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3 years ago

New alleles arising from mutations in a population will

Physics

1 answer:

ivanzaharov [21]3 years ago

8 0

Increase in frequency over time until they reach fixation, replacing the ancestral allele in the population

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Define thermopile and peltiers??

NikAS [45]

Answer:

Thermopile is an electronic device that converts thermal energy into electrical energy.

A Peltier cooler, heater, or thermoelectric heat pump is a solid-state active heat pump which transfers heat from one side of the device to the other, with consumption of electrical energy, depending on the direction of the current.

3 0

3 years ago

A concrete piling of 50 kg is suspended from a steel wire of diameter 1.0 mm and length 11.2 m. How much will the wire stretch?

pentagon [3]

Explanation:

It is given that,

Mass of concrete pilling, m = 50 kg

Diameter of wire, d = 1 mm

Radius of wire, r = 0.0005 m

Length of wire, L = 11.2

Young modulus of steel, $Y=20\times 10^{10}\ N/m^2$

The young modulus of a wire is given by :

$Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta L}{L}}$

$Y=\dfrac{F.L}{A\Delta L}$

$\Delta L=\dfrac{F.L}{A.Y}$

$\Delta L=\dfrac{50\ kg\times 9.8\ m/s^2\times 11.2\ m}{\pi (0.0005\ m)^2\times 20\times 10^{10}\ N/m^2}$

$\Delta L=0.034\ m$

So, the wire will stretch 0.034 meters. Hence, this is the required solution.

8 0

3 years ago

An electron (restricted to one dimension) is trapped between two rigid walls 1.40 nm apart. The electron's energy is approximate

Bumek [7]

Answer:

a) n = 9.9 b) E₁₀ = 19.25 eV

Explanation:

Solving the Scrodinger equation for the electronegative box we get

Eₙ = (h² / 8m L²2) n²

where l is the distance L = 1.40 nm = 1.40 10⁻⁹ m and n the quantum number

In this case En = 19 eV let us reduce to the SI system

En = 19 eV (1.6 10⁻¹⁹ J / 1 eV) = 30.4 10⁻¹⁹ J

n = √ (In 8 m L² / h²)

let's calculate

n = √ (8 9.1 10⁻³¹ (1.4 10⁻⁹)² 30.4 10⁻¹⁹ / (6.63 10⁻³⁴)²

n = √ (98) n = 9.9

since n must be an integer, we approximate them to 10

b) We substitute for the calculation of energy

In = (h² / 8mL2² n²

In = (6.63 10⁻³⁴) 2 / (8 9.1 10⁻³¹ (1.4 10⁻⁹)² 10²

E₁₀ = 3.08 10⁻¹⁸ J

we reduce eV

E₁₀ = 3.08 10⁻¹⁸ j (1ev / 1.6 10⁻¹⁹J)

E₁₀ = 1.925 101 eV

E₁₀ = 19.25 eV

the result with significant figures is

E₁₀ = 19.25 eV

3 0

3 years ago

What is the magnitude of the magnetic dipole moment of the bar magnet

Annette [7]

The magnitude of the magnetic dipole moment of the bar magnet is 1.2 Am²

<h3> Magnetic dipole moment of the bar magnet</h3>

The magnitude of the magnetic dipole moment of the bar magnet at distance from its axis is calculated as follows;

$B = \frac{2\mu_0m}{4\pi r^3} \\\\m = \frac{4\pi r^3 B}{2\mu_0}$

where;

B is magnetic field
m is dipole moment
μ is permeability of free space

m = (4π x 0.1³ x 2.4 x 10⁻⁴)/(2 x 4π x 10⁻⁷)

m = 1.2 Am²

The complete question is below:

What is the magnitude of the magnetic dipole moment of the bar magnet from 0.1 m of its axis and magnetic field strength of 2.4 x 10⁻⁴ T.

Learn more about dipole moment here: brainly.com/question/27590192

#SPJ11

6 0

1 year ago

As shown in the figure below, a bullet is fired at and passes through a piece of target paper suspended by a massless string. Th

NikAS [45]

Answer:

M = 0.730*m

V = 0.663*v

Explanation:

Data Given:

$v_{bullet, initial} = v\\v_{bullet, final} = 0.516*v\\v_{paper, initial} = 0\\v_{paper, final} = V\\mass_{bullet} = m\\mass_{paper} = M\\Loss Ek = 0.413 Ek$

Conservation of Momentum:

$P_{initial} = P_{final}\\m*v_{i} = m*0.516v_{i} + M*V\\0.484m*v_{i} = M*V .... Eq1$

Energy Balance:

$\frac{1}{2}*m*v^2_{i} = \frac{1}{2}*m*(0.516v_{i})^2 + \frac{1}{2}*M*V^2 + 0.413*\frac{1}{2}*m*v^2_{i}\\\\0.320744*m*v^2_{i} = M*V^2\\\\M = \frac{0.320744*m*v^2_{i} }{V^2} ....... Eq 2$

Substitute Eq 2 into Eq 1

$0.484*m*v_{i} = \frac{0.320744*m*v^2_{i} }{V^2} *V \\0.484 = 0.320744*\frac{v_{i} }{V} \\\\V = 0.663*v_{i}$

Using Eq 1

$0.484m*v_{i} = M* 0.663v_{i}\\\\M = 0.730*m$

7 0

3 years ago

New alleles arising from mutations in a population will​

New alleles arising from mutations in a population will