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3 years ago

When there is a large unbalanced charge in a conductor, positive charges will move through the conductor when another

Physics

1 answer:

GuDViN [60]3 years ago

7 0

This is true; it needs to balance it’s charge. However, it could also be negative charges as well.

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Classify Cd2+ Mn2+ Kr Zr as Paramagnetic or diamagnetic

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3 years ago

Read 2 more answers

Assume air resistance is negligible unless otherwise stated. Calculate the displacement in m and velocity in m/s at the followin

tankabanditka [31]

Answer: 1) t=0.5 s; S=6.225 m and v=14.9 m/s

2) t=1 s; S=14.9 m and v=19.8 m/s

3) t=1.5 s; S=26.05 m and v=24.7 m/s

Explanation:

The displacement $S$ is given by

$S=ut+\frac{1}{2} at^{2}$

and final velocity $v$ is given by

$v=u+at$

where $u$ is the initial velocity

$a$ is acceleration

$t$ is time taken

Case 1: when time is 0.5 s

The displacement is

$S=ut+\frac{1}{2} at^{2}\\S=10\times 0.5 +\frac{1}{2}\times 9.8\times 0.5^{2}\\\\S=6.225 m$

the velocity is

$v=u+at\\v=10+9.8\times 0.5\\v=14.9 m/s$

Case 2: when t=1 sec

The displacement is

$S=ut+\frac{1}{2} at^{2}\\S=10\times 1 +\frac{1}{2}\times 9.8\times 1^{2}\\\\S=14.9 m$

the velocity is

$v=u+at\\v=10+9.8\times 1\\v=19.8 m/s$

Case 3: t=1.5 s

The displacement is

$S=ut+\frac{1}{2} at^{2}\\S=10\times 1.5 +\frac{1}{2}\times 9.8\times 1.5^{2}\\\\S=26.05 m$

the velocity is

$v=u+at\\v=10+9.8\times 1.5\\v=24.7 m/s$

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3 years ago

A long coaxial cable consists of an inner cylindrical conductor with radius a and an outer coaxial cylinder with inner radius b

Natasha_Volkova [10]

Answer:

Part a)

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part b)

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part d)

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.

Explanation:

Part a)

By Guass law we know that

$\int E. dA = \frac{q}{\epsilon_0}$

$E. 2\pi rL = \frac{\lambda L}{\epsilon_0}$

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part b)

Outside the outer cylinder we will again use Guass law

$\int E. dA = \frac{q}{\epsilon_0}$

$E. 2\pi rL = \frac{\lambda L}{\epsilon_0}$

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Part d)

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.

4 0

3 years ago