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3 years ago

I have to write an essay about Quantum Mechanics. I need to know everything about it. Please Help!

Physics

2 answers:

Studentka2010 [4]3 years ago

6 0

ananı var ya ananı

kodumun abd lisi aptal cajhil özürlü

Neporo4naja [7]3 years ago

5 0

Quantum mechanics means dealing with physics at a very small level. Or u can say microscopic level. Studying physics at sub atomic level

Study regarding protons, electrons and quarks include in this study.

Also the laws of physics at normal level fail in quantum mechanics.

Laws like uncertainty principal of Heisenberg, wave nature, are there

(So if u want to write an essay, first u can tell about) the difference between normal sized objects and the subatomic ones. And how physics act upon them

Then u can write about how laws fail for microscopic things so we need to study quantum mechanics.

And elaborate every paragraph like i ve written in points.

hope that will make ur essay complete ♥

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2 years ago

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Stells [14]

Answer:

422.36 N

Explanation:

given,

time of rotation = 4.30 s

T = 4.30 s

Assuming the diameter of the ring equal to 16 m

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$v = \dfrac{2\pi\times 8}{4.30}$

v = 11.69 m/s

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$- N - m g = \dfrac{-mv^2}{R}$

$N + m g = \dfrac{mv^2}{R}$

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6 0

2 years ago

8. two +1 C charges are separated by 3000m. What is the magnitude of the electric force between them?

Sidana [21]

Answer:

1000 N

Explanation:

The magnitude of the electrostatic force between two charged object is given by

$F=k\frac{q_1 q_2}{r^2}$

where

k is the Coulomb constant

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r is the distance between the two objects

Moreover, the force is:

- Attractive if the two forces have opposite sign

- Repulsive if the two forces have same sign

In this problem:

$q_1=q_2=+1C$ are the two charges

r = 3000 m is their separation

Therefore, the electric force between the charges is:

$F=(9\cdot 10^9)\frac{(1)(1)}{3000^2}=1000 N$

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3 years ago

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Lapatulllka [165]

Answer:

The speed of the electron is $2.55\times 10^3\ m/s$ .

Explanation:

Given that,

The magnitude of electric field, $E=1.32\ kV/m=1.32\times 10^3\ V/m$

The magnitude of magnetic field, B = 0.516 T

Both the magnetic and electric fields are acting on the moving electron. Then, the magnitude of electric field and magnetic field is balanced such that :

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$v=2.55\times 10^3\ m/s$

So, the speed of the electron is $2.55\times 10^3\ m/s$ . Hence, this is the required solution.

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3 years ago

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2 years ago