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:
32.3 m/s
Explanation:
The ball follows a projectile motion, where:
- The horizontal motion is a uniform motion at costant speed
- The vertical motion is a free fall motion (constant acceleration)
We start by analyzing the horizontal motion. The ball travels horizontally at constant speed of
and it covers a distance of
d = 165 m
So, the total time of flight of the ball is
In order to find the vertical velocity of the ball, we have now to analyze its vertical motion.
The vertical motion is a free-fall motion, so the ball is falling at constant acceleration; therefore we can use the following suvat equation:
where
is the vertical velocity at time t
is the initial vertical velocity
is the acceleration of gravity (taking downward as positive direction)
Substituting t = 3.3 s (the time of flight), we find the final vertical velocity of the ball:
Answer: elastic potential energy = 20.27 J
Explanation:
Given that the
Mass M = 0.470 kg
Height h = 4.40 m
Spring constant K = 85 N/m
The maximum elastic potential will be equal to the maximum kinetic energy experienced by the block.
But according to conservative of energy, the maximum kinetic energy is equal to the maximum potential energy experienced by the block of mass M.
That is
K .E = P.E = mgh
Where g = 9.8m/s^2
Substitutes all the parameters into the formula
K.E = 0.470 × 9.8 × 4.4
K.E = 20.27 J
Where K.E = maximum elastic potential energy stored in the spring during the motion of the blocks after the collision which is 20.27J.
Answer:
5.77×10¹⁰ m
Explanation:
From the question,
Applying
Kepler's third law
P² = d³...................... Equation 1
Where P = Planet's period, d = distance between the center of the planet and the sun.
make d the subject of the formula n equation 1
d = ![\sqrt[3]{P^{2} }](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7BP%5E%7B2%7D%20%20%7D)
.................. Equation 2
Given: P = 0.24 sidereal.
Substitute the value of P into equation 2
d = ![\sqrt[3]{0.24^{2} }](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B0.24%5E%7B2%7D%20%7D)
d = 0.386 Au
d = 0.386×1.496×10¹¹
d = 5.77×10¹⁰ m
