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4 years ago

6

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

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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3 years ago

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Ecology is the study of

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Organisms and their environment

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3 years ago

Who serves as the presiding officer of the Senate?

Cerrena [4.2K]

Who serves as the presiding officer of the Senate?

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The $\sf\purple{Vice\:President}$ of the United States serves as the presiding officer of the Senate.

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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