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1 year ago

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

Physics

1 answer:

Vikki [24]1 year 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

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I what is the resistance of

Lynna [10]

The resistance of a conductor is given by:
$R= \frac{\rho L}{A}$
where
$\rho$ is the resistivity of the material
L is the length of the conductor
A is its cross-sectional area

We can use this formula to solve both parts of the problem.

a) The length of the copper wire is L=1.0 m. Its diameter is d=0.50 mm, so its radius is
$r= \frac{d}{2}=0.25 mm=0.25 \cdot 10^{-3} m$
And its cross-sectional area is
$A=\pi r^2 = \pi (0.25 \cdot 10^{-3}m)^2 = 1.96 \cdot 10^{-7} m^2$
The copper resistivity is $\rho=1.68 \cdot 10^{-8} \Omega m$ , therefore the resistance of this piece of wire is
$R= \frac{\rho L}{A}= \frac{(1.68 \cdot 10^{-8} \Omega m)(1.0 m)}{1.96 \cdot 10^{-7} m^2}= 8.57 \cdot 10^{-2} \Omega$

b) The length of this piece of iron is L=10 cm=0.10 m. Its cross-sectional size is L=1.0 mm=0.001 m, so its cross-sectional area is
$A=L^2 = (0.001 m)^2 =1 \cdot 10^{-6}m^2$
The iron resistivity is $\rho = 9.71 \cdot 10^{-8} \Omega m$ , therefore the resistance of this piece of wire is
$R= \frac{\rho L}{A}= \frac{(9.71 \cdot 10^{-8} \Omega m)(0.10 m)}{1.0 \cdot 10^{-6} m^2}=9.71 \cdot 10^{-3} \Omega$

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3 years ago

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Answer:

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2 years ago

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Read 2 more answers

g n diffraction, the formula for minima is given by a times s i n (theta )equals m lambda, where a is the width of the slit, the

ki77a [65]

Answer:

θ = 22.2

Explanation:

This is a diffraction exercise

a sin θ = m λ

The extension of the third zero is requested (m = 3)

They indicate the wavelength λ = 630 nm = 630 10⁻⁹ m and the width of the slit a = 5 10⁻⁶ m

sin θ = m λ / a

sin θ = 3 630 10⁻⁹ / 5 10⁻⁶

sin θ = 3.78 10⁻¹ = 0.378

θ = sin⁻¹ 0.378

to better see the result let's find the angle in radians

θ = 0.3876 rad

let's reduce to degrees

θ = 0.3876 rad (180º /π rad)

θ = 22.2º

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