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Snowcat [4.5K]
3 years ago
13

How are vibrations different between bigger sizes rubber bands and smaller sized rubber bands?

Chemistry
1 answer:
arlik [135]3 years ago
5 0
Assuming the kind of vibration you are talking about is the kind where you stretch the rubber band between two points and then "twang" it, then the answer is fairly complex. What happens when you cause the vibrations to start is you make something called a "standing wave". In a standing wave, each particle in the rubber band has a certain amount of energy which causes it to move backwards and forwards, the particles with more energy have a larger "amplitude" (how much they move), and of course the particles with less energy have a smaller amplitude. Now a standing wave has two main components: The amplitude, and the frequency. The amplitude of the whole wave refers to the largest amplitude any particles has. The frequency refers to how often it takes for one of the particles to move between the two furthest away points it can be.
To compare rubber bands, you must remember to keep certain things constant. If you're looking at their vibrations, the amount of energy you use to "twang" the rubber band should be the same each time you twang it (which is the same as applying the same force each time you twang it).

A larger rubber band has more area over which to spread the energy, as well as it has more mass for the energy to move, so the vibrations will have smaller amplitudes, and smaller frequencies, overall vibrating less and with smaller vibrations. 
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What is bias in an experiment?
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○ c. ○

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Predict the products for the following equation.
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A 250 mL sample of gas is collected over water at 35°C and at a total pressure of 735 mm Hg. If the vapor pressure of water at 3
ELEN [110]

Answer:

The volume of the gas sample at standard pressure is <u>819.5ml</u>

Explanation:

Solution Given:

let volume be V and temperature be T and pressure be P.

V_1=250ml

V_2=?

P_{total}=735 mmhg

1 torr= 1 mmhg

42.2 torr=42.2 mmhg

so,

P_{water}=42.2mmhg

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Now

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Now

By using combined gas law equation:

\frac{P_1*V_1}{T_1} =\frac{P_2*V_2}{T_2}

V_2=\frac{P_1*}{P_2}*\frac{T_2}{T_1} *V_1

V_2=\frac{P_gas}{P_2}*\frac{T_2}{T_1} *V_1

Here P_2 \:and\: T_2 are standard pressure and temperature respectively.

we have

P_2=750mmhg \:and\: T_2=273K

Substituting value, we get

V_2=\frac{692.8}{750}*\frac{273}{308} *250

V_2= 819.51 ml

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1 year ago
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