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Y_Kistochka [10]
3 years ago
8

The magnitude of the electric current is directly proportional to the _____ of the electric field.

Physics
2 answers:
OLEGan [10]3 years ago
8 0

Answer: The magnitude of the electric current is directly proportional to the potential difference of the electric field.

Explanation:

According to Ohm's law, the current flowing between two points in the circuit is directly proportional to the potential difference at a constant temperature.

The electric potential is the amount of work done in moving a unit positive charge from reference point to a particular point in a electric field. And the electric field is equal to the negative potential gradient.

Therefore, the magnitude of the electric current is directly proportional to the potential difference of the electric field.  

Sedbober [7]3 years ago
3 0
Is proportional to the CHANGE.

hope this helps
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makkiz [27]
Some guidance notes which may help.To calculate the current flow, Ohm's law can be used. This can be written as current=voltage/resistance, or I=V/R. V is 1.5V.R for the copper wire quoted would be calculated as R = resistivity x length/cross sectional area. The area would be calculated from the formula area = pi x diameter squared/4So, R=resistivity x length divided by (pi x diameter squared/4)Until is the resistivity of copper is known, that's about as far as can be gone.Any further questions, please ask.
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3 years ago
Se lanza verticalmente una esfera con una rapidez de 30m/se. Determinar la rapidez de la esfera a una altura de 40m (g=10m/s2)
sergeinik [125]

v^2-{v_0}^2=2a(x-x_0)

dónde v es la velocidad de la esfera, v_0 es suya velocidad inicial, a=-g la aceleración debida a la gravedad, x la posición, y x_0 la posición inicial. Tomamos x_0=0\,\mathrm m a referirse a la posición de la esfera en el momento que la esfera fue lanzada.

Entonces

v^2-\left(30\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-10\,\dfrac{\mathrm m}{\mathrm s^2}\right)(40\,\mathrm m)

\implies v^2=100\,\dfrac{\mathrm m^2}{\mathrm s^2}\implies v=\pm10\,\dfrac{\mathrm m}{\mathrm s}

Esto nos dice que la esfera alcanza una altura de 40 m en dos momentos - una vez hacia arriba y una vez hacia abajo. Sin embargo, independientemente del signo de la velocidad, sabemos que suya magnitud es 10 m/s, y así tenemos una rapidez de 10 m/s también en ambos momentos.

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djverab [1.8K]
<h2>Answer: about the same size of the gap  or slit</h2>

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<h2>Waves diffract the most when their wavelength is <u>about the same size of the gap </u></h2>

<u />

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