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3 years ago

11

Una carga de 4 nC es transportada desde el suelo hasta la superficie de una esfera cargada con un trabajo de 7 X 10-5 j. Determi

nar el potencial eléctrico de la esfera

1 answer:

ollegr [7]3 years ago

3 0

Respuesta:

2 × 10⁴ V

Explicación:

Paso 1: Información provista

Carga transportada (q): 4 nC

Trabajo realizado (W): 7 × 10⁻⁵ J

Paso 2: Convertir q a Coulomb

Usaremos el factor de conversión 1 C = 10⁹ nC.

4 nC × 1 C/10⁹ nC = 4 × 10⁻⁹ C

Paso 3: Calcular el potencial eléctrico (V) de la esfera

Usaremos la siguiente fórmula.

V = W/q

V = 7 × 10⁻⁵ J/4 × 10⁻⁹ C = 2 × 10⁴ V

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Particle 1 and particle 2 have masses of m1 = 2.2 × 10-8 kg and m2 = 4.8 × 10-8 kg, but they carry the same charge q. The two pa

Lorico [155]

Answer: $r_2=11.81 cm$

Explanation:

Given

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$m_2=4.8\times 10^{-8} kg$

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potential Energy due to Magnetic Field =Kinetic Energy of Particle

$qV=\frac{mv^2}{2}$

$v=\sqrt{\frac{2qV}{m}}$

and we know

Force due to magnetic field will Provide centripetal Force

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and B is equal for both particles

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2 years ago

A playground merry-go-round has a mass of 115 kg and a radius of 2.50 m and it is rotating with an angular velocity of 0.520 rev

tatuchka [14]

Answer:

$W_f = 2.319 rad/s$

Explanation:

For answer this we will use the law of the conservation of the angular momentum.

$L_i = L_f$

so:

$I_mW_m = I_sW_f$

where $I_m$ is the moment of inertia of the merry-go-round, $W_m$ is the initial angular velocity of the merry-go-round, $I_s$ is the moment of inertia of the merry-go-round and the child together and $W_f$ is the final angular velocity.

First, we will find the moment of inertia of the merry-go-round using:

I = $\frac{1}{2}M_mR^2$

I = $\frac{1}{2}(115 kg)(2.5m)^2$

I = 359.375 kg*m^2

Where $M_m$ is the mass and R is the radio of the merry-go-round

Second, we will change the initial angular velocity to rad/s as:

W = 0.520*2 $\pi$ rad/s

W = 3.2672 rad/s

Third, we will find the moment of inertia of both after the collision:

$I_s = \frac{1}{2}M_mR^2+mR^2$

$I_s = \frac{1}{2}(115kg)(2.5m)^2+(23.5kg)(2.5m)^2$

$I_s = 506.25kg*m^2$

Finally we replace all the data:

$(359.375)(3.2672) = (506.25)W_f$

Solving for $W_f$ :

$W_f = 2.319 rad/s$

7 0

2 years ago