Answer:
BO : OO
2 : 2
Explanation:
The results of mating between a man of blood group B heterozygous and a woman of blood group O produces children that will either have blood group B or O thus the probability of a child being either blood group B or O is 50%.
B O
O BO OO
O BO OO
Therefore genotype BO:OO
Phenotypes 2:2
<span>Little to No Global Warming, Emissions Improved Public Health, and Environmental Quality.</span>
A) The answer is 5.2 · 10⁻¹⁴ kg
It is known that the mass (m) is density (D) multiplied by volume (V):
m = D · V
D = 1.0 · 10³ kg/m³
m = ?
V = ?
Let's calculate the volume.
The volume (V) of a spherical cell is V = 4/3 π r³ (r - radius of a sphere)
r = ?
d = 1.0 um = 1.0 · 10⁻⁶ m
Since radius is a half of the diameter, then:
r = d ÷ 2 = 1.0 · 10⁻⁶ m ÷ 2 = 0.5 · 10⁻⁶ m
It is known than π = 3.14
Therefore:
V = 4/3 π r³ = 4/3 · 3.14 · (0.5 · 10⁻⁶)³ = 4/3 · 3.14 · 0.5³ · (10⁻⁶)³
= 4/3 · 3.14 · 0.125 · 10⁻⁶*³ = 4/3 · 3.14 · 0.125 · 10⁻¹⁸ = 0.52 · 10⁻¹⁸
= 5.2 · 10⁻¹⁷ m₃
So, now when we know D and V, it is easy to calculate m:
m = D · V = 1.0 · 10³ kg/m³ · 5.2 · 10⁻¹⁷ m³ = 5.2 · 10⁻¹⁷⁺³ kg = 5.2 · 10⁻¹⁴ kg
b) The answer is 12.56 · 10⁻⁶ kg
It is known that the mass (m) is density (D) multiplied by volume (V):
m = D · V
D = 1.0 · 10³ kg/m³
m = ?
V = ?
Let's calculate the volume.
The volume (V) of a fly in the shape of cylinder is V = h π r²
h = 4.0 mm = 4.0 · 10⁻³ m
r = ?
d = 2.0 mm = 2.0 · 10⁻³ m
Since radius is a half of the diameter, then:
r = d ÷ 2 = 2.0 · 10⁻³ m ÷ 2 = 1.0 · 10⁻³ m
It is known than π = 3.14
Therefore:
V = h π r³ = 4.0 · 10⁻³ · 3.14 · (1.0 · 10⁻³ )² = 4.0 · 10⁻³ · 3.14 · 1.0² · (10⁻³ )² =
= 4.0 · 10⁻³ · 3.14 · 1.0 · 10⁻³*² = 4.0 · 10⁻³ · 3.14 · 1.0 · 10⁻⁶ =
= 12.56 · 10⁻³⁻⁶ = 12.56 · 10⁻⁹ m³
So, now when we know D and V, it is easy to calculate m:
m = D · V = 1.0 · 10³ kg/m³ · 12.56 · 10⁻⁹ m³ = 12.56 · 10⁻⁹⁺³ kg = 12.56 · 10⁻⁶ kg
Answer: Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune
Explanation:
Answer:
here
Explanation:
There will be the nitrogenous bases (adenosine, cytosine, guanine, thymine), held in complementary pairs (A T, C G) by hydrogen bonds. Weak van der Waal's forces exists between the 'rungs'.
