Answer:
Option (B)
Explanation:
In the stabilizing natural selection, the extreme traits from both the ends are eliminated by natural selection and natural selection favors the intermediate trait. So over time individuals having the intermediate traits are selected over the individuals having extreme traits.
So here the population of the bat which possesses moderate wing length is selected over the individual with extreme traits like individuals with short wings and long wings. As a result, the population of moderate length wing bats increased.
Therefore the correct answer is (B)- stabilizing natural selection.
Answer:
I would say it's B. But just in case here is some information if I'm wrong.
Explanation:
Igneous rocks are very dense and hard. They may have a glassy apprearance. Metamorphic rocks may also have a glassy appearance. You can distinguish these from igneous rocks based on the fact that metamorphic rocks tend to be brittle, lightweight, and an opaque black color.
Hope this helps!
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Answer: To focus on a near object – the lens becomes thicker, this allows the light rays to refract (bend) more strongly. To focus on a distant object – the lens is pulled thin, this allows the light rays to refract slightly.
Explanation:
Answer:
T = 2.82π s
x = Acos(0.71t)
Explanation:
This problem can be solved by using the expressions
( 1 )
( 2 )
where T is the period of oscillation of the system, m is the mass of the object attached to the spring, k is the spring constant and x is the position of the object.
By replacing in the expression (1):

Taking 6 inches as the amplitude of the motion, we have

I hope this is useful for you
Regards
Answer:
a) 2.693*10^-4 C
b) 8.875*10^-5 s
c) 2.96 W
Explanation:
Given that
Inductance of the circuit, L = 4.24 mH
Capacitance of the circuit, C = 3.02 μF
Current in the circuit, I = 2.38 A
See attachment for calculations