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Lady_Fox [76]
3 years ago
11

A cube with sides of area 32 cm2 contains a 35.9 nanoCoulomb charge. Find the flux of the electric field through the surface of

the cube in unis of Nm2/C.

Physics
1 answer:
enot [183]3 years ago
3 0

Answer:

The electric flux through the surface is equal to 3.878 x 10³ Nm²/C

The field distance r is equal to half the length of each side of the cube.

From the area the length of each size was calculated and the field distance and charge were used in calculating the magnitude of the electric field vector which was found to be 202 x 10³ N/C

The total flux area available to this electric field is 6x32cm²

Explanation:

The full solution can be found in the attachment below.

Thank you for reading this post and I hope it is helpful to you.

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alexandr402 [8]
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1 year ago
Un movil viaja a 40km/h y comienza a reducir su velocidad a partir del instante t=0. Al cabo de 6 segundo se detiene completamen
aleksklad [387]

Answer:

1,85 m / s²

Explanation:

De la pregunta anterior, se obtuvieron los siguientes datos:

Velocidad inicial (u) = 40 km / h

Hora inicial (t₁) = 0

Tiempo final (t₂) = 6 s

Velocidad final (v) = 0

Aceleración (a) =?

A continuación, convertiremos 40 km / ha m / s. Esto se puede obtener de la siguiente manera:

1 km / h = 0,2778 m / s

Por lo tanto,

40 km / h = 40 km / h × 0,2778 m / s / 1 km / h

40 km / h = 11,11 m / s

Por tanto, 40 km / h equivalen a 11,11 m / s.

Finalmente, determinaremos la aceleración del móvil durante el período en el que desaceleró. Esto se puede obtener de la siguiente manera:

Velocidad inicial (u) = 11,11 m / s

Hora inicial (t₁) = 0

Tiempo final (t₂) = 6 s

Velocidad final (v) = 0

Aceleración (a) =?

a = (v - u) / (t₂ - t₁)

a = (0 - 11,11) / (6 - 0)

a = - 11,11 / 6

a = –1,85 m / s²

Por tanto, la aceleración del móvil durante el período en el que se ralentizó es de –1,85 m / s²

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3 years ago
A block of mass 10 kg slides down an inclined plane that has an angle of 30. If the inclined plane has no friction and the block
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No friction present means: Ek = Ep

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a.

The work done by a constant force along a rectilinear motion when the force and the displacement vector are not colinear is given by:

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where F is the magnitude of the force, theta is the angle between them and d is the distance.

The problen gives the following data:

The magnitude of the force 750 N.

The angle between the force and the displacement which is 25°

The distance, 26 m.

Plugging this in the formula we have:

\begin{gathered} W=\left(750\right)\left(\cos25\right)\left(26\right) \\ W=17673 \end{gathered}

Therefore the work done is 17673 J.

b)

The power is given by:

P=\frac{W}{t}

the problem states that the time it takes is 6 s. Then:

\begin{gathered} P=\frac{17673}{6} \\ P=2945.5 \end{gathered}

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Answer:

Explanation:

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