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Verizon [17]
3 years ago
10

Un neumático sin cámara, soporta una presión de 1.5 atm cuando la temperatura ambiente es de 300°K. ¿Qué presión llegará a sopor

tar dicho neumático si, en el transcurso de un viaje, las ruedas alcanzan una temperatura de 67ºC? Atm
Physics
1 answer:
arlik [135]3 years ago
4 0

Answer:

El neumático soportará una presión de 1.7 atm.

Explanation:

Podemos encontrar la presión final del neumático usando la ecuación del gas ideal:

PV = nRT

En donde:

P: es la presión

V: es el volumen

n: es el número de moles del gas

R: es la constante de gases ideales

T: es la temperatura

Cuando el neumático soporta la presión inicial tenemos:

P₁ = 1.5 atm

T₁ = 300 K

V_{1} = \frac{nRT_{1}}{P_{1}}  (1)  

La presión cuando T = 67 °C es:

P_{2} = \frac{nRT_{2}}{V_{2}}   (2)

Dado que V₁ = V₂  (el volumen del neumático no cambia), al introducir la ecuación (1) en la ecuación (2) podemos encontrar la presión final:

P_{2} = \frac{nRT_{2}}{V_{2}} = \frac{nRT_{2}}{\frac{nRT_{1}}{P_{1}}} = \frac{P_{1}T_{2}}{T_{1}} = \frac{1.5 atm*(67 + 273)K}{300 K} = 1.7 atm  

Por lo tanto, si en el transcurso de un viaje las ruedas alcanzan una temperatura de 67 ºC, el neumático soportará una presión de 1.7 atm.

Espero que te sea de utilidad!

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Sort the sentences based on whether they describe radio waves, visible light waves, or both.
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1) Initial upward acceleration: 6.0 m/s^2

2) Mass of burned fuel: 0.10\cdot 10^4 kg

Explanation:

1)

There are two forces acting on the rocket at the beginning:

- The force of gravity, of magnitude F_g = mg, in the downward direction, where

m=1.9\cdot 10^4 kg is the rocket's mass

g=9.8 m/s^2 is the acceleration of gravity

- The thrust of the motor, T, in the upward direction, of magnitude

T=3.0\cdot 10^5 N

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:

T-mg=ma (1)

where a is the acceleration of the rocket.

Solving for a, we find the initial acceleration:

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2)

When the rocket reaches an altitude of 5000 m, its acceleration has increased to

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The reason for this increase is that the mass of the rocket has decreased, because the rocket has burned some fuel.

We can therefore rewrite eq.(1) as

T-m'g=m'a'

where

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And since the initial mass of the rocket was

m=1.9 \cdot 10^4 kg

This means that the mass of fuel burned is

\Delta m = m-m'=1.9\cdot 10^4 - 1.80\cdot 10^4 = 0.10\cdot 10^4 kg

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