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miv72 [106K]
3 years ago
13

1. En un diagrama PV trace las trayectorias para cada uno de los siguientes procesos que ocurren en forma sucesiva en un sistema

cerrado consistente en 2 moles de aire a condiciones estándar de presión y temperatura.
Proceso 1: isobárico hasta duplicar la temperatura inicial
Proceso 2: isotérmico hasta triplicar el volumen del estado inicial
Proceso 3: isocórico hasta reducir la temperatura al valor del estado inicial
Proceso 4: isotérmico hasta reducir el volumen al valor inicial.
Physics
1 answer:
goblinko [34]3 years ago
8 0
Usando la ecuación PV = nRT 
n = 2
P = 101.3kPa / 1 atm 
T = 0 grados centígrados 
1. en el primer proceso hay un sistema isobárico y eso nos dice que la presión no cambia asi que la trayectoria va a ser una línea horizontal. 
2. en un proceso isotérmico la temperatura no cambia asi que la trayectoria va a ser una función exponencial. Aquí el volumen aumenta así que la función va a empezar de el lado izquierdo a el derecho.  
3. en un isocórico el volumen no cambia entonces la trayectoria es una linea vertical 
4. la trayectoria se ve igual que la del proceso 2 pero va de derecha a izquierda. 



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Part a) Force on car = 2833.84 Newtons

Part b) Time to stop the car = 3.8 seconds

Part c) Factor for stopping distance is 4.

Part d) Factor for stopping time is 1.

Explanation:

The deceleration produced when the car is brought to rest in 20 meters can be found by third equation of kinematics as

v^2=u^2+2as

where

v = final speed of the car ( = 0 in our case since the car stops)

u = initial speed of the car = 38 km/hr =\frac{38\times 1000}{3600}=10.56m/s

a = deceleration produces

s = distance in which the car stops

Applying the given values we get

0^2=10.56^2+2\times a\times 20\\\\a=\frac{0-10.56^2}{2\times 20}\\\\\therefore a=-2.78m/s^2

Now the force can be obtained using newton's second law as

Force=\frac{Weight}{g}\times a

Applying values we get

Force=\frac{1.0\times 10^4}{9.81}\times -2.78\\\\\therefore F=-2833.84Newtons

The negative direction indicates that the force is opposite to the motion of the object.

Part b)

The time required to stop the car can be found using the first equation of kinematics as

v=u+at with symbols having the same meanings

Applying values we get

0=10.56-2.78\times t\\\\\therefore t=\frac{10.56}{2.78}=3.8seconds

Part c)

From the developed relation of stopping distance we can see that the for same force( Same acceleration) the stopping distance is proportional to the square of the initial speed thus doubling the initial speed increases the stopping distance 4 times.

Part d)

From the relation of stopping time and the initial speed we can see that the stopping distance is proportional initial speed thus if we double the initial speed the stopping time also doubles.

8 0
3 years ago
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