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mr_godi [17]
3 years ago
10

In a popular classroom demonstration, a cotton ball is placed in the bottom of a strong test tube. A plunger fits inside the tub

e and it makes an air-tight seal. It is then pushed down very rapidly, and the cotton flashes and burns. This happens because ______.
Physics
1 answer:
mario62 [17]3 years ago
3 0

Answer:

An increase in air temperature because of its compression.

Explanation:

The Gay-Lussac's Law states that a gas pressure is directly proportional to its temperature in an enclosed system to constant volume.  

P = kT  

<em>where P: is the gas pressure, T: is the gas temperature and k: is a constant.</em>

Therefore, due to Gay-Lussac's Law, when the plunger is pushed down very rapidly, the pressure of the air increase, which leads to its temperature increase. That is why cotton flashes and burns.      

I hope it helps you!

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Si el coeficiente de fricción cinética entre los neumáticos y el pavimento seco es de 0.80. ¿Cuál es la distancia mínima para de
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Answer: 52.9 metros.

Explanation:

Podemos escribir la fuerza de fricción cinética como

F = μ*N

donde N es la fuerza normal entre el coche y el suelo, cuya magnitud es igual al peso en esta situación.

F = μ*m*g

donde m es la masa del coche y g es 9.8m/s^2

y sabemos que μ = 0.8

Por la segunda ley de Newton, sabemos que:

F = m*a

fuerza es igual a masa por aceleración.

a = F/m

entonces la aceleración causada por la fuerza de rozamiento es:

F = 0.8*m*g

a = F/m = (0.8*m*g)/m = 0.8*g.

Entonces ya encontramos la aceleración, hay que recordar que esta aceleración es en sentido opuesto a la sentido de movimiento, entonces podemos escribir la aceleración como:

a(t) = -0.8*g

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v(t) = -0.8*g*t + 28.8m/s

Ahora podemos encontrar el tiempo necesario para que la velocidad del coche sea cero, en ese momento, como deja de moverse, ya no tendremos rozamiento cinético, entonces no habrá aceleración y el coche se detendrá completamente.

v(t) = 0m/s = -0.8*9.8m/s^2*t + 28.8m/s

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                 t   = (28.8m/s)/(7.84m/s^2) = 3.63 segundos.

Ahora vamos a la ecuación de movimiento, donde asumimos que la posición inicial del coche es 0m, así que no tendremos constante de integración.

p(t) = -(1/2)*(0.8*9.8m/s^2)*t^2 + 28.8m/s*t

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p(3.63s) = -(1/2)*(0.8*9.8m/s^2)*(3.63s)^2 + 28.8m/s*(3.63s) = 52.9 metros.

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