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Olegator [25]
3 years ago
7

An ideal gas is enclosed in a cylinder that has a movable piston on top. The piston has a mass m and an area A and is free to sl

ide up and down, keeping the pressure of the gas constant. How much work is done on the gas as the temperature of n mol of the gas is raised from T1 to T2? (Use any variable or symbol stated above along with the following as necessary: R.)
Physics
1 answer:
elena-14-01-66 [18.8K]3 years ago
4 0

Answer:

Explanation:

Given mass of piston \left ( m\right )

no. of moles =n

Given Pressure remains same

Temperature changes from T_1 to T_2

Work done\left ( W\right ) is given by=\int_{V_1}^{V_2}PdV

W=P\left ( V_2-V_1\right )

also PV_1=nRT_1

PV_2=nRT_2

W=nR\left ( T_2-T_1\right )

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ou have just moved into a new apartment and are trying to arrange your bedroom. You would like to move your dresser of weight 3,
Dimas [21]

So far, since you moved into the apartment until the end of this much of the story, you haven't done ANY work on the dresser yet.

I'll admit that you pushed, groaned and grunted, sweated and strained plenty.  You're physically and mentally exhausted, you're not interested in the dresser at the moment, and right now you just want to snappa cappa brew, crash on the couch, and watch cartoons on TV.  But if you've done your Physics homework, you know you haven't technically done any <u><em>work</em></u> yet.

In Physics, "Work" is the product of Force times Distance.

Since the dresser hasn't budged yet, the Distahce is zero.  So no matter how great the Force may be, it's multiplied by zero, so the <em>Work is zero</em>.

5 0
3 years ago
Which of the following can be thought of as either a wave or a particle?
IrinaK [193]

Both matter and light have been demonstrated to exhibit wave-like and particle-like behavior.

Light as a wave: light can diffract & refract

Light as a particle: photoelectric effect, Compton scattering

Matter as a wave: Davisson-Germer experiment

Matter as a particle: find a picture of any kinematics problem in a high school physics textbook

Choice D

7 0
3 years ago
An athlete at high performance inhales 4.0L of air at 1 atm and 298 K. The inhaled and exhaled air contain 0.5% and 6.2% by volu
LenaWriter [7]

To solve this problem we will calculate the total volume of inhaled and exhaled water. From the ideal gas equation we will find the total number of moles of water.

An athlete at high performance inhales 4.0L of air at 1atm and 298K.

The inhaled and exhaled air contain 0.5% and 6.2% by volume of water, respectively.

During inhalation, volume of water taken is

V_i = (4L)(0.5\%)

V_i = 0.02L

During exhalation, volume of water expelled is

V_e = (4L)(6.2\%)

V_e = 0.248L

During 40 breathes, total volume of water taken is

V_{it} = (40L)(0.02L) = 0.8L

During 40 breathes, total volume of water expelled out is

V_{et} = (40L)(0.248L) = 9.92L

Therefore resultant volume of water expelled out from the lung is

\Delta V = 9.92L-0.8L = 9.12

From the body through the lung we have that

n = \frac{PV}{RT}

Here,

P = Pressure

R= Gas ideal constant

T= Temperature

V = Volume

Replacing,

n = \frac{(1atm)(9.12L)}{(8.314J/mol \cdot K)(298K)}

n = 0.373mol/min

Therefore the moles of water per minute are expelled from the body through the lungs is 0.373mol/min

8 0
3 years ago
Una carga positiva de 4 x 10-5 C, se encuentra a 0.05 m de otra carga positiva de 2 x 10-5 C. Calcular la fuerza que se ejerce e
KATRIN_1 [288]

Answer:

La fuerza que se ejerce entre las dos cargas es 2880 N.

Explanation:

La ley de Coulomb indica que los cuerpos cargados sufren una fuerza atractiva o repulsiva al acercarse. La fuerza es atractiva si las cargas son del signo opuesto y repulsión si son del mismo signo. El valor de la fuerza es proporcional al producto del valor de sus cargas e inversamente proporcional al cuadrado de la distancia que los separa. Esto se expresa matemáticamente como:

F=k*\frac{Q*q}{r^{2} }

donde:

  • F es la fuerza eléctrica de atracción o repulsión. Se mide en Newtons (N).
  • Q y q son los valores de las dos cargas puntuales. Se miden en culombios (C).
  • r es el valor de la distancia que los separa. Se mide en metros (m).
  • k es una constante de proporcionalidad llamada constante de la ley de Coulomb.

En este caso:

  • F= ?
  • Q= 4*10⁻⁵ C
  • q= 2*10⁻⁵ C
  • r= 0.05 m
  • k= 9*10⁹ \frac{N*m^{2} }{C^{2} }

Reemplazando:

F=9*10^{9} \frac{N*m^{2} }{C^{2} }*\frac{4*10^{-5} C*2*10^{-5}C }{(0.05 m)^{2} }

F= 2880 N

<u><em>La fuerza que se ejerce entre las dos cargas es 2880 N.</em></u>

7 0
3 years ago
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Sati [7]

Answer:

the second one

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6 0
3 years ago
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