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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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7 0
3 years ago
Runner A is initially 5.7 km west of a flagpole and is running with a constant velocity of 8.9 km/h due east. Runner B is initia
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Answer:

B meet A 0.01 km east of flagpole

Explanation:

given data

distance A = 5.7 km  west

velocity V1 = 8.9 km/h

distance B = 4.5 km  east

velocity V2 = 7 km/h

to find out

How far runners from the flagpole, when paths cross

solution

we know A and B are 5.7 + 4.5 = 10.2 km apart

and we consider here B will run distance x km for meet

so time will be for B is

time B = distance / velocity

time B = x / 7     ...................1

and

for A distance for meet = ( 10.2 - x ) km

so time A = distance / velocity

time A = ( 10.2 - x )  / 8.9      .............2

now equating equation 1 and 2

time A = time B

x / 7  =   ( 10.2 - x )  / 8.9

x = 4.490

so distance of B run for meet is 4.490 km

so distance from the flagpole when their paths cross is 4.5 - 4.490 = 0.01 km

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If a 51kg snowboarder falls of a cliff, and is falling 15 m/s when they impact the snow, what is the average force of the snow o
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A football player kicks a football downfield. The height of the football increases until it reaches a maximum height of 15 yards
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