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kirill115 [55]
11 months ago
7

As additional resistors are connected in parallel to a constant voltage source, how is the power supplied by the source affected

?.
Physics
1 answer:
Mars2501 [29]11 months ago
3 0

As additional resistors are included lined up across a consistent voltage source, there are more ways for current to take.

So more current spills out of the source, and the all out current provided by the source increments.

The power supplied by the battery is (voltage) x (current).  So if the voltage is constant and the current increases, the power being supplied must also increase.

<h3>What is a resistor?</h3>

A resistor is an electrical part that cutoff points or manages the progression of electrical flow in an electronic circuit. Resistors can likewise be utilized to give a particular voltage to a functioning gadget like a semiconductor.

To learn more about resistors from the given link

brainly.com/question/12864050

#SPJ4

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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.

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PLEASE HELP URGENT: Please see attachment for problem! AP physics projectile question. I'd REALLY appreciate any help.
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I’m so sorry, I need more information. Good luck and I’m sorry I couldn’t help you :(
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