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tankabanditka [31]
3 years ago
15

In which situation is no work considered to be done by a force?

Chemistry
2 answers:
Masja [62]3 years ago
7 0

If the angle is either 0 or 180, that means that there is either negative or positive work, so A and D are not correct.

If the angle is 45, then there is still some work involved.

The only option where there is no work done by a force is B. when the angle is between the force and displacement is 90. 

san4es73 [151]3 years ago
6 0

Answer:

The correct answer is option B.

Explanation:

Work done is defined as product of applied force to an object and displacement that is distance covered by an object when force applied.

W=Force\times distance\times \cos\theta

  • When \theta =0^o or 180^o. positive or negative work will be done.
  • When \theta =45^o s, some amount wok will be done
  • When \theta =90^o no work will be done.
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En condiciones normales 1g de aire ocupa un volumen de 773 mL ¿ qué volumen ocupará la misma masa de aire a 0 ºC y la presión a
emmainna [20.7K]

Answer:

El volumen que ocupará la misma masa de aire es 839.49 mL.

Explanation:

Las condiciones normales de presión y temperatura (abreviado CNPT) o presión y temperatura normales (abreviado PTN o TPN), son términos que implican que la temperatura referenciada es de 0ºC (273,15 K) y la presión de 1 atm (definida como 101.325 Pa).

La ley de Boyle dice que “El volumen ocupado por una determinada masa gaseosa a temperatura constante, es inversamente proporcional a la presión”  y se matemáticamente como

Presión*Volumen=constante

o P*V=k

La ley de Charles es una ley que dice que cuando la cantidad de gas y de presión se mantienen constantes, el cociente que existe entre el volumen y la temperatura siempre tendrán el mismo valor:  

\frac{V}{T} =k

La ley de Gay-Lussac​ establece que la presión de un volumen fijo de un gas, es directamente proporcional a su temperatura. Se expresa matemáticamente como:

\frac{P}{T} =k

Combinando estas tres leyes se obtiene:

\frac{P*V}{T} =k

Siendo un estado inicial 1 y un estado final 2, la expresión anterior queda determinada como:

\frac{P1*V1}{T1} =\frac{P2*V2}{T2}

En este caso:

  • P1=  101325 Pa
  • V1= 773 mL
  • T1= 273.15 K
  • P2= 93,3 kPa= 93300 Pa
  • V2= ?
  • T2= 0°C= 273.15 K

Reemplazando:

\frac{101325 Pa*773 mL}{273,15 K} =\frac{93300 Pa*V2}{273.15 K}

y resolviendo obtenes:

V2=\frac{273.15 K}{93300 Pa} *\frac{101325 Pa*773 mL}{273,15 K}

V2= 839,49 mL

<u><em>El volumen que ocupará la misma masa de aire es 839.49 mL.</em></u>

8 0
3 years ago
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