Answer:
Polygenic traits basically mean the traits that are a result of multiple intertwined traits like for example skin colour. When two people mate their traits get intertwined and the resultant trait of the baby is a result of the polygenic traits.
Explanation:
Answer:
When sulfurous, sulfuric, and nitric acids in polluted air and rain react with the calcite in marble and limestone, the calcite dissolves. In exposed areas of buildings and statues, we see roughened surfaces, removal of material, and loss of carved details. Stone surface material may be lost all over or only in spots that are more reactive.
Explanation:
You might expect that sheltered areas of stone buildings and monuments would not be affected by acid precipitation. However, sheltered areas on limestone and marble buildings and monuments show blackened crusts that have peeled off in some places, revealing crumbling stone beneath. This black crust is primarily composed of gypsum, a mineral that forms from the reaction between calcite, water, and sulfuric acid. Gypsum is soluble in water; although it can form anywhere on carbonate stone surfaces that are exposed to sulfur dioxide gas (SO2), it is usually washed away. It remains only on protected surfaces that are not directly washed by the rain.
Answer:
Explanation:
- Anaphase
- Interphase
- Metaphase
- Prophase
- Cytokinesis
In order, they would be 2, 4, 3, 1, 5.
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
