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2 years ago

Why does The human population continue to grow exponentially

2 answers:

6 0

Global human population growth is around 75 million annually, or 1.1% per year. The global population has grown from 1 billion in 1800 to 7 billion in 2012. Globally, the growth rate of the human population has been declining since 1962 and 1963, when it was 2.20% per annum. In 2009, the estimated annual growth rate was 1.1%. The CIA World Factbook gives the world annual birthrate, mortality rate, and growth rate as 1.89%, 0.79%, and 1.096% respectively. The last 100 years have seen a rapid increase in population due to medical advances and massive increase in agricultural productivity.

Simora [160]2 years ago

3 0

Hello!

The human population is increasing so much because the death rate has gown down. The farther we go, the more technology we have available. Which in turn helps people live. So rapid birth rates and decreasing death rate is causing the rate to grow.

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to 10 Hz. Superimposed on this signal is 60-Hz noise with an amplitude of 0.1 V. It is desired to attenuate the 60-Hz signal to

givi [52]

Answer:

$G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1$

If we square both sides we got:

$G^2 (1+\frac{f}{f_c})^{2n}= 1$

We divide both sides by $G^2$ and we got:

$(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}$

Now we can apply log on both sides and we got:

$2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})$

And solving for n we got:

$n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}$

And replacing we got:

$n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}$

$n = \frac{4.60517}{3.8918}=1.18$

And since n needs to be an integer the correct answer would be n=2 for the filter order.

Explanation:

For this case we can use the formula for the Butterworth filter gain given by:

[tec] G = \frac{1}{\sqrt{1 +(\frac{f}{f_c})^{2n}}}[/tex]

Where:

G represent the transfer function and we want that G =0.1 since the desired signal is less than 10% of it's value

$f_c = 10 Hz$ represent the corner frequency

$f= 60 Hz$ represent the original frequency

n represent the filter order and that's the variable that we need to find

$G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1$

If we square both sides we got:

$G^2 (1+\frac{f}{f_c})^{2n}= 1$

We divide both sides by $G^2$ and we got:

$(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}$

Now we can apply log on both sides and we got:

$2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})$

And solving for n we got:

$n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}$

And replacing we got:

$n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}$

$n = \frac{4.60517}{3.8918}=1.18$

And since n needs to be an integer the correct answer would be n=2 for the filter order.

7 0

3 years ago

How is the energy of the wave affected if the amplitude of the wave increases from 2 meters to 4

Alex777 [14]

Answer:

Energy of wave will increase as the energy of wave is related to the amplitude of wave

5 0

2 years ago

To avoid using the variable "a"

sdas [7]

Answer:

option c .

Explanation:

..............................

6 0

2 years ago

At what speed does a typical electron pass by any given point in the wire?

attashe74 [19]

A 5.00 A current runs through a 12 gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5*10^28 free electrons per cubic metre.

a) How many electrons pass through the light bulb each second?

b) What is the current density in the wire? (answer in A/m^2)

<span>c) At what speed does a typical electron pass by any given point in the wire? (answer in m/s)

</span>a) 5.0 A = 5.0 C/s
. Number of electrons in 5.0C = 5.0 / 1.60^-19 = 3.125^19
. 5.0 A ►= 3.125^19 electrons/s

b) A/m² = 5.0 / π(1.025^-3 m)² .. .. ►= 1.52^6 A/m²

c) Charge density (q/m³) = 8.50^28 e/m³ x 1.60^-19 = 1.36^10 C/m³

(q/m³)(m²)(m/s) = q/s (current i in C/s [A])
(m²) = Area
(m/s) = mean drift speed

(q/m³)(A)(v) = i

v = i.[(q/m³)A]ˉ¹

<span>v = 5.0 [1.36^10 * π(1.025^-3 m)²]ˉ¹.. .. ►v = 1.10^-4 m/s</span>

7 0

3 years ago

Name two important differences between light waves and sound waves

emmasim [6.3K]

Answer:

1. Light is electromagnetic waves while sound is mechanical.

Light wave is transverse and sound is longitudinal.

Explanation:

<i = <r

when distance between source and obstacle and distance between obstacle and screen is finite then the diffraction is fresnal diffraction

when distance between source and obstacle and distance between obstacle and screen is infinite then the diffraction is Fraunhofer's diffraction

8 0

3 years ago