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IgorC [24]
3 years ago
8

What does the Pauli exclusion principle state? . . A.Electrons will be added to an atom at the lowest possible level or sublevel

. . . . . B. An orbital can contain two electrons only if all other orbitals at that sublevel are empty.. . . . C.An orbital can contain two electrons only if all other orbitals at that sublevel contain at least one electron.. . . . D.Two electrons occupy the same orbital only if they have opposite spins. . . . . E.An orbital can contain two electrons only if all other orbitals at that sublevel are empty..
Physics
1 answer:
BARSIC [14]3 years ago
5 0
The Pauli exclusion principle state that : D. Two electrons occupy the same orbital only if they have opposite spins

This happen because he stated that in an atom or molecule, two electrons CANNOT have same four electronic quantum numbers

hope this helps
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Concept Simulation 2.3 offers a useful review of the concepts central to this problem. An astronaut on a distant planet wants to
Alexandra [31]

Answer:

1.40 m/s^2

Explanation:

Given data

Velocity= 17.4 m/s

time= 12.4 seconds

We want to find the acceleration of the rock

We know that

acceleration = velocity/time

Substitute

acceleration= 17.4/12.4

acceleration=1.40 m/s^2

Hence the acceleration is 1.40 m/s^2

7 0
3 years ago
A runner starts at point A, runs around a 1-mile track and finishes the run back at point A. Which of the following statements i
Lyrx [107]

This question is incomplete because the options are missing; here is the complete question:

A runner starts at point A, runs around a 1-mile track, and finishes the run back at point A. Which of the following statements is true?

A. The runner's displacement is 1 mile.

B. The runner's displacement is zero.

C. The distance the runner covered is zero.

D. The runner's speed was zero.

The answer to this question is B. The runner's displacement is zero

Explanation:

Displacement always implies a change of position; this means an object or individual moves from point A to point B, and therefore the original position is different from the final position. Additionally, in displacement, other related factors such as the total distance the body moved and the direction of movement. In the case presented, it can be concluded there was no displacement or the displacement is zero because even when the runner moved and ran two miles, he returned to the initial position, and without a change in the position, there is no displacement.

4 0
3 years ago
A roller coaster is going through a loop with a circular radius of 9.1 meters, how fast must the roller coaster be moving at the
Verdich [7]

Answer:

at least 20mph

Explanation:

8 0
3 years ago
The inclusion of the scientific method in psychology is so important because it
Gekata [30.6K]

I know this question is from 2017 but I am doing this for anyone else that is looking for a answer. The answer should be C.) Allowed psychology to be considered a science.

4 0
3 years ago
Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L
antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}

Similarly, for rod 2

\alpha_2 = \dfrac{\tau_2}{I_2}.

Now, the moment of inertia for rod 1 is

I_1 = \dfrac{1}{3}ML^2,

and the torque acting on it is (about the center of mass)

\tau_1 = Mg\dfrac{L}{2};

therefore, the angular acceleration of rod 1 is  

\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},

\boxed{\alpha_1 = \dfrac{3g}{2L} }

Now, for rod 2 the moment of inertia is

I_2 = \dfrac{1}{3}(2M)(2L)^2

I_2 = \dfrac{8}{3} ML^2,

and the torque acting is (about the center of mass)

\tau _2 = (2M)g \dfrac{(2L)}{2}

\tau _2 = 2MgL;

therefore, the angular acceleration \alpha_2 is

\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.

\boxed{\alpha_2 = \dfrac{3g}{4L}}

We see here that

\dfrac{3g}{2L} > \dfrac{3g}{4L}

therefore

\boxed{\alpha_1 > \alpha_2.}

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

7 0
3 years ago
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