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Salsk061 [2.6K]
3 years ago
14

The second ionization energy is the energy required to remove the _______ Electron after a __________ one has been removed.

Physics
1 answer:
Sindrei [870]3 years ago
6 0

The second ionization energy is the energy required to remove the <u>second </u>electron after a <u>valence</u> one has been removed.

<h3><u>Explanation:</u></h3>

For an element, the first ionization energy is defined as the amount of energy required to remove one electron from the outermost valence shell of a neutral atom. Removing one electron increases the number of protons, making it a 1+ ion.  

The nucleus (protons) has more bonding to the electrons with negative charge and thus more energy is required if another electron needs to be removed. This higher energy required to remove second electron from a 1+ ion (after the first one has been removed) is termed as the second ionization energy. Second ionization energy leads to formation of a 2+ ion. Similarly, third ionization energy is higher than second ionization energy.

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The object's speed will not change.

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A 6.0-kilogram cart initially traveling at 4.0 meters per second east accelerates uniformly at 0.50 meter per second squared eas
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3 0
2 years ago
Using energy considerations, calculate the average force (in N) a 62.0 kg sprinter exerts backward on the track to accelerate fr
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Answer:

69.68 N

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Work done is equal to change in kinetic energy

W = ΔK = Kf - Ki = \frac{1}{2} mv^{2} _{f}  - \frac{1}{2} mv^{2} _{i}

W = F_{total} .d

where m = mass of the sprinter

vf = final velocity

vi = initial velocity

W  = workdone

kf = final kinetic energy

ki = initial kinetic energy

d = distance traveled

Ftotal = total force

vf = 8m/s

vi= 2m/s

d = 25m

m = 60kg

inserting parameters to get:

W = ΔK = Kf - Ki = \frac{1}{2} mv^{2} _{f}  - \frac{1}{2} mv^{2} _{i}

F_{total} .d =\frac{1}{2} mv^{2} _{f}  - \frac{1}{2} mv^{2} _{i}

F_{total} = \frac{\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}}{d}

F_{total=} \frac{\frac{1}{2} X 62 X6^{2} -\frac{1}{2} X 62 X2^{2} }{25}

= 39.7

we know that the force the sprinter exerted F sprinter, the force of the headwind Fwind = 30N

F_{sprinter} = F_{total} + F_{wind}  = 39.7 + 30 = 69.68 N

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