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kumpel [21]
2 years ago
10

How can urbanization influence the composition of natural systems?

Biology
1 answer:
DerKrebs [107]2 years ago
4 0

Answer:

Urbanization also affects the broader regional environments. Regions downwind from large industrial complexes also see increases in the amount of precipitation, air pollution, and the number of days with thunderstorms. Urban areas affect not only the weather patterns, but also the runoff patterns for water.

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Why did the xyy chromosome theory fall out of favor as an explanation for criminal behavior?
exis [7]
XYY disorder is a hereditary condition in which a human male has an additional male (Y) chromosome, giving an aggregate of 47 chromosomes rather than the more regular 46. The "XYY" chromosome hypothesis drop out of support as a clarification for criminal conduct is on account of thought most lawbreakers don't have XYY and everybody with XYY is not a criminal
4 0
3 years ago
The elements in the boron family of the periodic table are known as:
Norma-Jean [14]
Alkaline earth metals
Group 13 aka the boron family is categorized as alkaline earth metals
4 0
2 years ago
Which processes are involved in the chloroplast of a plant cell?
Ahat [919]

Answer:

Chloroplast absorbs sunlight and it is used to make feed for the plant together with water and carbon dioxide gas. Chloroplasts are used to generate the free energy stored in ATP and NADPH via a photosynthesis process.

Explanation:

The site of photosynthesis action is chloroplast within a plant cell consisting of two chlorophyll molecules (PS1 and PS2), which have been embedded in the thylakoid membranes. The chloroplast consists of two chlorophyll molecules (photosynthetic pigments responsible for the green color of chloroplast). Each chlorophyll molecule absorbs light, caused to depart the chlorophyll molecules. This absorbs two electrons from each phenotype. PS2 electrons pass through the transportation chain for electron carriers, a series of redox reactions that release the energy used to synthesize ATP via Photophosphorylation/Chemiosmose (as the H+ ions diffuse through the stalked particles ATP, which changes the shape and catalysts, the electrochemical gradient diffuses down through the stalky particle ATP synthase).

Then these electrons replace the electrons lost in PS1. PS2 electron is replaced by photolysis electron, which when light strikes chloroplast, splitting the water into oxygen gas, H+ ions, and electron enzymes in the thylakoid space are catalyzed. The PS1 electrons combine to create NADPH with H+ ions and NADP (reduced NADP). These are the light-dependent photosynthetic reactions in chloroplasts. In the light-independent reactions, the NADPH and ATP are created. A pile of thylakoids is known as granum.

The light-independent processes take happen in the stroma. This is the site of carbon fixation; CO2 reacts with RUBP to generate GP (glycerate-3-phosphate) which is catalyzed by the enzyme RUBISCO (the most abundant enzyme in the world) (the most abundant enzyme in the world). The NADPH and ATP from the light-dependent processes convert GP to GALP (glyceraldehyde 3-phosphate). Two out of every 12 GALP molecules produced are used to synthesize glucose that can be employed either in breathing or in cellulose-forming condensation polymerization to add extra strength to the planted cell wall. The other GALP molecules are returned to RUBP.

7 0
3 years ago
Day 1: 4 km Day 2: 10 km Day 3: 12 km Day 4: 12 km Day 5: 18 km Day 6: 24 km
yKpoI14uk [10]

Answer:

In order to find average speed during each interval, we need to divide the distance during those intervals with the period of time. So, for the first interval (day 0 to day 2) hawksbill started from 0 and reached 10 kilometers by the end of the second day. That means that it crossed 10 kilometers in 2 days, so the average speed is 10/2 which is 5 km/day. Similarly, we can calculate speed for other intervals:

• day 2 - day 3: it went from 10 to 12 km in one day, which means it crossed 2 km in one day, so the average speed is 2/1 = 2 km/day

• day 3 - day 4: at the end of the third day it reached 12 km and at the end of the day 4 it remained at 12 km. That means the hawksbill wasn't moving in that interval so the speed was 0

• day 4 - day 5: it went from 12 km to 18 km, which means it crossed 18-12=6 km in one day, so the average speed is 6/1=6 km/day

• day 5 - day 6: it went from 18 to 24 km, which means it crossed 24-18=6 km in one day, so the speed was 6/1=6 km/day

So, to summarize, during the first interval turtle was moving with average speed of 5 km/day, then 2 km/day, in the third interval it wasn't moving and in the last two intervals, it moved in average speed of 6 km/day.

6 0
3 years ago
When was Barack Obama born
Lemur [1.5K]
August 4th, 1961.........................................
3 0
4 years ago
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