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Alborosie
4 years ago
6

What do we call a part of the body with a special function?

Physics
2 answers:
Artemon [7]4 years ago
8 0
These certain parts of the body are known as organs.
murzikaleks [220]4 years ago
4 0
This is called an organ.  Examples of organs are: your heart, brain, lungs, liver, stomach etc.
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The picture shows an aerialist walking on a tightrope and holding a balancing bar.
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A. The aerialist’s feet and the rope
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Ecology is the study of
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Organisms and their environment

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Who serves as the presiding officer of the Senate?
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Who serves as the presiding officer of the Senate?

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3 0
3 years ago
The amount of energy necessary to remove an electron from an atom is a quantity called the ionization energy, Ei. This energy ca
larisa86 [58]

Answer:

The value is E_i  =  1.5596 *10^{-18} \  J

Explanation:

From the question we are told that

The wavelength is \lambda  =  48.2 nm  =  48.2  *10^{- 9 }\  m

The velocity is v = 2.371*10^6 \ m/s

The mass of electron is m_e  =  9.109*10^{-31} \  kg

Generally the energy of the incident light is mathematically represented as

E =  \frac{h *  c}{\lambda}

Here c is the speed of light with value c =  3.0 *10^{8} \  m/s

h is the Planck constant with value h = 6.62607015 *  10^{-34 }  J\cdot s

So

E =  \frac{6.62607015 *  10^{-34 }* 3.0 *10^{8}}{48.2  *10^{- 9 }}

=> E = 4.12 *10^{-18} \  J

Generally the kinetic energy is mathematically represented as

E_k  =  \frac{1}{2} *  m_e * v^2

=> E_k  =  \frac{1}{2} *  9.109*10^{-31} * (2.371*10^6 )^2

=> E_k  =  2.56 *0^{-18} \  J

Generally the ionization energy is mathematically represented as

E_i  =  4.12 *10^{-18} -   2.56 *0^{-18}

=>     E_i  =  1.5596 *10^{-18} \  J

8 0
4 years ago
A photographer uses his camera, whose lens has a 50 mm focal length, to focus on an object 1.5 m away. He then wants to take a p
allsm [11]

Answer:

The distance is z = 0.008 \ m

Explanation:

From the question we are told that

   The  focal length is  f =  50 \ mm  =  50*10^{-3} \ m

   

Generally the lens equation is mathematically represented as  

     \frac{1}{u}  + \frac{1}{v} =  \frac{1}{f}

At  image  distance  u =  1.5 m

       \frac{1}{1.5}  + \frac{1}{v} =  \frac{1}{50 *10^{-3}}

=>      \frac{1}{50 *10^{-3}}  - \frac{1}{1.5}    =  \frac{1}{v}

=>v  =  0.052 \ m

At  image  distance  u = 30\  cm  =  0.30 \ m

        \frac{1}{0.3}  + \frac{1}{v_1} =  \frac{1}{50 *10^{-3}}

=>     \frac{1}{50 *10^{-3}}  - \frac{1}{0.30 }    =  \frac{1}{v_1}

=>    v_1 = 0.06 \ m

The distance the lens need to move is evaluate as

   z =  |v - v_1|

   z =  |0.052 - 0.06|

   z = 0.009 \ m

8 0
3 years ago
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