It is very camouflaged this way it will survive longer then most other animals
Test his theory like a model theory or research more common ideas
Answer:
9.17 atm
Explanation:
To find the new pressure of the gas, you need to use the following manipulated formula:
P₁V₁ / T₁ = P₂V₂ / T₂
In this formula,
P₁ = initial pressure (atm) P₂ = new pressure (atm)
V₁ = initial volume (L) V₂ = new volume (L)
T₁ = initial temperature (K) T₂ = new temperature (K)
Because you have been given values for all of the variables except for the new pressure, you can substitute them into the equation and simplify.
P₁ = 4.0 atm P₂ = ? atm
V₁ = 5.5 L V₂ = 2.0 L
T₁ = 300 K T₂ = 250 K
P₁V₁ / T₁ = P₂V₂ / T₂ <----- Given formula
(4.0 atm)(5.5 L) / (300 K) = P₂(2.0 L) / (250 K) <----- Insert variables
0.073333 = P₂(2.0 L) / (250 K) <----- Simplify left side
18.33333 = P₂(2.0 L) <----- Multiply both sides by 250
9.17 = P₂ <----- Divide both sides by 2.0
Answer:
<em>By showing that changing the frequency of light causes the emission of faster electrons.
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Explanation:
<em>The photoelectric effect happens when light strikes a metal surface causing the emission of electrons from it (photoelectrons).
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<em>If you increase the intensity of the light you get, as acresult, more electrons emitted but their kinetic energy does not increase.
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<em>If you increase the frequency of the incident light the number of photoelectrons emitted does not increase while the velocity, and so their kinetic energy, increases...the emitted electrons are more...energetic!
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<em>This can be explained considering the incident light as a shower of particle-like packets of energy (photons); if you increase the intensity you simply increase the number of packets (all with the same energy) hitting the metal; these can be used by a lot of electrons to escape.
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<em>On the other hand if you increase the frequency the number of packets remains the same (emitting fewer electrons perhaps) but the energy carried by each of them increases.
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<em>Each packet carries an energy directly proportional to the frequency.</em>
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