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Marta_Voda [28]
3 years ago
13

How much work does it take to move a 50 μC charge against a 12 V potential difference?

Physics
1 answer:
lukranit [14]3 years ago
8 0
<span>work =V*Q =12*50*10^-6

The total work done will be equal to 

work = V.Q

which means 

w= 12 . 50.10^-6
Hence,
w= 0.0006 J</span>
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10. How much total work do you do when you lift a 50 kg microwave 1.0 m off the ground and then push it 1.0 m
Mariulka [41]

Work formula:

W = Fd\cos(\theta)

F = 50N, d = 1.0 m

When you lift something straight up, the angle of the force is 90º

cos(90º) is 0, so there's no work done when you lift the microwave off the ground

W = (50N)(1.0)(0) = 0

F = 50N, d = 1.0 m

When you push the microwave, the angle is 0º and cos(0º) is 1. So there is work done here:

W = (50 N)(1.0m)(1)

W = 50

total work = 50 joules

6 0
3 years ago
Explain how do winds cause surface currents?
igor_vitrenko [27]

Answer:

As wind or an ocean current moves, the Earth spins underneath it. ... The Coriolis effect bends the direction of surface currents to the right in the Northern Hemisphere and left in the Southern Hemisphere.

Explanation:

The Coriolis effect causes winds and currents to form circular patterns.

8 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:

\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
The instruction booklet for your pressure cooker indicates that its highest setting is 12.3 psi . you know that standard atmosph
zmey [24]
<span>118 C The Clausius-Clapeyron equation is useful in calculating the boiling point of a liquid at various pressures. It is: Tb = 1/(1/T0 - R ln(P/P0)/Hvap) where Tb = Temperature boiling R = Ideal Gas Constant (8.3144598 J/(K*mol) ) P = Pressure of interest Hvap = Heat of vaporization of the liquid T0, P0 = Temperature and pressure at a known point. The temperatures are absolute temperatures. We know that water boils at 100C at 14.7 psi. Yes, it's ugly to be mixing metric and imperial units like that. But since we're only interested in relative pressure differences, it's safe enough. So P0 = 14.7 P = 14.7 + 12.3 = 27 T0 = 100 + 273.15 = 373.15 And for water, the heat of vaporization per mole is 40660 J/mol Let's substitute the known values and calculate. Tb = 1/(1/T0 - R ln(P/P0)/Hvap) Tb = 1/(1/373.15 K - 8.3144598 J/(K*mol) ln(27/14.7)/40660 J/mol) Tb = 1/(0.002679887 1/K - 8.3144598 1/K ln(1.836734694)/40660) Tb = 1/(0.002679887 1/K - 8.3144598 1/K 0.607989372/40660) Tb = 1/(0.002679887 1/K - 5.055103194 1/K /40660) Tb = 1/(0.002679887 1/K - 0.000124326 1/K) Tb = 1/(0.002555561 1/K) Tb = 391.3034763 K Tb = 391.3034763 K - 273.15 Tb = 118.1534763 C Rounding to 3 significant figures gives 118 C</span>
3 0
3 years ago
The steps in the scientific process must be followed in order. True False
olya-2409 [2.1K]

Answer: True

Hope I helped :)

5 0
3 years ago
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