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3 years ago

The normal boiling point of cyclohexane is 81.0 ℃. what is the vapor pressure of cyclohexane at 81.0 ℃?

1 answer:

6 0

The boiling point is defined as the temperature at which the pressure of the vapor of the liquid is equivalent to the external atmospheric pressure surrounding the liquid. Therefore, the boiling point of the liquid is dependent on the atmospheric pressure.

Based on this, the vapor pressure of cyclohexane at 81 degrees celcius will be equal to atmospheric pressure (based on barometric readings)

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If the second harmonic of a certain string is 42 Hz, what is the fundamental frequency of the string?

sdas [7]

Data:
$f_{2} = 42 Hz$
n (Wave node)
V (Wave belly)
L (Wave length)
The number of bells is equal to the number of the harmonic emitted by the string.

$f_{n} = \frac{nV}{2L}$

Wire 2 → 2º Harmonic → n = 2

$f_{n} = \frac{nV}{2L}$
$f_{2} = \frac{2V}{2L} 
$
$2V = f_{2} *2L$
$V = \frac{ f_{2}*2L }{2}$
$V = \frac{42*2L}{2}$
$V = \frac{84L}{2}$
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Wire 1 → 1º Harmonic or Fundamental rope → n = 1

$f_{n} = \frac{nV}{2L}$
$f_{1} = \frac{1V}{2L}$
$f_{1} = \frac{V}{2L}$

If, We have:
V = 42L
Soon:
$f_{1} = \frac{V}{2L}$
$f_{1} = \frac{42L}{2L}$
$\boxed{f_{1} = 21 Hz}$

Answer:

The fundamental frequency of the string:
21 Hz

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3 years ago

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