The question is asking to states on how i earth's early lang environment different than today's land environment and base on my research, I would say that in the past centuries or early ages, the Earth's land environment are compost of tree and plants and the air is still fresh. I hope you are satisfied with my answer and feel free too ask for more
Answer:
Meat, fish and eggs.
Explanation:
Meat, fish and eggs are the foods that do not contain any or less carbohydrate, these foods are high in protein and fats. By looking on the nutrition label for the amounts per serving (in grams) of fiber and total carbohydrates, if there is no or less percentage of carbohydrates so by see the label we can say that whether the food has low or high amount of carbohydrates.
Differences in the density of water can also cause currents to form and move. Density is affected by temperature and salinity. Cold water or water with dissolved salts (higher salinity) is denser than warm water or water without dissolved salts (low or no salinity).
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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
