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2 years ago

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

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75 POINTS AND BRAINLY

kramer

Answer:

B=work=power/time

Explanation:

8 0

3 years ago

Read 2 more answers

A bucket of water of mass 14.0 kg is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.260

ruslelena [56]

Answer:

The tension in the rope is 41.38 N.

Explanation:

Given that,

Mass of bucket of water = 14.0 kg

Diameter of cylinder = 0.260 m

Mass of cylinder = 12.1 kg

Distance = 10.7 m

Suppose we need to find that,

What is the tension in the rope while the bucket is falling

We need to calculate the acceleration

Using relation of torque

$\tau=F\times r$

$I\times\alpha=F\times r$

Where, I = moment of inertia

$\alpha$ = angular acceleration

$\dfrac{Mr^2}{2}\times\dfrac{a}{r}=F\times r$

$F=\dfrac{M}{2}a$ ...(I)

Here, F = tension

The force is

$F=m(g-a)$ ...(II)

Where, F = tension

a = acceleration

From equation (I) and (II)

$\dfrac{M}{2}a=m(g-a)$

$a=\dfrac{g}{1+\dfrac{M}{2m}}$

Put the value into the formula

$a=\dfrac{9.8}{1+\dfrac{12.1}{2\times14.0}}$

$a=6.84\ m/s^2$

We need to calculate the tension in the rope

Using equation (I)

$F=\dfrac{M}{2}a$

Put the value into the formula

$F=\dfrac{12.1}{2}\times6.84$

$F=41.38\ N$

Hence, The tension in the rope is 41.38 N.

6 0

3 years ago

an object near the surface of a planet falls 54m in 3s. what is the acceleration due to gravity on that planet

Harman [31]

Answer:

12s

Explanation:

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3 years ago

A small object with momentum 7.0 kg∙m/s approaches head-on a large object at rest. The small object bounces straight back with a

EastWind [94]

Answer:

The magnitude of the large object's momentum change is 3 kilogram-meters per second.

Explanation:

Under the assumption that no external forces are exerted on both the small object and the big object, whose situation is described by the Principle of Momentum Conservation:

$p_{S,1}+p_{B,1} = p_{S,2}+p_{B,2}$ (1)

Where:

$p_{S,1}$ , $p_{S,2}$ - Initial and final momemtums of the small object, measured in kilogram-meters per second.

$p_{B,1}$ , $p_{B,2}$ - Initial and final momentums of the big object, measured in kilogram-meters per second.

If we know that $p_{S,1} = 7\,\frac{kg\cdot m}{s}$ , $p_{B,1} = 0\,\frac{kg\cdot m}{s}$ and $p_{S, 2} = 4\,\frac{kg\cdot m}{s}$ , then the final momentum of the big object is: