What is the least possible initial kinetic energy kmin the oxygen atom could have and still excite the cesium atom?

1 answer:

5 0

If perfectly elastic there would be no energy left over for exciting the atom. if the collision were partially elastic, then some of the initial kinetic energy would be converted into internal energy,

hope this help

You might be interested in

Which is the hottest planet in the solar system

polet [3.4K]

Answer:

venus

Explanation:

6 0

3 years ago

Read 2 more answers

Competition in the Los Angeles Flower District results in

stiks02 [169]

Competition in the Los Angeles Flower District results in better quality flowers.

<h3>Why quality is the standard in flower competition?</h3>

Competition results in better quality flowers because in the competition, best quality of flowers will be selected as a winner so the competitors produces best quality of flowers in order to claim the prize so we can conclude that Competition in the Los Angeles Flower District results in better quality flowers

Learn more about competition here: brainly.com/question/25605883

8 0

3 years ago

An exoplanet with one half of Earth's mass and 50% of Earth's radius is discovered.

Georgia [21]

Answer:

The space cadet that weighs 800 N on Earth will weigh 1,600 N on the exoplanet

Explanation:

The given parameters are;

The mass of the exoplanet = 1/2×The mass of the Earth, M = 1/2 × M

The radius of the exoplanet = 50% of the radius of the Earth = 1/2 × The Earth's radius, R = 50/100 × R = 1/2 × R

The weight of the cadet on Earth = 800 N

$The \ weight, W =G\dfrac{M \times m}{R^{2}} = 800 \ N$

Therefore, for the weight of the cadet on the exoplanet, W₁, we have;

$W_1 =G\dfrac{\dfrac{M}{2} \times m}{ \left ( \dfrac{R}{2} \right ) ^{2}} = G\dfrac{\dfrac{M}{2} \times m \times 4}{ R ^{2}} = 2 \times G \times \dfrac{M \times m}{R^{2}} = 2 \times 800 \, N = 1,600 \, N$

The weight of a space cadet on the exoplanet, that weighs 800 N on Earth = 1,600 N.

7 0

3 years ago

An Olympic swimmer swims 50.0 meters in 23.1 seconds . What is his average speed

Aneli [31]

If you take 50 meters and divide by 23.1 seconds, you will get 2.16 meters per second.

So his average speed is 2.16 m/s.

3 0

3 years ago

A surfer is moving at a constant velocity of 5.2 m/s north relative to a wave which is pushing him west at a constant velocity o

Vika [28.1K]

The speed and distances are directly proportional. Use ratios to solve for vertical y-distance. The ratio of x-distance west to y-distance north equals the x-velocity to y-velocity.

x/y = vx/vy
41/y = 8.6/5.2
41/y = 1.65
41/1.65 = y
24.8 m = y

5 0

3 years ago