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2 years ago

Wolfgang pauli hypothesized an exclusion principle. This principle says two electrons in an atom cannot have the same what?.

Physics

1 answer:

PilotLPTM [1.2K]2 years ago

5 0

No two electrons can have the same set of quantum numbers .

<h3>What is Wolfgang Pauli hypothesized an exclusion principle?</h3>

Pauli made a significant advance when he proposed the notion of adding a fourth quantum number to the three that were previously used to represent the quantum state of an electron. Physically speaking, the first three quantum numbers made sense since they had to do with how the electron moved about the nucleus.

The following rule was developed by Austrian physicist Wolfgang Pauli. The quantum numbers of any two electrons cannot be identical.

To put it another way, no two electrons can be in the same state. The Pauli exclusion principle is the name given to this proposition since it forbids electrons from being in the same state.

to learn more about exclusion principle go to - brainly.com/question/90573

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The New England Merchants Bank Building in Boston is 152 mm high. On windy days it sways with a frequency of 0.20 HzHz , and the

kkurt [141]

Answer:

The total distance, side to side, that the top of the building moves during such an oscillation = 31 cm

Explanation:

Let the total side to side motion be 2A. Where A is maximum acceleration.

Now, we know know that equation for maximum acceleration is;

A = α(max) / [(2πf)^(2)]

So 2A = 2[α(max) / [(2πf)^(2)] ]

α(max) = (0.025 x 9.81) while frequency(f) from the question is 0.2Hz.

Therefore 2A = 2 [(0.025 x 9.81) / [((2π(0.2)) ^(2)] ] = 2( 0.245 / 1.58) = 0.31m or 31cm

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3 years ago

how much voltage is in the primary coil if there are 3200 winding in the secondary coil 500 winding in the primary and 25v in th

Tema [17]

Answer:

Voltage in the primary winding of the coil is 3.9 V

Explanation:

As we know by the principle of transformer we have

$\frac{V_s}{V_p} = \frac{N_s}{N_p}$

here we know that

$V_s = 25 volts$

$N_s = 3200$

$N_p = 500$

now we have

$\frac{25}{V_p} = \frac{3200}{500}$

$V_p = 3.9 Volts$

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3 years ago

Charge Q is distributed uniformly throughout the volume of an insulating sphere of radius R = 4.00 cm. At a distance of r = 8.00

Elena L [17]

Answer:

$2.62898\times 10^{-6}\ C/m^3$

$1979.99974\ N/C$

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Q = Charge

r = Distance = 8 cm

R = Radius = 4 cm

Electric field is given by

$E=\dfrac{kQ}{r^2}\\\Rightarrow Q=\dfrac{Er^2}{k}\\\Rightarrow E=\dfrac{990\times 0.08^2}{8.99\times 10^{9}}\\\Rightarrow Q=7.04783\times 10^{-10}\ C$

Volume charge density is given by

$\sigma=\dfrac{Q}{\dfrac{4}{3}\pi R^3}\\\Rightarrow \sigma=\dfrac{7.04783\times 10^{-10}}{\dfrac{4}{3}\pi (0.04)^3}\\\Rightarrow \sigma=2.62898\times 10^{-6}\ C/m^3$

The volume charge density for the sphere is $2.62898\times 10^{-6}\ C/m^3$

$E=\dfrac{kQr}{R^3}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 7.04783\times 10^{-10}\times 0.02}{0.04^3}\\\Rightarrow E=1979.99974\ N/C$

The magnitude of the electric field is $1979.99974\ N/C$

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3 years ago

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Maksim231197 [3]

Explanation:

This is true because at maximum height, the velocity is 0

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3 years ago

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3 years ago