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Schach [20]
4 years ago
13

Changing the length of an air column will alter its

Physics
2 answers:
netineya [11]4 years ago
8 0

The change in the length of air column alters the natural frequency.

<u>Explanation: </u>

Various resonance is obtained by changing the length of air column.

The length of air column is in inverse relation with natural frequency, that is, when air column length is increased, the natural frequency decreases. And, when air column length is decreased, the natural frequency is increased. The source frequency will not change without physically altering the source.

AveGali [126]4 years ago
7 0

Answer:

i think its pressure

Explanation:

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Densidad= Masa / Volumen

Volumen =Masa / Densidad

Es conocido que la densidad del alcohol es 789 kg/m³, y como sabemos por dato que tiene 450gr el siguiente paso es la sustitución:

Volumen=0.45/ 789

Volumen = 0,00057 m³    

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3 years ago
Analyze the chemical reaction below
34kurt

Answer:

Please find the answer in the explanation.

Explanation:

Given that 16 g CH4 + 64 g 02 - 44 g CO2 + 36g H2O

To explain the law of conservation of mass and describe how the equation represents the law of conservation of mass, let me first start from law.

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The equation represents the law of conservation of mass because the mass of molecules at the right hand side is equal to or balance with the molecules at the left hand side. For example, the number of Oxygen, and othe elements are the at both side.

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3 years ago
What are two important uses of petroleum
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Some are fuel,chemicals,plastics,asphalt,and road oil.
5 0
3 years ago
Read 2 more answers
a 15 kg television sits on a shelf at a height of 0.3 m how much graviitional potential energy is added to the television when i
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15x 9.8 x .3 = 44.1
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102.9 J
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3 years ago
A gas is compressed from an initial volume of 5.55 L to a final volume of 1.22 L by an external pressure of 1.00 atm. During the
algol13

Answer:

Explanation:

change in the volume of the gas = 5.55 - 1.22

= 4.33 X 10⁻³ m³

external pressure ( constant ) P = 1 x 10⁵ Pa

work done on the gas

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= 10⁵ x  4.33 X 10⁻³

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433 J

Using the formula

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Given

Q = - 124 J ( heat is released so negative )

W = - 433 J . ( work done by gas is negative, because it is done on gas  )

- 124  = ΔE - 433

ΔE = 433  - 124

= 309 J

There is increase of 309 J in the internal energy of the gas.

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