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masha68 [24]
3 years ago
14

Pyrolobus fumarii, an extreme thermophile, can survive at a temperature as high as 113°C. This microorganism could be categorize

d as which of the following?
a. Eubacteria
b. Archaeabacteria
c. Cyanobacteria
Biology
2 answers:
MrRa [10]3 years ago
6 0
The answers is B. 
Pyrolobus fumarii, an extreme thermophile, can survive at a temperature as high as 113°C. This microorganism could be categorized as <span>Archaeabacteria.
</span>Pryrolobus fumarii is a species of <span>Archaeabacteria which is </span><span>known for its ability to live at extremely high temperatures that kill most </span>organisms. 
Hope it helped.
pshichka [43]3 years ago
3 0

Answer: b. Archaeabacteria

Pyrolobus fumarii is the species of archeabacteria. These organisms are known for their ability to survive in the extreme high temperatures. They are called as hyperthermophiles. They are found in the deep sea vent, and beneath the earth crust. The cells of these organisms are circular. They have lobed shaped cocci cells. These organisms are used in processing of agricultural products, animal feeding and other human consumable products.

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A fish maintains its depth in fresh water by adjusting the air content of porous bone or air sacs to make its average density th
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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

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