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1 year ago

Assortative mating changes frequencies but does not change frequencies.

Physics

1 answer:

andrew-mc [135]1 year ago

4 0

Allele frequencies are unaffected by assortative mating, but genotype frequencies .

<h3>Assortative mating: </h3>

Individuals with similar phenotypes and genotypes mate with others more frequently than is anticipated under a random mating pattern in assortative mating, which is a mating pattern and a type of sexual selection.

<h3>Frequencies of genotypes:</h3>

A population's genotype frequency is calculated by dividing the number of people having a particular genotype by the overall population size. The genotype frequency in population genetics is the frequency or ratio (i.e., 0 f 1) among genotypes inside a population.

<h3>The frequency for alleles in biology:</h3>

The term "allele frequency" describes the prevalence of an allele in a population. It is calculated by calculating the number of times the allele occurs in the population and dividing by the sum of all the gene copies.

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A filing cabinet weighing 556 N rests on the floor. The coefficient of static friction between it and the floor is 0.68, and the

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Explanation:

"Static friction is a force that keeps an object at rest. It must be overcome to start moving the object."

(556 x 0.68) = static friction of 378.08N. before movement occurs.

The forces (a) and (b) will not move it.

Each will incur a frictional force preventing movement equal to itself, = 222N. and 334N. respectively.

Forces (c) and (d) will move it, and accelerate it.

Forces (c) and (d) will both encounter friction of (556 x 0.56) = 311.36N. when the cabinet is moving.

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3 years ago

A charge of 8.4 × 10–4 C moves at an angle of 35° to a magnetic field that has a field strength of 6.7 × 10–3 T.

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Answer:10842.33m/s

Explanation:

F=qvBsine

V=f/(qBsine)

V=(3.5×10^-2)÷(8.4×10^-4×6.7×10^-3×sin35)

V=10842.33m/s

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3 years ago

The electromagnetic waves that have the lowest frequencies are called

o-na [289]

A. Radio waves
Have the lowest frequencies

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3 years ago

A 4000 kg satellite is placed 2.60 x 10^6 m above the surface of the Earth.

mash [69]

a) The acceleration of gravity is $4.96 m/s^2$

b) The critical velocity is 6668 m/s (24,006 km/h)

c) The period of the orbit is 8452 s

d) The satellite completes 10.2 orbits per day

e) The escape velocity of the satellite is 9430 m/s

f) The escape velocity of the rocket is 11,191 m/s

Explanation:

The acceleration of gravity for an object near a planet is given by

$g=\frac{GM}{(R+h)^2}$

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

h is the height above the surface

In this problem,

$M=5.98\cdot 10^{24} kg$ (mass of the Earth)

$R=6.37\cdot 10^6 m$ (Earth's radius)

$h=2.60\cdot 10^6 m$ (altitude of the satellite)

Substituting,

$g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)^2}=4.96 m/s^2$

The critical velocity for a satellite orbiting around a planet is given by

$v=\sqrt{\frac{GM}{R+h}}$

where we have again:

$M=5.98\cdot 10^{24} kg$ (mass of the Earth)

$R=6.37\cdot 10^6 m$ (Earth's radius)

$h=2.60\cdot 10^6 m$ (altitude of the satellite)

Substituting,

$v=\sqrt{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=6668 m/s$

Converting into km/h,

$v=6668 m/s \cdot \frac{3600 s/h}{1000 m/km}=24,006 km/h$

The period of the orbit is given by the circumference of the orbit divided by the velocity:

$T=\frac{2\pi (R+h)}{v}$

where

$R=6.37\cdot 10^6 m$

$h=2.60\cdot 10^6 m$

v = 6668 m/s

Substituting,

$T=\frac{2\pi (6.37\cdot 10^6 + 2.60\cdot 10^6)}{6668}=8452 s$

One day consists of:

$t = 24 \frac{hours}{day} \cdot 60 \frac{min}{hours} \cdot 60 \frac{s}{min}=86400 s$

While the period of the orbit is

T = 8452 s

So, the number of orbits completed by the satellite in one day is

$n=\frac{t}{T}=\frac{86400}{8452}=10.2$

The escape velocity for an object in the gravitational field of a planet is given by

$v=\sqrt{\frac{2GM}{R+h}}$

where here we have:

$M=5.98\cdot 10^{24} kg$

$R=6.37\cdot 10^6 m$

$h=2.60\cdot 10^6 m$

Substituting, we find

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=9430 m/s$

We can apply again the formula to find the escape velocity for the rocket:

$v=\sqrt{\frac{2GM}{R+h}}$

Where this time we have:

$M=5.98\cdot 10^{24} kg$

$R=6.37\cdot 10^6 m$

$h=0$ , because the rocket is located at the Earth's surface, so its altitude is zero.

And substituting,

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6)}}=11,191 m/s$

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3 years ago

Which of the following Resistors A, B, C or D, would use the least power? Make

Westkost [7]

i think the answer is D 10.0

4 0

3 years ago

Assortative mating changes ______ frequencies but does not change ______ frequencies.

Assortative mating changes frequencies but does not change frequencies.