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3 years ago

13

When an electron in a certain excited energy level in a one-dimensional box of length 2.00 Å makes a transition to the ground st

ate, a photon of wavelength 8.79 nm is emitted. Find the quantum number of the initial state?

1 answer:

Fiesta28 [93]3 years ago

5 0

Answer:

Calculate the wavelength associated with an electron with energy 2000 eV.

Sol: E = 2000 eV = 2000 × 1.6 × 10–19 J

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How much work does a 65 kg gymnast have to do to climb a rope to a height of 7? meters? PLEASE HURRY ITS A TEST

Karo-lina-s [1.5K]

How much force did the gymnast use to climb the rope?

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3 years ago

Questions 14 out of 20

KiRa [710]

Answer:

B

Explanation:

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4 years ago

A ball is dropped from the rest from a high window of a tall building and called for 4 seconds. Neglecting air resistance, the f

diamong [38]

Answer:

-39.2m/s

Explanation:

Given that :

t = 4secs

g = -9.8m/s^2

v = ?

u = 0m/s ( since it was at rest )

V = u +at............. 1

Where v is the final velocity

a = -g = -9.8m/s^2 since the ball was dropped from a height which will eventually make it move against gravity

t = 4secs

Substitute the values into 1

v = 0 - 9.8×4

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4 years ago

Emily holds a banana of mass m over the edge of a bridge of height h. She drops the banana and it falls to the river below. Use

Sladkaya [172]

Answer:

The speed of the banana just before it hits the water is:

√(2 · g · h) = v

Explanation:

Hi there!

Before Emily throws the banana, its potential energy is:

PE = m · g · h

Where:

PE = potential energy.

m = mass of the banana.

g = acceleration of the banana due to gravity.

h = height of the bridge (distance from the bridge to the ground).

When the banana reaches the water, all its potential energy will have converted to kinetic energy. The equation for kinetic energy is as follows:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass of the banana.

v = speed.

Then, when the banana hits the water:

m · g · h = 1/2 · m · v²

multiply by 2 and divide by m both sides of the equation:

2 · g · h = v²

√(2 · g · h) = v

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3 years ago

A ball is projected into the air with 100 j of kinetic energy which is transformed to gravitational potential energy at the top

raketka [301]

<span>when it returns to its original level after encountering air resistance, its kinetic energy is decreased.
In fact, part of the energy has been dissipated due to the air resistance.

The mechanical energy of the ball as it starts the motion is:
</span> $E=K = 100 J$
<span>where K is the kinetic energy, and where there is no potential energy since we use the initial height of the ball as reference level.
If there is no air resistance, this total energy is conserved, therefore when the ball returns to its original height, the kinetic energy will still be 100 J. However, because of the presence of the air resistance, the total mechanical energy is not conserved, and part of the total energy of the ball has been dissipated through the air. Therefore, when the ball returns to its original level, the kinetic energy will be less than 100 J.</span>

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4 years ago