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Evgen [1.6K]
3 years ago
5

A 06-C charge and a 07-C charge are apart at 3 m apart. What force attracts them?​

Physics
1 answer:
Vlad [161]3 years ago
5 0

Answer:

Force of 37.8 × 10^(6) N attracts the two charges

Explanation:

The force between two charges is given by

F = k*q1*q2/r²

Where q1 and q2 are 0.06 C and 0.07 C.

r is the distance between q1 and q2 which is equal to 3 m

k is a constant = 9 × 10^(9) N.m²/C²

F = (9 × 10^(9) × 0.06 × 0.07)/3²

F = 37.8 × 10^(6) N

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Cuanto cambia la entropía de 0.50 kg de vapor de mercurio [Lv: 2.7 x 10⁵ j/kg ] al calentarse en su punto de ebullición de 357°
lord [1]

Answer:

La entropía del vapor de mercurio cambia en 214.235 joules por Kelvin.

Explanation:

Por definición de entropía (S), medida en joules por Kelvin, tenemos la siguiente expresión:

dS = \frac{\delta Q}{T} (1)

Donde:

Q - Ganancia de calor, en joules.

T - Temperatura del sistema, en Kelvin.

Ampliamos (1) por la definición de calor latente:

dS = \frac{L_{v}}{T}\cdot dm (1b)

Donde:

m - Masa del sistema, en kilogramos.

L_{v} - Calor latente de vaporización, en joules

Puesto que no existe cambio en la temperatura durante el proceso de vaporización, transformamos la expresión diferencial en expresión de diferencia, es decir:

\Delta S = \frac{\Delta m \cdot L_{v}}{T}

Como vemos, el cambio de la entropía asociada al cambio de fase del mercurio es directamente proporcional a la masa del sistema. Si tenemos que m = 0.50\,kg,L_{v} = 2.7\times 10^{5}\,\frac{J}{kg} and T = 630.15\,K, entonces el cambio de entropía es:

\Delta S = \frac{(0.50\,kg)\cdot \left(2.7\times 10^{5}\,\frac{J}{kg} \right)}{630.15\,K}

\Delta S = 214.235 \,\frac{J}{K}

La entropía del vapor de mercurio cambia en 214.235 joules por Kelvin.

3 0
3 years ago
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Eddi Din [679]

Answer:

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