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marishachu [46]
3 years ago
12

A gas is compressed at a constant pressure of 0.800 atm from 12.00 L to 3.00 L. In the process, 390 J of energy leaves the gas b

y heat.
a) What is the work done on the gas?
b) What is the change in its internal energy?
Physics
1 answer:
Andru [333]3 years ago
5 0

Answer:

a)W= - 720 J

b)ΔU= 330 J

Explanation:

Given that

P = 0.8 atm

We know that 1 atm = 100 KPa

P = 80 KPa

V₁ = 12 L = 0.012 m³       ( 1000 L = 1 m³)

V₂ = 3 L = 0.003 m³

Q= - 390 J ( heat is leaving from the system )

We know that work done by gas given as

W = P (V₂ -V₁ )

W= 80 x ( 0.003 - 0.012 ) KJ

W= - 0.72 KJ

W= - 720 J    ( Negative sign indicates work done on the gas)

From first law of thermodynamics

Q = W + ΔU

ΔU=Change in the internal energy

Now by putting the values

- 390 = - 720 + ΔU

ΔU= 720 - 390  J

ΔU= 330 J

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What are some actions that can be taken to remedy air pollution? Select all that apply. Run your air conditioner at a higher set
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4 0
2 years ago
If a jet travels 350 m/s, how far will it travel each second?
77julia77 [94]

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6 0
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Calcula la energía cinética de un coche de 500kg de masa que se mueve a la velocidad de 100 km/h. Pasamos la velocidad a las uni
castortr0y [4]

Responder:

192,900.64 Julios

Explicación:

La energía cinética es la energía que posee un cuerpo en virtud de su movimiento. Sea la masa del cuerpo m, su velocidad v.

Energía cinética = 1/2 mv²

Parámetros dados

masa del coche = 500 kg

velocidad = 100 km / h

La velocidad debe estar en m / s según la unidad internacional estándar. Al convertir;

100 km / h = 100 * 1000/3600 m / s

100 km / h = 27,8 m / s

velocidad del cuerpo = 27,8 m / s

Necesario

Energía cinética del coche = 1/2 * 500 * 27,8²

Energía cinética del automóvil = 1/2 * 500 * 771.6

Energía cinética del automóvil = 1/2 * 385,801.28

Energía cinética del automóvil = 192,900.64 Julios

<em>Por lo tanto, la energía cinética del automóvil es 192,900.64 julios.</em>

5 0
3 years ago
The only force acting on a 3.1 kg body as it moves along the positive x axis has an x component fx = -8x n, where x is in meters
My name is Ann [436]
<span>(a) -9.97 m/s
 (b) x = 2.83

   This is a simple problem in integral calculus. You've been given part of the 2nd derivative (acceleration), but not quite. You've been given the force instead. So let's setup a function for acceleration.

   f''(x) = -8x N / 3.1 kg= -8x kg*m/s^2 / 3.1 kg = -2.580645161x m/s^2

   So the acceleration of the body is now expressed as
 f''(x) = -2.580645161x m/s^2

    Let's calculate the anti-derivative from that.
 f''(x) = -2.580645161x m/s^2
 f'(x) = -1.290322581x^2 + C m/s

   Now let's use the known velocity value at x = 2.0 to calculate C
 f'(x) = -1.290322581x^2 + C 1
1 = -1.290322581*2^2 + C
 11 = -1.290322581*4 + C
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 16.161290323 = C

   So the velocity function is
  f'(x) = -1.290322581x^2 + 16.161290323

   (a) The velocity at x = 4.5
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 f'(4.5) = -1.290322581*4.5^2 + 16.161290323
 f'(4.5) = -1.290322581*20.25 + 16.161290323
  f'(4.5) = -26.12903227 + 16.161290323
 f'(4.5) = -9.967741942

   So the velocity is -9.97 m/s

   (b) we want a velocity of 5.8 m/s
5.8 = -1.290322581x^2 + 16.161290323
 0 = -1.290322581x^2 + 10.36129032
 1.290322581x^2 = 10.36129032
 x^2 = 8.029999998
 x = 2.833725463</span>
4 0
3 years ago
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