The natural process by which the ozone layer get depleted are sun spots and atmospheric winds. The volcanic eruptions are also responsible for the depletion of the ozone layer.
The thinning of the ozone layer, which is present in the upper atmosphere is called ozone layer depletion. These natural processes causes and some chemicals compounds releases chlorine and bromine, which in exposure of high ultraviolet light causes the depletion of ozone.
The ozone layer found in the stratosphere of the earth atmosphere. This region protects the earth from harmful ultraviolet radiation of the sun. It has potential to absorb around 99% of the harmful ultraviolet radiation.
The main causes of ozone depletion are chloroflorocarbons, carbon tetrachloride, methylbromide. The natural processes does not cause more that 1 to 2 % of the ozone depletion.
To learn more about stratosphere here
brainly.com/question/13497783
#SPJ4
Answer: If there is competition for food and nesting areas.
Explanation: Barred owls and spotted owls are well known as rivals. Barred owls are known to be aggressive towards spotted owls and are both bigger and more aggressive than them.
Answer:
There are many points at which eukaryotic gene expression can be controlled, through pretranscriptional control, transcriptional control, and posttranscriptional control
Explanation:
The pretranscriptional control determines the accessibility of chromatin to the transcription machinery. It is affected by supercoiling and methylation. It is also known as epigenetic regulation, and it does not depend on the sequence but on the conformation of the DNA.
While transcriptional control determines the frequency and / or speed of transcription initiation through the accessibility of the start sites, the availability of transcription factors and the effectiveness of promoters.
The post-transcriptional control is the one that is exercised once the transcript has finished synthesizing. It can be of several types:
• Maturation control: As the RNA adjustment can be made.
• Transport control: Most RNA has to go out to the cytoplasm to perform its function. For this they have to cross the pores of the nuclear membrane, where you can select the RNAs that will be transported and those that will not.
• Stability control: The half-life of RNA can be regulated by the expression of RNAs or mRNA stabilizing proteins in the cytoplasm.
• Translational control: It is exercised on the frequency with which the mRNAs begin to be translated. It can also affect the frequency with which proteins mature and the availability of enzymatic effectors.
Answer:
0.153
Explanation:
We know the up-thrust on the fish, U = weight of water displaced = weight of fish + weight of air in air sacs.
So ρVg = ρ'V'g + ρ'V"g where ρ = density of water = 1 g/cm³, V = volume of water displaced, g = acceleration due to gravity, ρ'= density of fish = 1.18 g/cm³, V' = initial volume of fish, ρ"= density of air = 0.0012 g/cm³ and V" = volume of expanded air sac.
ρVg = ρ'V'g + ρ"V"g
ρV = ρ'V'g + ρ"V"
Its new body volume = volume of water displaced, V = V' + V"
ρ(V' + V") = ρ'V' + ρ"V"
ρV' + ρV" = ρ'V' + ρ"V"
ρV' - ρ"V' = ρ'V" - ρV"
(ρ - ρ")V' = (ρ' - ρ)V"
V'/V" = (ρ - ρ")/(ρ' - ρ)
= (1 g/cm³ - 0.0012 g/cm³)/(1.18 g/cm³ - 1 g/cm³)
= (0.9988 g/cm³ ÷ 0.18 g/cm³)
V'/V" = 5.55
Since V = V' + V"
V' = V - V"
(V - V")/V" = 5.55
V/V" - V"/V" = 5.55
V/V" - 1 = 5.55
V/V" = 5.55 + 1
V/V" = 6.55
V"/V = 1/6.55
V"/V = 0.153
So, the fish must inflate its air sacs to 0.153 of its expanded body volume
