tool business. These primitive tools were straightforward, often created from one or a few flakes that were scraped off with another stone. Oldowan tools are often shaped by hitting a roughly spherical hammerstone on the edge, or hitting platform, of a competent core rock to create a conchoidal crack with sharp edges used for many uses. The procedure is frequently referred to as lithic reduction. The flake is the piece of the chip that was broken off by the blow.

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Why water saturation of shale is 0 related to interconnected
pores, effective porosity and isolated pores

Answers

When water saturation of shale is 0, it is related to the presence of interconnected pores, effective porosity, and isolated pores. These factors contribute to the inability of water to saturate the shale.

Water saturation refers to the proportion of pore space within a rock or sediment that is filled with water. In the context of shale, which is a type of fine-grained sedimentary rock, the water saturation is 0 when there are no water-filled pores present. This can occur due to the presence of interconnected pores, effective porosity, and isolated pores.

Interconnected pores in shale allow for the movement and flow of fluids, including water. When these pores are absent or limited, water cannot penetrate the shale effectively, resulting in a water saturation of 0.

Shale typically has low effective porosity, which refers to the volume of interconnected pores that contribute to fluid flow. Effective porosity is an important factor in determining the ability of water to saturate a rock.

Isolated pores in shale refer to pore spaces that are not interconnected and are difficult for water to access. These isolated pores can contribute to a lower overall water saturation since water cannot easily fill these isolated spaces.

Therefore, when water saturation of shale is 0, it indicates a lack of interconnected pores, limited effective porosity, and the presence of isolated pores, which collectively prevent water from saturating the shale.

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Europa is one of the moons of the planet Jupiter. This moon is believed to have an ocean of liquid water below it's icy surface, making it one of the places in the Solar System most like Earth's oceans in terms of habitable conditions. This makes Europa a very important place to study. Europa has a radius of 1561 km. Europa has a mass of 4.800×10^22 kg. a) Determine the acceleration due to gravity at the surface of Europa. b) Determine the escape velocity of Europa. c) What would be the orbital period of a satellite sent to study Europa if it were to orbit a distance of 45.0 km above the surface of that moon?

Answers

To determine the acceleration due to gravity at the surface of Europa, we can use the formula for gravitational acceleration.

acceleration due to gravity (g) = GM/r²

where G is the gravitational constant (approximately 6.67430 × 10^-11 m³/kg/s²), M is the mass of Europa, and r is the radius of Europa.

Given:

Radius of Europa (r) = 1561 km = 1,561,000 meters

Mass of Europa (M) = 4.800 × 10^22 kg

Substituting the values into the formula, we get:

g = (6.67430 × 10^-11 m³/kg/s²) × (4.800 × 10^22 kg) / (1561,000 meters)²

Calculating the value, we find:

g ≈ 1.315 m/s²

Therefore, the acceleration due to gravity at the surface of Europa is approximately 1.315 m/s².

b) The escape velocity of Europa can be calculated using the formula:

escape velocity = √(2GM/r)

Using the same values for G and M as in part a, and the radius of Europa (r), we can calculate the escape velocity:

escape velocity = √(2 × 6.67430 × 10^-11 m³/kg/s² × 4.800 × 10^22 kg / 1,561,000 meters)

Calculating the value, we find:

escape velocity ≈ 2,025 m/s

Therefore, the escape velocity of Europa is approximately 2,025 m/s.

c) The orbital period of a satellite around Europa can be determined using Kepler's third law:

orbital period (T) = 2π√(r³/GM)

Given:

Distance above Europa's surface (r) = 45.0 km = 45,000 meters

Gravitational constant (G) = 6.67430 × 10^-11 m³/kg/s²

Mass of Europa (M) = 4.800 × 10^22 kg

Substituting the values into the formula, we get:

T = 2π√((45,000 + 1,561,000 meters)³ / (6.67430 × 10^-11 m³/kg/s² × 4.800 × 10^22 kg))

Calculating the value, we find:

T ≈ 38,164 seconds

Therefore, the orbital period of a satellite sent to study Europa, orbiting a distance of 45.0 km above its surface, is approximately 38,164 seconds.

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