Answer: Natural selection is taking place.
Explanation:
As you can see, the lighter colored mice are more visible than their surroundings, so the hawk picks them off one by one. the brown mice on the other hand are less visible, blending in with their surroundings, so they are successful, and pass on the genes that allow them to survive better.
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Answer \|/
Ice is less dense than water.
Reason why \|/
When water freezes the molecules inside completely stop moving (They still vibrate but don't change their position much). In doing so, they spread out a touch which makes it less dense than liquid water. So ice floats
Kinematics : Study of motion
Fluid kinematics : study of how fluid flows and how to describe its motion.
There are two ways to describe fluid motion
one is Eulerian, where the variations are described at all fixed stations as a function of time.
the other is Lagrangian, in which one follows all fluid particles and describes the variations around each fluid particle along its trajectory.
<u>DIFFRENCE BETWEEN LAGRANGIAN AND EULERIAN:</u>
1.Both Lagrangian and Eulerian describes time variation.
2. Eulerian describes the rate of change in one point of space
Lagrangian descries rate of change of a property of material system.
To know more about the Lagrangian and Eulerian :\brainly.com/question/14944792
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The average power output is the ratio between the work done to compress the spring, W, and the time taken, t:

(1)
The work done is equal to the elastic energy stored by the compressed spring:

where

is the spring constant and

is the compression of the spring. If we substitute the numbers, we find:

And now we can use eq.(1) to calculate the average power output:
Answer:
a) x = ⅔ d
, b) the charge must be negative, c) Q
Explanation:
a) In this exercise the force is electric between the charges, we are asked that the system of the three charges is in equilibrium, we use Newton's second law. Balance is on the third load that we are placing
∑ F = 0
-F₁₂ + F₂₃ = 0
F₁₂ = F₂₃
let's replace the values
k Q Q / r₁₂² = k Q 4Q / r₂₃²
Q² / r₁₂² = 4 Q² / r₂₃²
suppose charge 3 is placed at point x
r₁₂ = x
r₂₃ = d-x
we substitute
1 / x² = 4 / (d-x) 2
1 / x = 2 / (d-x)
x = 2 (x-d)
x = 2x -2d
3x = 2d
x = ⅔ d
b) The sign of the charge must be negative, to have an attractive charge on the two initial charges
c) Q