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seraphim [82]
3 years ago
10

Speculate on how higher sea-surface temperatures (sst) might impact earth's greenhouse effect.

Biology
1 answer:
Aleks [24]3 years ago
6 0

The SST or sea surface temperature refers to the temperature of the water close to the surface of the ocean. Sea surface temperature influences the behavior of the atmosphere of the Earth, so their initialization into the atmospheric model is essential. While SST plays a vital role for tropical cyclogenesis, it is also essential in predicting the formation of sea breezes and sea fog.  

The changes in SST reflects the overall trend of warming of the climatic system, and in turn, affects the climate and weather patterns globally. If the SST is too warm, the humidity upsurges and the greenhouse effect enhances.  


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please help me name 5 each of producers, primary consumers (which eat only plants), and top predators (which eat only meat, or m
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Should glucose erythrocytes leukocytes and protein appear in the urine
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3 0
3 years ago
In a species of bird, the allele for the striped trait is recessive to the allele for the speckled trait. Suppose that a coastal
mylen [45]

Answer:

The answer is "0.42"

Explanation:

Please find the complete question in the attached file.

Coastal community striped allele intensity = q = 0.62

Landlocked community, strip intensity allele=q'=0.40

Its eliminated allele frequency for movement = q"=0.42 \ in \ indoorscommunity And now in the sturdy vineyard optimum communities, genotype frequency of stripped characteristic (heterozygous device) in coasts=q_2=0.39

Heritable (striped) allele frecency:

\to q=\sqrt{q_2}

      =\sqrt{0.39} \\\\ =0.624 \\\\ =0.62

(round up to the closest cent) is therefore the only frequency of coastal people.

The genome of pulling function in host population= q'_2=0.16 Heterozygous feature regularly.

Therefore the inland community rate of recessed (striped), allele rate:

\to q'= \sqrt{q'_{2}}= \sqrt{0.16}=0.40

Following migration;

Its percentage of coastal migrants:

m=10\% \\\\

   = \frac{10}{100} \\\\= 0.1

Coastal population Non-immigrant percentage of coastal residents:

=1- m\\\\=1-0.1\\\\ =0.9

Stripping coastal community of allele rate after immigration:

(q")=\text{(stripping of migrant only freq)}\times (m)+\text{(stripping allele baud rate for non-immigrants)} \times ( 1-m)

       = [q \times m]+[[q' \times(1-m)]\\\\=(0.62 \times 0.1)+(0.40 \times 0.9)\\\\=0.062+ 0.36\\\\=0.422\\\\=0.42

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Answer:it c or b

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