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4 years ago

The process of change in a population over time is called_____

Physics

2 answers:

m_a_m_a [10]4 years ago

7 0

Answer:

micro evolution

Explanation:

Changes within a gene pool occurring from generation to generation is called micro evolution. Allele frequencies in a population may change due to gene flow, genetic drift, natural selection and mutation. These are referred to as the four fundamental forces of evolution.

zmey [24]4 years ago

5 0

Evolution should be the answer

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Evaluate the use of hydrogen as a fuel. Explain how hydrogen is produced. Write about the environmental impact of using diesel a

posledela

Explanation:

Hydrogen is used as a fuel because it has a high calorific value per molecule when burned in air and also burns explosively. When burned in air, it reacts with oxygen to produce water which is safe for the environment. This same property of hydrogen of high explosiveness makes it difficult to store safely.

Hydrogen can be produced through the electrolysis of water. However this is high energy consuming. The most common method is called steam-methane reforming. Methane from hydrocarbon is mixed with steam under high pressure and reacts to produce hydrogen and carbon monoxide.

Diesel and petrol and other petroleum products when burned produce carbon dioxide which is a greenhouse gas. Greenhouse gases when increasing in the atmosphere, cause greenhouse effect that results in global warming.

Learn More:

For more on hydrogen fuel check out;

brainly.com/question/2666329

#LearnWithBrainly

8 0

4 years ago

What is the period and frequency of the second hand on a clock?

Allushta [10]

Answer:

Periods are 60 seconds, 60 minutes and 12 hours frequency are their reciprocal.

Explanation:

For second hand, period = 1 min = 3660 s; frequency = 0.0166

For minute hand, period = 1 hr = 3600 s; frequency = 2.77}

For hour hand, period = 12 hrs = 4300 s; frequency = 2.3}

4 0

3 years ago

The earth’s radius is 6.37 × 10 6 m ; it rotates once every 24 hours. What is the earth’s angular speed? Viewed from a point abo

Advocard [28]

Answer:

a) the angular speed of the Earth's rotation is 7.272 × 10⁻⁵ rad/s.

b) Viewed from the North Pole, Earth's angular speed rotates in an anticlockwise direction. Therefore, the angular velocity is positive.

c) Earth's speed at a point on the equator is 463.23 m/s.

d) the speed of a point on the earth’s surface halfway between the equator and the pole is 231.61 m/s.

Explanation:

a) The angular speed of the Earth's rotation is:

ω = 2π / T

where

T is the period

ω = 2π / (24 hr × (3600 s / 1 hr))

ω = 7.272 × 10⁻⁵ rad/s

b) Viewed from the North Pole, Earth's angular speed rotates in an anticlockwise direction. Therefore, the angular velocity is positive.

c) Earth's speed at a point on the equator is:

v = r × ω

v = (6.37 × 10⁶ m) × (7.272 × 10⁻⁵ rad/s)

v = 463.23 m/s

d) The radius of the circle in which the point moves is half of Earth's radius. Therefore,

r = 1/2(6.37 × 10⁶ m)

r = 3.19 × 10⁶ m

Therefore, the speed of a point on the earth’s surface halfway between the equator and the pole is:

v = r × ω

v = (3.19 × 10⁶ m) × (7.272 × 10⁻⁵ rad/s)

v = 231.61 m/s

6 0

3 years ago

Which element has an atomic number of 9 and an atomic mass of 19?

yan [13]

Answer:

fluorine

Explanation:

3 0

3 years ago

Read 2 more answers

A) A real object 4.0 cm high stands 30.0 cm in front of a converging lens of focal length 23 cm. Find the image distance, the im

vfiekz [6]

Explanation:

Given that,

Height of object = 4.0 cm

Distance of the object u= -30.0 cm

Focal length = 23 cm

We need to calculate the image distance

Using lens's formula

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

Where, u = object distance

v = image distance

f = focal length

Put the value into the formula

$\dfrac{1}{v}-\dfrac{1}{-30}=\dfrac{1}{23}$

$\dfrac{1}{v}=\dfrac{1}{23}-\dfrac{1}{30}$

$\dfrac{1}{v}=\dfrac{7}{690}$

$v=98.57\ cm$

We need to calculate the height of the image

Using formula of height

$\dfrac{h'}{h}=\dfrac{-v}{u}$

Where, h' = height of image

h = height of object

$\dfrac{h'}{4}=\dfrac{-98.57}{-30}$

$h'=\dfrac{98.57\times4}{30}$

$h'=13.14\ cm$

The image is real and inverted.

(b). Now, object distance u = 13.0 cm

We need to calculate the image distance

Using lens's formula

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

Put the value into the formula

$\dfrac{1}{v}=\dfrac{1}{23}-\dfrac{1}{13.0}$

$\dfrac{1}{v}=-\dfrac{10}{299}$

$v=-29.9\ cm$

We need to calculate the height of the image

$\dfrac{h'}{4}=\dfrac{-(-29.9)}{-13.0}$

$h=-\dfrac{29.9\times4}{13.0}$

$h=-9.2\ cm$

The image is virtual and erect.

Hence, This is required solution.

6 0

3 years ago