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3 years ago

What emerging technologies will make hydroelectric energy source safer, more usable, more efficient, cleaner, etc?

Physics

1 answer:

vitfil [10]3 years ago

3 0

<h2>Emerging Technologies that will make Hydroelectric Energy Source Safer</h2>

The variation in technology to labor lower loads and higher peaks make the hydroelectric energy source safer, more usable, more efficient and cleaner. The plants can also adapt speed drives changing the capability to suffice various sorts of demands which includes great speed response. Such variation can alter the income rate by 85%. The spread of operating range of the plant can increase income rate by 61%.

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Two identical point charges are 3.00 cm apart. find the charge on each of them if the force or repulsion is 4.00 x 10^-7. (Use C

DanielleElmas [232]

Answer:

Charge on each is 2 x 10⁻¹⁰.

Explanation:

We know that Force between two point charges is given b the Coulomb's law as:

F = kq₁q₂/r^2

k = 9 x 10^9

r = 3.00 cm

= 0.03 m

q₁ = q₂

F = 4.00 x 10^-7

Rearranging the formula, we get:

F = k q²/r²

q² = Fr²/k

q² = 4 x 10⁻⁷ x 0.03²/(9x10⁹)

q² = 4 x 10⁻²⁰

q = 2 x 10⁻¹⁰

As there is force of repulsion between the charges, the charges must be both positive or both negative.

3 0

3 years ago

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: