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3 years ago

A. What is an electric field line?

Physics

1 answer:

Lorico [155]3 years ago

7 0

Answer:

Please see below as the answers are self-explanatory

Explanation:

A electric field line is an imaginary line, which has the property that the electric field vector is tangent to it at any point. It starts from positive charges (since the electric field by convention it has the direction of the trajectory that would take a positive test charge, so it always goes away from positive charges) and ends in negative charges.

Since the potential difference between two points represents the work per unit charge needed for a charge to move between these points, a equipotential surface is the one over which it is not needed to do work to move a charge from any point on the surface to any other point, which means that all points are at the same potential.

Equipotential surfaces are not necessarily physical surfaces, they can be defined in vaccum for instance.
As an example, any spherical surface concentric with a point charge, is an equipotential surface, and it can be a real surface or a fictitious one.

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Which of the following is an example of rotation

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I believe it is D. Earth spinning on it's axis.

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3 years ago

The magnitude E of an electric field depends on the radial distance r according to E = A/r4, where A is a constant with unit vol

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Answer:

$\Delta V = 0.053 A$

Explanation:

Electric field in a given region is given by equation

$E = \frac{A}{r^4}$

as we know the relation between electric field and potential difference is given as

$\Delta V = -\int E. dr$

so here we have

$\Delta V = - \int (\frac{A}{r^4}) .dr$

$\Delta V = \frac{A}{3r_1^3} - \frac{A}{3r_2^3}$

here we know that

$r_1 = 1.71 m$ and $r_2 = 2.89 m$

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$\Delta V = \frac{A}{3}(\frac{1}{1.71^3} - \frac{1}{2.89^3})$

so we will have

$\Delta V = 0.053 A$

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3 years ago

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2 years ago

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3 years ago

A neutron star is an extremely dense, rapidly spinning object that results from the collapse of a massive star at the end ofits

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,

Mass of neutron $M_{n}= 13M_{s}$

Density of neutron $\rho=4.8\times10^{17}\ kg/m^3$

(a). We need to calculate the rotational inertia

Using formula of rotational inertia for sphere

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

3 years ago