Answer:
They provide energy, protection and insulation for the organs in the body. They're also very important in the cell membranes.
Explanation:
<span>Catalase catalyzes the decomposition of hydrogen peroxide to water and oxygen. It is a very </span>important<span> enzyme in protecting the </span>cell<span> from oxidative damage by reactive oxygen species. Hope this answers the question. Have a nice day.</span>
Answer:
Insufficient information to generate conclusion
Explanation:
It is really difficult to answer this question based on provided data. Rather speculations can be made. This is because of several reasons. The height of plant varies for different plant species. Likewise, it is really difficult for any plant species to the height of 5 inches (the lowest given option) within period of 8 hours only. It needs 8 hours light for several days to reach this much of height even for studied model plants such as arabidopsis. Thereon, each plant species would need different number of days.
On the other hand, it is important to consider that sunlight is a vital parameter for plant cells to elongate and multiply. This is because of the fact that plant use sunlight to fix carbon dioxide into the carbohydrates which is used in many cellular processes including cell division and growth. However, metabolic machinery in different plants works at different rates. Based on these reasons, it is super difficult to pick one option out of the given information to answer this question. Hope this helps.
<span>Animal-like protists are called protozoa. Protozoa are single-celled eukaryotes that share some traits with animals. Like animals, they can move, and they are heterotrophs.
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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
