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3 years ago

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A metal spherical shell with inner radius 14 cm and outer radius 24 cm has a net charge of 1= -3 nC. At the center of the shell

is a small particle with charge Q2= -7 nC. What is the charge density in (C/m2) on the outer surface of the spherical shell?

1 answer:

CaHeK987 [17]3 years ago

6 0

Answer:

I don't no the answer sorry

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What are the basic si units for the wavelength of light?

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Westkost [7]

Answer:

(a). The rotational inertia is $5.72\times10^{39}\ kg m^2$

(b). The magnitude of the magnetic torque is $3.20\times10^{35}\ N-m$

Explanation:

Given that,

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$I=\dfrac{2}{5}MR^2$ ...(I)

We know that,

$\rho=\dfrac{M}{V}$

Put the value of volume

$\rho=\dfrac{3M_{n}}{4\pi R^3}$

$R^2=(\dfrac{3M_{n}}{4\pi\rho})^{\frac{2}{3}}$

Put the value of R in equation (I)

$I=\dfrac{2}{5}\times M_{n}\times(\dfrac{3M_{n}}{4\pi\rho})^{\frac{2}{3}}$

Put the value into the formula

$I=\dfrac{2}{5}\times(13\times2\times10^{30})^{\frac{5}{3}}\times(\dfrac{3}{4\pi\times(4.8\times10^{17})})^{\frac{2}{3}}$

$I=5.72\times10^{39}\ kg m^2$

The rotational inertia is $5.72\times10^{39}\ kg m^2$ .

(b). We need to calculate the magnitude of the magnetic torque

Using formula of torque

$\tau=I\times \alpha$

Put the value into the formula

$\tau=5.72\times10^{39}\times5.6\times10^{-5}$

$\tau=3.20\times10^{35}\ N-m$

The magnitude of the magnetic torque is $3.20\times10^{35}\ N-m$

Hence, (a). The rotational inertia is $5.72\times10^{39}\ kg m^2$

(b). The magnitude of the magnetic torque is $3.20\times10^{35}\ N-m$

4 0

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