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lions [1.4K]
1 year ago
5

Consider the two electron arrangements for neutral atoms a and b. Are atoms a and b the same element? a - 1s2, 2s2, 2p6, 3s1 b -

1s2, 2s2, 2p6, 5s1.
Physics
1 answer:
Agata [3.3K]1 year ago
4 0

The two neutral atoms A and B have the same number of electrons and atomic number 11. So, the two elements are said to be same.

The electronic configuration of the element is the arrangement of the electrons in the atom of the element in energy levels, orbitals around the nucleus.

The electrons in the atoms of the element with lowest energy are written first before those with higher energy levels. Thus, the electronic configuration shows the electrons in the atoms of the element arranged in order of increasing energies.

The electronic configuration of atoms are given as

A = 1s² 2s² 2p⁶ 3s¹

B = 1s² 2s² 2p⁶ 5s¹

The number of electrons in both the elements is 11. Therefore, their atomic number is also the same i.e, 11. So, both the elements are the same.

To know more about atomic number:

brainly.com/question/8645622

#SPJ4

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3 years ago
008 (part 1 of 3) 10.0 points A 3.2 kg object is subjected to two forces, F~ 1 = (1.9 N) ˆı + (−1.9 N) ˆ and F~ 2 = (3.8 N) ˆı
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Answer:

a' =4.15 m/s²

Explanation:

Given that

m= 3.2 kg

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From second law of Newton's

F(net) = m a

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1.9  i −1.9 j + 3.8 i −10.1 j  = 3.2 a

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4 years ago
Consider the following situation, in which a compass is influenced by not only the force of Earth's magnetic field, but also an
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Answer:

1) The angle of deflection will be less than 45° ( C )

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Explanation:

1) Assuming that the force applied  has a direction which is perpendicular to the Earth's magnetic field

∴ Fearth > Fapplied   hence the angle of deflection will be < 45°

2) when the Fearth < Fapplied

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Which form of energy is related to the movement of charged particles?
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In a ballistic pendulum an object of mass m is fired with an initial speed v0 at a pendulum bob. The bob has a mass M, which is
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Answer:

The expression for the initial speed of the fired projectile is:

\displaystyle v_0=\frac{M+m}{m}(2gL[1-cos(\theta)]^{\frac{1}{2}})

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Explanation:

For the expression for the initial speed of the projectile, we can separate the problem in two phases. The first one is the moment before and after the impact. The second phase is the rising of the ballistic pendulum.

First Phase: Impact

In the process of the impact, the net external forces acting in the system bullet-pendulum are null. Therefore the linear momentum remains even (Conservation of linear momentum). This means:

P_0=P_f\\v_0m=v_i(m+M)\\v_0=v_i\frac{m+M}{m}  (1)

Second Phase: pendular movement

After the impact, there isn't any non-conservative force doing work in al the process. Therefore the mechanical energy remains constant (Conservation Of Mechanical Energy). Therefore:

Em_i=Em_f\\\frac{1}{2}mv^2_i=mgH\\v_i=[2gH]^\frac{1}{2}  (2)

The height of the pendulum respect L and θ is:

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Using equations (1),(2) and (3):

\displaystyle v_0=\frac{M+m}{m}(2gL[1-cos(\theta)]^{\frac{1}{2}}) (4)

The initial speed ratio for the 9.0mm/44-caliber bullet is obtained using equation (4):

\displaystyle \frac{v_{9mm}}{v_{44}} =\frac{(M+m_{9mm})m_{44}}{(M+m_{44})m_{9}}(\frac{1-cos(\theta_{9mm})}{1-cos(\theta_{44})} )^{\frac{1}{2}}=1.773

5 0
3 years ago
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