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Nana76 [90]
3 years ago
7

Un recipiente con una capacidad de 25L contiene un gas a una presión de 7,5 atm. Calcula la nueva presión a la que se verá somet

ido el gas si lo comprimimos hasta un volumen de 10L sin cambiar la temperatura.
Physics
1 answer:
amid [387]3 years ago
5 0

Responder:  

18.75 atmósferas  

Explicación:  

Paso uno:  

datos dados

volumen inicial V1 = 25L

Presión inicial P1 = 7.5 atm

volumen final V2 = 10L

presión final P2 = ??

Segundo paso:

Aplicando la expresión de gas que relaciona el volumen y la presión, es decir

P1V1 = P2V2

sustituyendo nuestros datos tenemos

7.5 * 25 = P2 * 10

187.5 = P2 * 10

divide ambos lados por 10

P2 = 187.5 / 10

P2 = 18.75 atmósferas

<em><u>La presión final es de 18.75 atm.</u></em>

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Answer:

5.15348 Beats/s

4.55 mm

Explanation:

v_1 = Velocity of sound = 342 m/s

v_2 = Velocity of sound = 346 m/s

f_1 = First frequency = 440 Hz

Frequency is given by

f_2=\frac{v_2}{2L_1}\\\Rightarrow f_2=\frac{346}{2\times 0.38863}\\\Rightarrow f_2=445.15348\ Hz

Beat frequency is given by

|f_1-f_2|=|440-445.15348|=5.15348\ Beats/s

Beat frequency is 5.15348 Hz

Wavelength is given by

\lambda_1=\frac{v_1}{f}\\\Rightarrow \lambda_1=\frac{342}{440}\\\Rightarrow \lambda_1=0.77727\ m

Relation between length of the flute and wavelength is

\lambda_1=2L_1\\\Rightarrow L_1=\frac{\lambda_1}{2}\\\Rightarrow L_1=\frac{0.77727}{2}\\\Rightarrow L_1=0.38863\ m

At v = 346 m/s

\lambda_2=\frac{v_2}{f}\\\Rightarrow \lambda_2=\frac{346}{440}\\\Rightarrow \lambda_1=0.78636\ m

L_2=\frac{\lambda_2}{2}\\\Rightarrow L_2=\frac{0.78636}{2}\\\Rightarrow L_2=0.39318\ m

Difference in length is

\Delta L=L_2-L_1\\\Rightarrow \Delta L=0.39318-0.38863\\\Rightarrow \Delta L=0.00455\ m=4.55\ mm

It extends to 4.55 mm

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