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GaryK [48]
3 years ago
14

The addition of electron shells results in shielding of electrons from the nucleus.

Physics
1 answer:
Pani-rosa [81]3 years ago
4 0

Answer:

Addition of shells increases the distance of outer electrons from the nucleus.

Explanation:

Shielding effect is known as the attraction between the nucleus and an electron of any atom. In other words, it is the reduction in effective nuclear charge on an electron cloud.

Addition of electron shells results in the shielding of electron from nucleus.  As the number of electron shells increases then farther will be the electrons placed from the nucleus and hence it will become easier to expel the electrons from outer shells with only little amount of ionization energy.

So, the amount of ionization energy require will be indirectly proportional to the shielding effect because more the shielding of electrons from the nucleus less will be the ionization energy require to expel the electrons.

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Two cars, one of mass 1300 kg, and the second of mass 2400 kg, are moving at right angles to each other when they collide and st
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To solve this problem we will apply the momentum conservation theorem, that is, the initial momentum of the bodies must be the same final momentum of the bodies. The value that will be obtained will be a vector value of the final speed of which the magnitude will be found later. Our values are given as,

m_1 = 1300kg

m_2 = 2400kg

u_1 = 12m/s i

u_2 = 18m/s j

Using conservation of momentum,

m_1u_1+m_2u_2 = (m_1+m_2)v_f

1300*12i-2400*18j = (1300+2400)v_f

Solving for v_f

v_f = 4.2162i-11.6756j

Using the properties of vectors to find the magnitude we have,

|v| = \sqrt{(4.2162^2)+(-11.6756)^2}

|v| = 12.4135m/s

Therefore the magnitude of the velocity of the wreckage of the two cars immediately after the collision is 12.4135m/s

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2 years ago
While standing at the edge of the roof of a building, a man throws a stone upward with an initial speed of 6.63 m/s. The stone s
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Time=Distance/Speed

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An 80- kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15m/s . Sort the fol
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Answer:

1. The mass of the quarterback = known = 80 kg

2. The mass of the football = known = 0.43 kg

3. The horizontal velocity of quarterback before throwing the ball = known = 0

4. The horizontal velocity of football before being thrown = known = 0

5. The horizontal velocity of quarterback after throwing the ball, = unknown quantity and it can be calculated using the conservation of linear momentum.

6. The horizontal velocity of football after being thrown = known = 15 m/s

Explanation:

Given that,

Mass of the quarterback, m = 80 kg

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Speed of the football, v' = 15 m/s

We need to sort the following quantities as known or unknown.

1. The mass of the quarterback = known = 80 kg

2. The mass of the football = known = 0.43 kg

3. The horizontal velocity of quarterback before throwing the ball = known = 0

4. The horizontal velocity of football before being thrown = known = 0

5. The horizontal velocity of quarterback after throwing the ball, = unknown quantity and it can be calculated using the conservation of linear momentum.

6. The horizontal velocity of football after being thrown = known = 15 m/s

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