(a)
Electronic configuration is given as follows:
![[Kr]4d^{3}](https://tex.z-dn.net/?f=%5BKr%5D4d%5E%7B3%7D)
Since, this is the electronic configuration of ion with+3 that means 3 electrons are removed. On adding the 3 electrons, the electronic configuration of neutral atom can be obtained.
Thus, electronic configuration of neutral atom is ![[Kr]4d^{5}5s^{1}](https://tex.z-dn.net/?f=%5BKr%5D4d%5E%7B5%7D5s%5E%7B1%7D)
.
The atomic number of Kr is 36, thus, total number of electrons become 36+6=42.
This corresponds to element: molybdenum. Thus, the tripositive atom will be 
.
(b) The given electronic configuration is ![[Kr]5s^{2}4d^{2}](https://tex.z-dn.net/?f=%5BKr%5D5s%5E%7B2%7D4d%5E%7B2%7D)
.
The atomic number of Kr is 36, thus, total number of electrons become 36+4=40.
This corresponds to element zirconium, represented by symbol Zr.
5.625 hours and it is 450 divided by 80
Have A Good Day
Answer:
v = -v₀ / 2
Explanation:
For this exercise let's use kinematics relations.
Let's use the initial conditions to find the acceleration of the electron
v² = v₀² - 2a y
when the initial velocity is vo it reaches just the negative plate so v = 0
a = v₀² / 2y
now they tell us that the initial velocity is half
v’² = v₀’² - 2 a y’
v₀ ’= v₀ / 2
at the point where turn v = 0
0 = v₀² /4 - 2 a y '
v₀² /4 = 2 (v₀² / 2y) y’
y = 4 y'
y ’= y / 4
We can see that when the velocity is half, advance only ¼ of the distance between the plates, now let's calculate the velocity if it leaves this position with zero velocity.
v² = v₀² -2a y’
v² = 0 - 2 (v₀² / 2y) y / 4
v² = -v₀² / 4
v = -v₀ / 2
We can see that as the system has no friction, the arrival speed is the same as the exit speed, but with the opposite direction.
A, it's better to take a break to let your body rejuvenate.
Let us consider two bodies having masses m and m' respectively.
Let they are separated by a distance of r from each other.
As per the Newtons law of gravitation ,the gravitational force between two bodies is given as - 
where G is the gravitational force constant.
From the above we see that F ∝ mm' and 
Let the orbital radius of planet A is 
= r and mass of planet is 
.
Let the mass of central star is m .
Hence the gravitational force for planet A is 
For planet B the orbital radius 
and mass 
Hence the gravitational force 
![f_{2} =G\frac{m*3m_{1} }{[2r_{1}] ^{2} }](https://tex.z-dn.net/?f=f_%7B2%7D%20%3DG%5Cfrac%7Bm%2A3m_%7B1%7D%20%7D%7B%5B2r_%7B1%7D%5D%20%5E%7B2%7D%20%7D)
Hence the ratio is 
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