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Usimov [2.4K]
3 years ago
14

A sealed container holds 0.020 moles of ideal nitrogen (n2) gas, at a pressure of 1.5 atmospheres and a temperature of 290k. the

atomic mass of nitrogen is 14.0g/mol. what is approimate translational kinetic energy of nitrogen molecuole, in si units
Physics
1 answer:
avanturin [10]3 years ago
6 0
Translational kinetic energy is the kinetic energy of an object of finite size. It is equal to one half the product of the mass of the object and the square of the velocity. For an ideal gas, it is simplified into the expression:

Kinetic energy = 3nRT / 2   where n is the number of moles, R is the universal gas constant and T is the temperature of the gas.

We calculate as follows:
Kinetic energy = 3nRT / 2 
Kinetic energy = 3 (0.020 moles ) (8.314 J / mol-K) (290 K) / 2 
Kinetic energy = 72.33 J

Therefore, the translational kinetic energy of the ideal nitrogen gas is 72.33 J.
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El motor de una licuadora gira a 3600 rpm, disminuye su velocidad angular hasta 2000 rpm realizando 120 vueltas. Calcular: a) La
allsm [11]

Answer:

a) α = -65,2 rad/s².

b) t = 2,57 s.

Explanation:

a) La aceleración angular se puede calcular usando la siguiente ecuación:

\omega_{f}^{2} = \omega_{0}^{2} + 2\alpha \theta

En donde:

\omega_{f}: es la velocidad angular final =  2000 rpm = 209,4 rad/s

\omega_{0}: es la velocidad angular inicial = 3600 rpm = 377,0 rad/s

α: es la aceleración angular=?

θ: es el desplazamiento o número de vueltas = 120 rev = 754,0 rad

Las conversiones de unidades se hicieron sabiendo que 1 revolución = 2π radianes y que 1 minuto = 60 segundos.  

Resolviendo la ecuación (1) para α, tenemos:

\alpha = \frac{\omega_{f}^{2} - \omega_{0}^{2}}{2\theta} = \frac{(209,4 rad/s)^{2} - (377,0 rad/s)^{2}}{2*754,0 rad} = -65,2 rad/s^{2}  

Entonces, la aceleración angular es -65,2 rad/s². El signo negativo se debe a que el motor está desacelerando.  

b) El tiempo transcurrido se puede encontrar como sigue:

\omega_{f} = \omega_{0} + \alpha t

Resolviendo para t, tenemos:

t = \frac{\omega_{f} - \omega_{0}}{\alpha} = \frac{209,4 rad/s - 377,0 rad/s}{-65,3 rad/s^{2}} = 2,57 s

Por lo tanto, el tiempo transcurrido fue 2,57 s.

Espero que te sea de utilidad!

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3 years ago
A yo‑yo with a mass of 0.0800 kg and a rolling radius of =2.70 cm rolls down a string with a linear acceleration of 5.70 m/s2.
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Explanation:

Given that,

Mass, m = 0.08 kg

Radius of the path, r = 2.7 cm = 0.027 m

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The net force acting on the string is :

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Putting all the values,

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I\alpha =Fr\\\\I=\dfrac{Fr}{\alpha }\\\\I=\dfrac{0.328\times 0.027}{211.12}\\\\=4.19\times 10^{-5}\ kg-m^2

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