The mechanisms of reproductive isolation are:
1) Habitat isolation
2) Behavioral isolation
3) Mechanical isolation
4) Gametic isolation
The isolation happening here is behavioral isolation, because their different behavior patterns prevent the frogs from breeding.
Answer:
b) Vectors A and B are in the same direction.
Explanation:
To understand this problem we will say that vector A has a magnitude of 5 units and vector B a magnitude of 3 units. In the subtraction of vectors the initial parts of vectors always bind together. And the vector resulting from the subtraction is traced from the end of the second vector (B) to the end of the first vector (A).
The length of the resultant vector will be 5 - 3 = 2
In the attached image, we analyze case a), b), and d)
For a)
As we can see in the attached image the resultant vector has a length of 8 units.
For d)
As we can see in the attached image the resultant vector has a length of 5.83 units.
For b)
The resultant vector has a length of 2 units.
Therefore the case given in b) is true
Answer:
Explanation:
In case of oil slick a thin layer of oil is formed on water . This thin layer creates a rainbow of colour . The phenomenon is due to interference of light waves , one reflected from the upper surface of oil and the other reflected from the lower surface of the oil.
For formation of bright colour
2 μ t = ( 2n + 1 ) λ / 2
μ is refractive index of oil , t is thickness of oil layer λ is wave length of light falling on the layer .
given μ = 1.2 , λ = 750 x 10⁻⁹ ,
2 x 1.2 t = ( 2n + 1 ) 750 x 10⁻⁹ / 2
For minimum thickness n = 0
2.4 t = 375 x 10⁻⁹
t = 156.25 n m
B ) If the refractive index of layer of medium below oil is less than that of oil , the condition of formation of colour changes
The new condition is
2 μ t = n λ
2 x 1.5 t = 750 nm , n = 1 for minimum wavelength .
t = 250 nm
C ) Light mostly transmitted means dark spot is formed at that point .
For that to be observed from water side , the condition is
2 μ t = ( 2n + 1 ) λ / 2
λ = 4μ t / ( 2n + 1 )
For maximum wavelength n = 0
λ = 4μ t
= 4 x 1.5 x 200 nm
= 1200 nm .
Answer: 361° C
Explanation:
Given
Initial pressure of the gas, P1 = 294 kPa
Final pressure of the gas, P2 = 500 kPa
Initial temperature of the gas, T1 = 100° C = 100 + 273 K = 373 K
Final temperature of the gas, T2 = ?
Let us assume that the gas is an ideal gas, then we use the equation below to solve
T2/T1 = P2/P1
T2 = T1 * (P2/P1)
T2 = (100 + 273) * (500 / 294)
T2 = 373 * (500 / 294)
T2 = 373 * 1.7
T2 = 634 K
T2 = 634 K - 273 K = 361° C
