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irinina [24]
2 years ago
11

Erwin Schrödinger developed the quantum model of the atom. What scientific knowledge was necessary for Schrödinger’s work?

Physics
1 answer:
Vikki [24]2 years ago
6 0

The scientific knowledge was necessary for Schrödinger’s work is the discoverd of electron.

<h3>What is electron and how was it discovered?</h3>

Thomson carried out new experiments that led him to conclude that cathode rays were formed by particles that have a negative charge. Later, Thomson proved that these rays were deflected by applying an electric field. Thus, these particles were called electrons.

The discovery of the electron allowed Schrodinger to explain quantum models that are based on how electrons behave.

See more about quantum model at brainly.com/question/13352063

#SPJ1

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An apple with a mass of 0.95 kilograms hangs from a tree branch 3.0 meters above the ground. If it falls to the ground, what is
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Ke = pe
pe = mgh
= 0.95 x 9.8 x 3
= 27.93 J
6 0
3 years ago
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Forces normal to a particle's displacement do no work.
snow_tiger [21]

Answer:

Explanation:

The work done is defined as the product of force applied in the direction of displacement and the displacement.

W = F x d x Cosθ

where, F is the force applied, d be the displacement and θ be the angle between the displacement and force.

For the normal forces, the angle between the displacement and the force applied is 90 degree, and the value of Cos 90 is zero, so the work done is zero.

3 0
3 years ago
The masses are m1 = m, with initial velocity 2v0, and m2 = 7.4m, with initial velocity v0. Due to the collision, they stick toge
lesya [120]

Answer:

Loss, \Delta E=-10.63\ J

Explanation:

Given that,

Mass of particle 1, m_1=m =0.66\ kg

Mass of particle 2, m_2=7.4m =4.884\ kg

Speed of particle 1, v_1=2v_o=2\times 6=12\ m/s

Speed of particle 2, v_2=v_o=6\ m/s

To find,

The magnitude of the loss in kinetic energy after the collision.

Solve,

Two particles stick together in case of inelastic collision. Due to this, some of the kinetic energy gets lost.

Applying the conservation of momentum to find the speed of two particles after the collision.

m_1v_1+m_2v_2=(m_1+m_2)V

V=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}

V=\dfrac{0.66\times 12+4.884\times 6}{(0.66+4.884)}

V = 6.71 m/s

Initial kinetic energy before the collision,

K_i=\dfrac{1}{2}(m_1v_1^2+m_2v_2^2)

K_i=\dfrac{1}{2}(0.66\times 12^2+4.884\times 6^2)

K_i=135.43\ J

Final kinetic energy after the collision,

K_f=\dfrac{1}{2}(m_1+m_2)V^2

K_f=\dfrac{1}{2}(0.66+4.884)\times 6.71^2

K_f=124.80\ J

Lost in kinetic energy,

\Delta K=K_f-K_i

\Delta K=124.80-135.43

\Delta E=-10.63\ J

Therefore, the magnitude of the loss in kinetic energy after the collision is 10.63 Joules.

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The acceleration of a vertically thrown ball at the top of its path is?
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C between zero and g
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Every 6 seconds a pendulum completes 1 cycle. What are the period and frequency of this pendulum?
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Multiply 6 x 60 to get the frequency.
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