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3 years ago

In an electron dot diagram, the symbol for an element is used to represent

Physics

2 answers:

SIZIF [17.4K]3 years ago

7 0

A the nucleus bruh it can't be anything else

Ede4ka [16]3 years ago

4 0

Answer : The correct option is, (A) the nucleus.

Explanation:

Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.

In the Lewis-dot structure the valance electrons are shown by 'dot'.

The representation of valence electrons of an atom by dots around the symbol of an element is said to be the electron dot symbol.

In an electron dot diagram, the symbol for an element is used to represent the nucleus.

For example :

As we know that arsenic has '5' valence electrons. So, the symbol (As) is used to represent the nucleus and valance electrons around the is represented by the 'dot'.

The Lewis-dot structure of As (arsenic) is shown below.

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Explain how work is related to energy

Dahasolnce [82]

Work is closely related to energy. The work-energy principle states that an increase in the kinetic energy of a rigid body is caused by an equal amount of positive work.

So they are both closely related to each other.

HOPE THIS HELPS

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3 years ago

A boy jumps from a wall 3m high. What is an estimate of the change in momentum of the boy when he lands without rebounding?

EastWind [94]

The answer is D. Because the boy jump 3 high

8 0

3 years ago

A typical helium-neon laser found in supermarket checkout scanners emits 633-nm-wavelength light in a 1.0-mm-diameter beam with

Fofino [41]

Answer:

Eo = 9.796 x 10^2 N/C

Bo = 3.266 x 10^-6 T

Explanation:

Given

Wavelength λ = 633 nm

Diameter of the beam D = 1.0 mm

Power P = 1.0 mW

Solution

Radius of the beam r = D/2 = 0.5 mm = 0.0005 m

Area of cross section

$A = \pi r^{2} \\A = 3.15 \times 0.0005^{2}\\A = 7.58 \times 10^{-7} m^{2}\\$

Intensity

$I = \frac{P}{A} \\I = \frac{0.001}{7.85\times 10^{-7}} \\I = 1273.885 {W}/{m^{2} }$

Amplitude of Electric Field

$E_{o} = \sqrt{\frac{2I}{ \epsilon_{o}c } } \\E_{o} = \sqrt{\frac{2 \times 1273.88}{ 8.85 \times 10^{-12} \times 3 \times 10^{8} } }\\E_{o} = 9.796 \times 10^{2}N/C$

Amplitude of Magnetic Field

$B_{o} = \sqrt{\frac{2 \mu_{o}I}{c } } \\B_{o} = \sqrt{\frac{2 \times 4 \times \pi \times 10^{-7} \times 1273.88}{ 3 \times 10^{8} } }\\B_{o} = 3.266 \times 10^{-6} T$

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4 years ago

A positively charged glass rod attracts object X. The net charge of object X:A.may be zero or negative.B.may be zero or positive

Damm [24]

Answer:A

Explanation:

A positively charged glass rod attracts object x. So, object x must be negatively charged or uncharged.

This occurs because opposite charges attract each other or either object x is uncharged and a negative charge is induced in it as glass rod approach the object x.

So option A is correct

8 0

3 years ago

What is the pressure of the gas in this mercury manometer if h = 89 mm and atmospheric pressure is 775 mmhg?

ludmilkaskok [199]

The pressure of the gas is 686 mmhg.
If h = 89 mm
and atmospheric pressure = 775 mmhg
Pressure of the gas = ?
We can find the pressure of the gas by finding the difference between both values.
pressure of the gas = 775 mmhg - 89 mm = 686 mmhg

6 0

4 years ago

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