Answer:
negative, inhibition
Explanation:
The experiment in the illustration shows that early succession plants have a <u>negative </u>effect on the late succession plants, and the observation is most consistent with the <u>inhibition</u> model of succession.
<em>That the late succession plant thrives better in the absence of the early succession plants means that the early succession plant has been impacting the growth of the late succession plant negatively. This is consistent with the inhibition model of succession.</em>
There are 3 different models of succession. These include;
- Facilitation model in which colonists modify the environment to favor the growth of later successional species.
- Tolerance model in which early colonists' modification of the environment has no positive or negative impact on the growth of later successional species.
- Inhibition model in which early colonists modify the environment to inhibit the growth of later successional species.
I assume that in this item, we are asked to solve for the force exerted during the heartbeat. Force is the product of the mass and acceleration. To solve for the acceleration in this item, we divide the velocity by time.
a = 1 m/s / 0.2 s = 5 m/s²
Then, we multiply this by the mass (in kg)
F = (80 g / 1000 g/kg) x (5 m/s²)
= 0.4 kg m/s² = 0.4 J
Therefore, every hearbeat will take 0.4 J of force.
Answer:
The micrographs of cells shown in figure 8-3. what information about cells do these micrographs suggest is explained below in details.
Explanation:
Micrographs are the intensified images collected from small microscopes. Cell micrographs are often obtained from tissue specimens and show a constant mass of cells and inside compositions that are difficult to distinguish individually. It provides accurate images of the exteriors of cells and whole plants that are not imaginable by TEM. It can also be practiced for particle counting and size resolution, and method control.
The answer is 49.92%
Let's use the <span>Hardy-Weinberg principle:
p + q = 1
p</span>² + 2pq + q² = 1
<span>
where:
p - the frequency of dominant allele G
q - </span>the frequency of recessive allele g
p² - the frequency of homozygous dominant individuals GG with colour green
2pq - the frequency of heterozygous individuals Gg with colour green
p² - the frequency of homozygous recessive individuals gg with color brown
23% of the population is brown: p² = 23% = 0.23
p = √(p²) = √0.23 = 0.48
p = 0.48
p + q = 1
0.48 + q = 1
q = 1 - 0.48 = 0.52
<span>The percentage of the population that is expected to be heterozygous is 2pq:
2pq = 2 * p * q = 2 * 0.48 * 0.52 = 0.4992 = 49.92%</span>
