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3 years ago

12

Copper is used in wiring because its metallic bonding gives its _______ a high degree of freedom and the material a high conduct

ivity.
A) electrons

B) neutrons

C) protons

2 answers:

Alecsey [184]3 years ago

4 0

Answer:

electrons

Explanation:

Alexxx [7]3 years ago

4 0

Answer:

Electrons

Explanation:

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Explanation:

2 m/s²
0.19 m/s²
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0.04 m/s²
100.39 m/s²

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3 years ago

Is a switch essential for a circuit to operate

antiseptic1488 [7]

Answer:

yes it is essential

Explanation:

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3 years ago

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BabaBlast [244]

Let say the point is inside the cylinder

then as per Gauss' law we have

$\int E.dA = \frac{q}{\epcilon_0}$

here q = charge inside the gaussian surface.

Now if our point is inside the cylinder then we can say that gaussian surface has charge less than total charge.

we will calculate the charge first which is given as

$q = \int \rho dV$

$q = \rho * \pi r^2 *L$

now using the equation of Gauss law we will have

$\int E.dA = \frac{\rho * \pi r^2* L}{\epcilon_0}$

$E. 2\pi r L = \frac{\rho * \pi r^2* L}{\epcilon_0}$

now we will have

$E = \frac{\rho r}{2 \epcilon_0}$

Now if we have a situation that the point lies outside the cylinder

we will calculate the charge first which is given as it is now the total charge of the cylinder

$q = \int \rho dV$

$q = \rho * \pi r_0^2 *L$

now using the equation of Gauss law we will have

$\int E.dA = \frac{\rho * \pi r_0^2* L}{\epcilon_0}$

$E. 2\pi r L = \frac{\rho * \pi r_0^2* L}{\epcilon_0}$

now we will have

$E = \frac{\rho r_0^2}{2 \epcilon_0 r}$

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Lina20 [59]

Answer:

Hmm i need more info

Explanation:

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3 years ago