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coldgirl [10]
3 years ago
6

A charge +1.9 μC is placed at the center of the hollow spherical conductor with the inner radius 3.8 cm and outer radius 5.6 cm.

Suppose the conductor initially has a net charge of +3.8 μC instead of being neutral. What is the total charge (a) on the interior and (b) on the exterior surface?
Physics
1 answer:
Archy [21]3 years ago
5 0

To solve this problem we will apply the concepts related to load balancing. We will begin by defining what charges are acting inside and which charges are placed outside.

PART A)

The charge of the conducting shell is distributed only on its external surface. The point charge induces a negative charge on the inner surface of the conducting shell:

Q_{int}=-Q1=-1.9*10^{-6} C. This is the total charge on the inner surface of the conducting shell.

PART B)

The positive charge (of the same value) on the external surface of the conducting shell is:

Q_{ext}=+Q_1=1.9*10^{-6} C

The driver's net load is distributed through its outer surface. When inducing the new load, the total external load will be given by,

Q_{ext, Total}=Q_2+Q_{ext}

Q_{ext, Total}=1.9+3.8

Q_{ext, Total}=5.7 \mu C

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ExtremeBDS [4]

Answer:

a. S(t)=350−1t

Explanation:

To determine the equation of motion you take into account the general form of motion with constant velocity:

S(t)=S_o+vt    ( 1 )

So is the initial position from a specific reference frame. In this case is 350 m.

v is the speed of the motion, in this case is 1m/s. However, the motion is forward the zero point of the reference frame, hence, the speed is - 1m/s.

You replace the values of So and v in the equation ( 1 ) and you obtain:

S(t)=350-(1m/s)t

Hence, the answer is:

a. S(t)=350−1t

- - - - - - - - - - - - - - - - - - - - -

Para determinar a equação do movimento, você leva em consideração a forma geral do movimento com velocidade constante:

             (1)

Assim é a posição inicial de um quadro de referência específico. Neste caso, é de 350 m.

v é a velocidade do movimento, neste caso é de 1m / s. No entanto, o movimento é avançar o ponto zero do quadro de referência, portanto, a velocidade é de - 1m / s.

Você substitui os valores de So ev na equação (1) e obtém:

Portanto, a resposta é:

uma. S (t) = 350-1t, movimento retrógrado

4 0
3 years ago
Density of water is 1000kg/m3. Find out the amount of mass of 5m3 water​
Licemer1 [7]

Answer:

5000 m³

Explanation:

density = mass / volume

mass = density × volume

mass = 1000 × 5

mass = 5000 ( kg )

4 0
3 years ago
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Answer:

800pa

Explanation:

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6 0
3 years ago
A mass m = 1.1 kg hangs at the end of a vertical spring whose top end is fixed to the ceiling. The spring has spring constant k
adell [148]

Answer:

a) = 10.22 rad/s

b) = 0.35 m

Explanation:

Given

Mass of the particle, m = 1.1 kg

Force constant of the spring, k = 115 N/m

Distance at which the mass is released, d = 0.35 m

According to the differential equation of s Simple Harmonic Motion,

ω² = k / m, where

ω = angular frequency in rad/s

k = force constant in N/m

m = mass in kg

So,

ω² = 115 / 1.1

ω² = 104.55

ω = √104.55

ω = 10.22 rad/s

If y(0) = -0.35 m and we want our A to be positive, then suffice to say,

The value of coefficient A in meters is 0.35 m

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If there is no current in the wire .....the direction of magnetic field remains unchanged
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