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Nookie1986 [14]
3 years ago
5

The graph in the figure shows the position of a particle as it travels along the x-axis. What is the magnitude of the instantane

ous velocity of the particle when t=1.0 s?
Physics
1 answer:
alexandr402 [8]3 years ago
6 0

Answer:

pls first attach the fig.

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Assortative mating changes ______ frequencies but does not change ______ frequencies.
andrew-mc [135]

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.

To know more about Assortative mating visit:

brainly.com/question/28238408

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4 0
1 year ago
What is the change in momentum of a 50-kg rock that falls freely for 3 seconds?
Elan Coil [88]

Answer:

1470kgm/s

Explanation:

Given parameters:

Mass of the rock = 50kg

Time taken for the free fall  = 3s

Unknown:

Change in momentum = ?

Solution:

The change in momentum will be difference between the ending momentum and finishing momentum.

  Momentum is the product of mass and velocity

       Momentum  = mass x velocity

Initial momentum  = 0, the velocity is 0

Final momentum = mass x final velocity

      let us find the final velocity;

                V = U + gt

V is the final velocity

U is the initial velocity

g is the acceleration due to gravity = 9.8m/s²

t is the time

                 V  = 0 + 9.8x3 = 29.4m/s

So;

 Change in momentum  = 50 x 29,4  = 1470kgm/s

6 0
3 years ago
Help me!! giving out brainlest
Stolb23 [73]
Asteroid 1 has more mass because the same force exerted caused this asteroid to move less therefore having more mass
3 0
3 years ago
Read 2 more answers
Ultraviolet light having a wavelength of 97 nm strikes a metallic surface. Electrons leave the surface with speeds up to 3.48 ×
irga5000 [103]

Answer:

<h2><em>12.45eV</em></h2>

Explanation:

Before calculating the work function, we must know the formula for calculating the kinetic energy of an electron. The kinetic energy of an electron is the taken as the difference between incident photon energy and work function of a metal.

Mathematically, KE =  hf - Ф where;

h is the Planck constant

f is the frequency = c/λ

c is the speed of light

λ is the wavelength

Ф is the work function

The formula will become KE =  hc/λ - Ф. Making the work function the subject of the formula we have;

Ф = hc/λ - KE

Ф = hc/λ - 1/2mv²

Given parameters

c = 3*10⁸m/s

λ = 97*10⁻⁹m

velocity of the electron v = 3.48*10⁵m/s

h = 6.62607015 × 10⁻³⁴

m is the mass of the electron = 9.10938356 × 10⁻³¹kg

Substituting the given parameters into the formula Ф = hc/λ - 1/2mv²

Ф =  6.63 × 10⁻³⁴*3*10⁸/97*10⁻⁹ -  1/2*9.11*10⁻³¹(3.48*10⁵)²

Ф = 0.205*10⁻¹⁷ - 4.555*10⁻³¹*12.1104*10¹⁰

Ф = 0.205*10⁻¹⁷ - 55.163*10⁻²¹

Ф = 0.205*10⁻¹⁷ - 0.0055.163*10⁻¹⁷

Ф = 0.1995*10⁻¹⁷Joules

Since 1eV = 1.60218*10⁻¹⁹J

x = 0.1995*10⁻¹⁷Joules

cross multiply

x = 0.1995*10⁻¹⁷/1.60218*10⁻¹⁹

x = 0.1245*10²

x = 12.45eV

<em>Hence the work function of the metal in eV is 12.45eV</em>

6 0
3 years ago
An oscillator consists of a block of mass 0.373 kg connected to a spring. When set into oscillation with amplitude 33 cm, the os
aniked [119]

Answer:

(a)  T = 0.412s

(b)  f = 2.42Hz

(c)  w = 15.25 rad/s

(d)  k = 86.75N/m

(e)  vmax = 5.03 m/s

Explanation:

Given information:

m: mass of the block = 0.373kg

A: amplitude of oscillation = 22cm = 0.22m

T: period of oscillation = 0.412s

(a) The period is the time of one complete oscillation = 0.412s

The period is 0.412s

(b) The frequency is calculated by using the following formula:

f=\frac{1}{T}=\frac{1}{0.412s}=2.42Hz

The frequency is 2.42 Hz

(c) The angular frequency is:

\omega=2\pi f=2\pi (2.42Hz)=15.25\frac{rad}{s}

The angular frequency is 15.25 rad/s

(d) The spring constant is calculated by solving the following equation for k:

\omega=\sqrt{\frac{k}{m}}\\\\k=m\omega^2=(0.373kg)(15.25rad/s)^2=86.75\frac{N}{m}

The spring constant is 86.75N/m

(e) The maximum speed is:

v_{max}=\omega A=(15.25rad/s)(0.33m)=5.03\frac{m}{s}

(f) The maximum force applied by the spring if for the maximum elongation, that is, the amplitude:

F=kA=(86.75N/m)(0.2m)=17.35N

The maximum force that the spring exerts on the block is 17.35N

3 0
3 years ago
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