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2 years ago

The ionization energy of singly charged anions _ as electron affinity of the neutral atom _

Physics

1 answer:

eimsori [14]2 years ago

3 0

Answer:

The ionization energy of an atom is the required amount of energy which is provided to the atom, so, that the required number of electrons are removed from outermost shells of the electronic configuration of the atom. It is minimum amount of energy that is required by the atom, so, that it can remove the charged particle from its outer shell.
The ionization energy of singly charged anions <u>less same</u> as electron affinity of the neutral atom for <u>the electrons.</u>

As, for the anion having the net charge on its body will have larger size then any other normal atom, as the atom will have large and it will be difficult for the nucleus to hold off its whole electronic configuration into one place. This process makes it more easy for the foreign source or agent to remove the required number of charged particles from the atom's outer region.
While it may be same as the phenomenon which occurs inside the neutral atom's whole structure, as there is less affinity for the electrons and that is how the two of them are termed as inter related to one another in the mechanism of having the less affinity for the electrons.

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evaluate the numerical value of the vertical velocity of the car at time t=0.25 s using the expression from part d, where y0=0.7

likoan [24]

Given :

Displacement , y = 0.75 m .

Angular acceleration , $\alpha=0.95\ s^{-2}$ .

Initial angular velocity , $\omega_o=6.3\ s^{-1}$ .

To Find :

The value of vertical velocity after time t = 0.25 s .

Solution :

By equation of circular motion is given by :

$\omega=\omega_o+\alpha t$

Putting all given values we get :

$\omega=6.3+0.95\times 0.25\\\\\omega= $$6.5375\ s^{-1}$

Now , vertical velocity is given by :

$v=y\omega\\\\v=0.75\times 6.5375\ m/s\\\\v=4.90\ m/s$

Therefore , the numerical value of the vertical velocity of the car at time t=0.25 s is 4.90 m/s .

Hence , this is the required solution .

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2 years ago

An object on a number line moved from x = 15 cm to x = 165 cm and then

olasank [31]

Answer:

v_avg = 2.9 cm/s

Explanation:

The average velocity of the object is the sum of the distance of all its trajectories divided the time:

$v_{avg}=\frac{x_{all}}{t}$

x_all is the total distance traveled by the object. In this case you have that the object traveled in the first trajectory 165cm-15cm = 150cm, and in the second one, 165cm - 25cm = 140cm

Then, x_all = 150cm + 140cm = 290cm

The average velocity is, for t = 100s

$v_{avg}=\frac{290cm}{100s}=2.9\frac{cm}{s}$

hence, the average velocity of the object in the total trajectory traveled is 2.9 cm/s

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1 year ago

Read 2 more answers

The fastest pitched baseball was clocked at 47 m/s. Assume that the pitcher exerted his force (assumed to be horizontal and cons

arsen [322]

Answer:

Explanation:

F ×1 = 0.5×0.145×47×47

F = 160.15 N

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2 years ago

Find the net electric force that the two charges would exert on an electron placed at point on the xx-axis at xx = 0.200 mm. Exp

UkoKoshka [18]

Answer:

The question has some details missing, here is the complete question ; A -3.0 nC point charge is at the origin, and a second -5.0nC point charge is on the x-axis at x = 0.800 m. Find the net electric force that the two charges would exert on an electron placed at point on the x-axis at x = 0.200 m.

Explanation:

The application of coulonb's law is used to approach the question as shown in the attached file.

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1 year ago

The mass of the Sun is 2 × 1030 kg, and the distance between Neptune and the Sun is 30 AU. What is the orbital period of Neptune

Veronika [31]

Kepler's third law states that, for a planet orbiting around the Sun, the ratio between the cube of the radius of the orbit and the square of the orbital period is a constant:
$\frac{r^3}{T^2}= \frac{GM}{4 \pi^2}$ (1)
where
r is the radius of the orbit
T is the period
G is the gravitational constant
M is the mass of the Sun

Let's convert the radius of the orbit (the distance between the Sun and Neptune) from AU to meters. We know that 1 AU corresponds to 150 million km, so
$1 AU = 1.5 \cdot 10^{11} m$
so the radius of the orbit is
$r=30 AU = 30 \cdot 1.5 \cdot 10^{11} m=4.5 \cdot 10^{12} m$

And if we re-arrange the equation (1), we can find the orbital period of Neptune:
$T=\sqrt{ \frac{4 \pi^2}{GM} r^3} = \sqrt{ \frac{4 \pi^2}{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2} )(2\cdot 10^{30} kg)}(4.5 \cdot 10^{12} m)^3 }= 5.2 \cdot 10^9 s$

We can convert this value into years, to have a more meaningful number. To do that we must divide by 60 (number of seconds in 1 minute) by 60 (number of minutes in 1 hour) by 24 (number of hours in 1 day) by 365 (number of days in 1 year), and we get
$T=5.2 \cdot 10^9 s /(60 \cdot 60 \cdot 24 \cdot 365)=165 years$

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1 year ago

The ionization energy of singly charged anions _____ as electron affinity of the neutral atom _____

The ionization energy of singly charged anions _ as electron affinity of the neutral atom _