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Alona [7]
3 years ago
5

Why are not all microorganism harmful?

Physics
1 answer:
son4ous [18]3 years ago
8 0


In nature there are two categories of microorganisms as relating to health. Microorganisms that are considered harmful to humans are called pathogens and these cause disease. Examples include bacteria such as streptococcus which cause sore throat and salmonella which cause typhoid disease.

 There are some microorganisms which are helpful to man and they live mostly on the skin of man or in his gut and are mostly bacteria. They are collectively called bacterial normal flora.

In man the normal bacterial flora of the skin include staphylococcus found on dry skin, cornybacteria found in moist skin sites and propionibacteria in the sebaceous sites (head, neck, trunk) of the body. Normal bacterial flora of the gut include Escherichia coli.

One of the major function of bacterial flora is actually to protect our bodies  by competing for space with pathogens preventing them from gaining a foothold in our bodies.



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Why does the modern atomic theory used today represent the best explanation scientists have to offer?
LenaWriter [7]

Answer:

D: Scientists have not been successful in disproving the atomic theory.

Explanation:

This theory has had great implications in the understanding of nature at the microscopic level, from the first atomic model to the quantum model of the atom. All modern physics, chemistry and biochemistry are based on the theory that matter is composed of atoms of different elements. It has not been possible to disprove this theory or replace it with a better one.

8 0
3 years ago
Which statement is true about an object moving in a circular motion due to centripetal force, F, when the mass is doubled? A. It
Allushta [10]
Force = [Mass x (velocity)^2]/Radius
Therefore,
The mass and the force are directly proportional.If you double the mass,the force will be doubled too.

A. Its centripetal force would <span>be doubled.</span>
3 0
3 years ago
Imagine that the satellite described in the problem introduction is used to transmit television signals. You have a satellite TV
Korolek [52]

Complete Question

A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,000 km above the surface of the earth, and we assume it has an isotropic power output of 1 kW (although, in practice, satellite antennas transmit signals that are less powerful but more directional).

Imagine that the satellite described in the problem introduction is used to transmit television signals. You have a satellite TV reciever consisting of a circular dish of radius R which focuses the electromagnetic energy incident from the satellite onto a receiver which has a surface area of 5 cm2.

How large does the radius R of the dish have to be to achieve an electric field vector amplitude of 0.1 mV/m at the receiver?

For simplicity, assume that your house is located directly beneath the satellite (i.e. the situation you calculated in the first part), that the dish reflects all of the incident signal onto the receiver, and that there are no losses associated with the reception process. The dish has a curvature, but the radius R refers to the projection of the dish into the plane perpendicular to the direction of the incoming signal.

Give your answer in centimeters, to two significant figures.

Answer:

 The radius  of  the dish is R = 18cm

Explanation:

  From the question we are told that

     The radius of the orbit is  = R = 35,000km = 35,000 *10^3 m

    The power output of the power is  P = 1 kW = 1000W

   The electric vector amplitude is given as E = 0.1 mV/m = 0.1 *10^{-3}V/m

    The area of thereciever  is   A_R = 5cm^2

Generally the intensity of the dish is mathematically represented as

         I = \frac{P}{A}

Where A is the area orbit which is a sphere so this is obtained as

          A = 4 \pi r^2

              = (4 * 3.142 * (35,000 *10^3)^2)

              =1.5395*10^{16} m^2

  Then substituting into the equation for intensity

          I_s  =  \frac{1000}{1.5395*10^{16}}

            = 6.5*10^ {-14}W/m2

 Now the intensity received by the dish can be mathematically evaluated as

              I_d = \frac{1}{2}  * c \epsilon_o E_D ^2

  Where c is thesped of light with a constant value  c = 3.0*10^8 m/s

              \epsilon_o is the permitivity of free space  with a value  8.85*10^{-12} N/m

              E_D is the electric filed on the dish

So  since we are to assume to loss then the intensity of the satellite is equal to the intensity incident on the receiver dish

      Now making the eletric field intensity the subject of the formula

                  E_D = \sqrt{\frac{2 * I_d}{c * \epsilon_o} }

substituting values

                 E_D = \sqrt{\frac{2 * 6.5*10^{-14}}{3.0*10^{8} * 8.85*10^{-12}} }

                       = 7*10^{-6} V/m

The incident power on the dish is what is been reflected to the receiver

                P_D = P_R

Where P_D is the power incident on the dish which is mathematically represented as

              P_D = I_d A_d

                   = \frac{1}{2}  c \epsilon_o E_D^2  (\pi R^2)

And  P_R is the power incident on the dish which is mathematically represented as

                 P_R = I_R A_R

                       = \frac{1}{2} c \epsilon_o E_R^2 A_R

Now equating the two

                \frac{1}{2}  c \epsilon_o E_D^2  (\pi R^2) =  \frac{1}{2} c \epsilon_o E_R^2 A_R

   Making R the subject we have

                   R = \sqrt{\frac{E_R^2 A_R}{\pi E_D^2} }

Substituting values

                   R = \sqrt{\frac{(0.1 *10^{-3})^2 * 5}{\pi (7*10^{-6})^ 2} }

                     R = 18cm

8 0
3 years ago
A transverse wave on a string is described by the wave function
fgiga [73]

The acceleration of the given wave is 2.5m/s^2

<h3>Wave property</h3>

The standard wave function is expressed according to the equation

y = Asin(2πft+2πx/λ)

where

λ is the wave length

Given the equation below

y = 0.120 sin((π/8)x + 4πt)

Compare both equation to have:

(π/8)x = 2πx/λ

8 = 2/λ

λ = 2/8

λ = 1/4

λ = 0.25m

For the frequency

Compare both equation to have:

4πt = 2πft

4 = 2f

4/2 = 2f/2

f = 2 Hz

Speed = fλ

speed = 2(0.25)

speed = 0.5m/s

Determine the acceleration

acceleration = change in velocity/time

acceleration = 0.5/ 0.200

acceleration = 2.5m/s^2

Hence the acceleration of the given wave is2.5m/s^2

Learn more on wave function here: brainly.com/question/25699025

#SPJ4

7 0
2 years ago
How does momentum play in tackling?
shusha [124]

Answer:

When two players are running full speed at each other on a football field they build up their momentum

5 0
3 years ago
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