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Serjik [45]
3 years ago
9

How long would it take for 3.5 C of charge to pass through a cross-sectional area of a wire in which a current of 5 mA passes?

Physics
1 answer:
son4ous [18]3 years ago
4 0

Answer:

<em>The time taken for the charge to pass through the wire = 700 s or 11.67 minutes or 0.194 hours</em>

Explanation:

Electric Charge: This is defined as the product of electric current to time in a an electric circuit. The S.I unit of charge is Coulombs (C).

Mathematically, it is represented as

Q = it.......................... Equation 1

Where Q =quantity of charge, I = current, t = time

making t the subject of formula in equation 1,

t = Q/I ....................... Equation 2

<em>Given: Q = 3.5 C, I = 5 mA.</em>

<em>Conversion: We convert from mA to A </em>

<em>I.e 5 mA = (5 × 10⁻³) A = 0.005 A.</em>

<em>Substituting these values into equation 2</em>

<em>t = 3.5/0.005</em>

<em>t = 700 seconds  </em>

<em>or</em>

<em>(700/60) minutes = 11.67 minutes</em>

<em>or</em>

<em>(700/3600) hours = 0.194 hours.</em>

<em>Therefore the time taken for the charge to pass through the wire = 700 s or 11.67 minutes or 0.194 hours</em>

<em></em>

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Answer:

374.39 J/K

Explanation:

Entropy: This can be defined as the degree of disorder or randomness of a substance.

The S.I unit of entropy is J/K

ΔS = ΔH/T ..................................... Equation 1

Where ΔS = entropy change, ΔH = Heat change, T = temperature.

ΔH = cm................................... Equation 2

Where,

c = specific latent heat of fusion of water = 333000 J/kg, m = mass of ice = 0.3071 kg.

Substitute into equation 2

ΔH = 333000×0.3071

ΔH = 102264.3 J.

Also, T = 273.15 K

Substitute into equation 1

ΔS = 102264.3/273.15

ΔS = 374.39 J/K

Thus, The change in entropy = 374.39 J/K

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Datos

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Sabemos que el peso aparente de un cuerpo que se sumerge en un fluido es:

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Teniendo en cuenta que:

Preal = mcubo⋅gPfluido=E= dfluido⋅Vfluido⋅g

Como el cuerpo se sumerge completamente en el fluido, el volumen de fluido desalojado es exactamente el volumen del cubo. Por lo tanto si sustituimos los datos que nos proporcionan en el enunciado en la primera ecuación:

Paparente=mcubo⋅g−dfluido⋅Vfluido⋅g ⇒Paparente=10 kg ⋅9.8 m/s2 − 1000 kg/m3 ⋅10−3 m ⋅9.8 m/s2 ⇒Paparente = 88.2 N

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