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Alecsey [184]
3 years ago
12

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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The surface tension of isopropanol in air has a value of 23.00 units and the
Y_Kistochka [10]

Answer:

It's A & C

Explanation:

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3 years ago
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Elements of Music<br>Timbre<br>Dynamics<br>Rhythm<br>Form<br>Texture<br>Harmony<br>Style​
rodikova [14]

Explanation:

Rhythm, Harmony, Timbre, Dynamics, Texture, and Form

7 0
2 years ago
The dimensions of a block that weighs 30 N are 0.23 mx 0.12 mx 0.075 m.
pishuonlain [190]

Answer:25000/23 N/M^2

Explanation:

P=f/A=30/(0.23×0.12)=25000/23

8 0
3 years ago
A balloon rubbed against denim gains a charge of −7.0 µc. what is the electric force between the balloon and the denim when the
zubka84 [21]
The solution for this problem is computed by through this formula, F = kQq / d²Plugging in the given values above, we can now compute for the answer.
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4 0
3 years ago
A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum
s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle=q=-2.74\times 10^{-6} C

Kinetic energy of particle=K_E=6.65\times 10^{-10} J

Initial time=t_1=6.36 s

Final potential difference=V_2=0.351 V

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

qV=K.E

Using the formula

2.74\times 10^{-6}V_1=6.65\times 10^{-10} J

V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V

Initial voltage=V_1=2.43\times 10^{-4} V

\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2

Using the formula

\frac{V_1}{V_2}=(\frac{6.36}{t})^2

\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}

t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}

t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}

t=241.7 s

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0
3 years ago
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