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2 years ago

Explain how forces of attraction and repulsion exist within an atom

Physics

2 answers:

Tamiku [17]2 years ago

8 0

Answer:

There are three kinds of forces within the atom:

i) Electromagnetic force of attraction between the electrons and protons

ii) Electromagnetic force of repulsion between the protons or weak nuclear force

iii) Strong nuclear force between the electrons and protons

Explanation:

Electromagnetic force of attraction:

Electrons revolve in the orbits outside the nucleus. There exists an electromagnetic force of attraction between the electrons and protons. That’s why, electrons do not leave the atom.

Weak nuclear force:

It is an electromagnetic force of repulsion between the protons in the nucleus of the atoms.

Strong nuclear force:

This force is strongest from all the fundamentals forces and exists between the protons and neutrons in the nucleus of atom. This force overcomes the weak nuclear force and does not allow protons to stray away.

Stels [109]2 years ago

7 0

Answer:

Oppositely charged particles attract each other, while like particles repel one another. Electrons are kept in the orbit around the nucleus by the electromagnetic force, because the nucleus in the center of the atom is positively charged and attracts the negatively charged electrons.

Explanation:

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Consider a loop of wire with a resistor. In the same plane, (with switch S closed) a long wire has a current I flowing from left

Ulleksa [173]

Solution :

The given figure is a loop of a wire with a resistor.

When the switch S is closed for long time and is suddenly opened, the direction of the induced current can be find out by using the rule of right hand screw. According to the right hand screw rule, the direction of the magnetic field at the loop is in the direction that points outwards. The strength of the current rapidly decreases as it is switch off and the magnetic flux that is linked with the loop wire will also decrease.

According to the Lenz's law, the direction of the induced current must be such $\text{that it opposes}$ the decrease in the magnetic flux. It means the direction of the magnetic field must be outwards and also normal to the plane of the screen. The direction of the induced anti clockwise or from right to left in the resistance.

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2 years ago

Physical values in the real world have two components: magnitude and

Nadusha1986 [10]

Answer:

Dimension.

Explanation:

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2 years ago

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Sergio039 [100]

The average rate of change of distance over the time interval
3 ≤ t ≤ 6 represents the coin's average velocity over that interval.

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2 years ago

In 1935, a French destroyer, La Terrible, attained one of the fastest speeds for any standard warship. Suppose it took 3.0 min a

REY [17]

Answer:

Explanation:

From newton's equation of motion of uniform acceleration

v = u + at

where v is final velocity , u is initial velocity , a is acceleration and time is t .

putting the values

v = 0 + .5 x 3 x 60 ( time in second = 3 x 60 s )

= 90 m /s

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3 years ago

(10%) Problem 5: The Earth spins on its axis and also orbits around the Sun. For this problem use the following constants. Mass

11Alexandr11 [23.1K]

Incomplete Question.The Complete question is

The Earth spins on its axis and also orbits around the Sun. For this problem use the following constants. Mass of the Earth: 5.97 × 10^24 kg (assume a uniform mass distribution) Radius of the Earth: 6371 km Distance of Earth from Sun: 149,600,000 km

(i)Calculate the rotational kinetic energy of the Earth due to rotation about its axis, in joules.

(ii)What is the rotational kinetic energy of the Earth due to its orbit around the Sun, in joules?

Answer:

(i) KE= 2.56e29 J

(ii) KE= 2.65e33 J

Explanation:

i) Treating the Earth as a solid sphere, its moment of inertia about its axis is

I = (2/5)mr² = (2/5) * 5.97e24kg * (6.371e6m)²

I = 9.69e37 kg·m²

About its axis,

ω = 2π rads/day * 1day/24h * 1h/3600s

ω= 7.27e-5 rad/s,

so its rotational kinetic energy

KE = ½Iω² = ½ * 9.69e37kg·m² * (7.27e-5rad/s)²

KE= 2.56e29 J

(ii) About the sun,

I = mR²

I= 5.97e24kg * (1.496e11m)²

I= 1.336e47 kg·m²

and the angular velocity

ω = 2π rad/yr * 1yr/365.25day * 1day/24h * 1h/3600s

ω= 1.99e-7 rad/s

KE = ½ * 1.336e47kg·m² * (1.99e-7rad/s)²

KE= 2.65e33 J

6 0

2 years ago