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3 years ago

Natural selection is part of which step of speciation? *

Physics

2 answers:

Blizzard [7]3 years ago

6 0

Answer:

D (Not too sure)

Explanation:

Natural selection favors the species that can work the best in the environment.

evablogger [386]3 years ago

6 0

Answer:

B. Reproduction

Explanation:

Ecological speciation: A speciation process in which divergent natural selection drives the evolution of <em>reproductive</em> incompatibility (isolation) between taxa.

One of these is natural selection, which is a process that increases the frequency of advantageous gene variants, called alleles, in a population. Natural selection can result in organisms that are more likely to survive and reproduce and may eventually lead to speciation.

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Listed following are locations and times atwhich different phases of the Moon are visible fromEarth’s....?

ad-work [718]

Answer:

Explanation:

*visible near western horizon an hour aftersunset WAXING CRESCENT

*rises about the same time the sun sets FULLMOON

*visible near eastern horizon just before sun rises WANING CRESCENT

*occurs about 3 days before new moon WANING CRESCENT

*visible due south at midnite FULL MOON

*occurs 14 days after new moon FULL MOON

5 0

4 years ago

Due process rights are __________.

natima [27]

I'm pretty sure the answer is A

3 0

3 years ago

Read 2 more answers

A 3.00 kg mass is pushed against a spring and released. If the spring constant of the

Phoenix [80]

Answer:

Energy stored in the compressed spring: $37.5\; \rm J$ before the mass is released.

Maximum horizontal speed of the mass: $5.0\; \rm m \cdot s^{-1}$ .

(Assumption: the surface between the mass and the ground is frictionless until the mass separates from the spring.)

(Maximum horizontal acceleration of the mass: $250\;\rm m \cdot s^{-2}$ .)

Work that friction did on the mass: $13.5\; \rm J$ .

Maximum height of the mass: approximately $0.459\; \rm m$ (assuming that $g = 9.8\; \rm m \cdot s^{-2}$ .)

Explanation:

<h3>Energy stored in the spring</h3>

When an ideal spring of spring constant $k$ is compressed by a displacement of $x$ from the equilibrium position, the elastic potential energy stored in the spring is:

$\displaystyle \frac{1}{2}\, k \cdot x^{2}$ .

The question states that for this spring, $k = 7500\; \rm N \cdot m^{-1}$ (Newtons per meters) whereas $x = 10.0\; \rm cm$ (centimeters.) Convert the unit of $x$ to meters to match the unit of $k$ :

$\begin{aligned} x &= 10.0\; \rm cm\\ &= 10.0\; \rm cm \times \frac{1\; \rm m}{100\;\rm cm} = 0.100\; \rm m\end{aligned}$ .

Calculate the energy stored in this spring:

$\begin{aligned}& \text{Elastic Potential Energy} \\ &= \frac{1}{2}\, k \cdot x^{2} \\ &= \frac{1}{2} \times 7500\; \rm N \cdot m^{-1} \times 0.100\; \rm m \\ &=37.5\; \rm J \end{aligned}$ .

<h3>Maximum horizontal speed of the mass</h3>

Assume that the surface under the spring is frictionless. All that $37.5\; \rm J$ of elastic potential energy would be converted to kinetic energy by the time the mass separates from the spring.

The horizontal speed of the mass is largest at that moment. The reason is that immediately after this moment, friction (between the surface and the mass) would start to slow the mass down immediately.

Calculate the speed $v$ of the mass $m$ at that moment.

$\displaystyle \text{Kinetic Energy} = \frac{1}{2}\, m \cdot v^{2}$ .

Therefore:

$\displaystyle v = \sqrt{\frac{2\cdot (\text{Kinetic Energy})}{m}}$ .

The question states that $m =3.00\; \rm kg$ . At that moment, the kinetic energy of the mass is $37.5\; \rm J$ (or equivalently, $37.5\; \rm kg\cdot m^{2} \cdot s^{-2}$ in base units.)

Calculate the speed of the mass at that moment from the kinetic energy of the mass:

$\begin{aligned}v &= \sqrt{\frac{2\cdot (\text{Kinetic Energy})}{m}} \\ &= \sqrt{\frac{2 \times 37.5\; \rm kg \cdot m^{2}\cdot s^{2}}{3.00\; \rm kg}} = 5.00\; \rm m \cdot s^{-1}\end{aligned}$ .

<h3>Work friction did on the mass</h3>

Assume that the rough surface is level. Therefore, all the energy loss of the mass should be attributed to friction.

Kinetic energy of the mass before coming onto that rough surface: $37.5 \; \rm J$ .

Kinetic energy of the mass before leaving the rough surface:

$\begin{aligned} &\frac{1}{2}\, m \cdot v^{2} \\ &= \frac{1}{2} \times 3.00\; \rm kg \times {\left(4.00\; \rm m \cdot s^{-1}\right)}^{2} \\ &= 24.0\; \rm J \end{aligned}$ .

If that rough surface is level, there would be no change to the gravitational potential energy of the mass. Calculate the change to the kinetic energy of the mass:

$37.5\;\rm J - 24.0\; \rm J = 13.5\; \rm J$ .

That should be equal to the size of the work that friction did on the mass.

<h3>Maximum height of the mass</h3>

Assume that the smooth ramp does not change the total energy of the block. Therefore, the total mechanical energy of the block would still be $24.0\; \rm J$ when the height of the block is maximized. However, all that energy would be in the form of gravitational potential energy.

Let $g$ denote the gravitational field strength ( $g \approx 9.8\; \rm m \cdot s^{2}$ near the surface of the earth.) The gravitational potential energy of an object of mass $m$ and height $h$ (relative to the surface of zero gravitational potential energy) would be:

$\begin{aligned}&\text{Gravitational Potential Energy} = m \cdot g \cdot h\end{aligned}$ .

Rewrite this equation to find an expression for $h$ :

$\displaystyle h = \frac{\text{Gravitational Potential Energy}}{m\cdot g}$ .

Assume that $g = 9.8\; \rm m \cdot s^{-2}$ . The question states that $m = 3.00\; \rm kg$ . $\text{Gravitational Potential Energy} = 24.0\; \rm J$ . Therefore:

$\begin{aligned} &h\\ &= \frac{\text{Gravitational Potential Energy}}{m\cdot g}\\ &= \frac{24.0\; \rm kg \cdot m^{2} \cdot s^{-2}}{3.00\; \rm kg \times 9.8\; \rm m \cdot s^{-2}} \approx 0.459\; \rm m\end{aligned}$ .

3 0

3 years ago

An apple is placed on a kitchen counter. How could the movement of the particles in the apple be decreased? A.Put it in the refr

Tanzania [10]

A.Put it in the refrigerator

6 0

4 years ago

Read 2 more answers

The centers of two15.0-kilogram spheres are separated by 3.00 meters. The magnitude of the gravitational force between the two s

yKpoI14uk [10]

It's gonna be 5.0 cause 3x5=15 and just add the .0

6 0

4 years ago