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cricket20 [7]
3 years ago
11

Unas niñas en el receso estaban jugando a derramar flatulencias en un frasco de forma cilíndrica, cuyo radio tenía 5" y una altu

ra de 7 in, el gas se mantenía constante a una presión de 540 mm de Hg. Si se baja la presión a 320 mm de Hg, ¿Cuál será el volumen que posea la flatulencia?
Physics
1 answer:
hram777 [196]3 years ago
3 0

Answer:

Los gases siempre se adaptaran al volumen del contenedor en el que estén.

Entonces, si baja la presión del gas, la temperatura se mantendrá constante, lo que implica que tiene que aumentar la temperatura, pues para un gas ideal tenemos:

P*V = n*k*T

donde si el frasco esta cerrado, tenemos que n, k y V son constantes, entonces:

P = (n*k/V)*T

P = constante*T

(Notar que si el frasco no estuviera cerrado, entonces el numero n variaría de forma incontrolable, y este problema no se podría resolver de forma sencilla)

Entonces, si la presión baja, también baja la temperatura, pero el volumen se mantendrá constante, eso es lo importante.

Como el volumen se mantiene constante, el volumen que tomara el gas va a ser igual al volumen del frasco, sabemos que el volumen de un cilindro es:

V = (pi*r^2*h)

donde:

r = radio

pi = 3.14

h = altura.

en este caso, r = 5'' = 5 in

                       h = 7 in

Entonces el volumen sera:

V = 3.14*(5in)^2*7in = 549.5 in^2

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A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(
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Answer:

a) t=2s

b) h_{max}=1024ft

c) v_{y}=-256ft/s

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2

To find what time does it take to reach maximum height we need to find how high will it go

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v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

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y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

h=-16t^2+64t+960

1024=-16t^2+64t+960

0=-16t^2+64t-64

Solving the quadratic equation

t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}

a=-16\\b=64\\c=-64

t=2s

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s

Since the projectile is going down the velocity at the instant it reaches the ground is:

v=-256ft/s

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