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3 years ago

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When electric current is flowing in a metal, the electrons are movinga. at nearly the speed of light.b. at the speed of sound in

air.c. at the speed of sound in the metal.d. at none of the above speeds.e. at the speed of light.

1 answer:

yaroslaw [1]3 years ago

8 0

Answer:

Option (d)

Explanation:

The electrons in a conductor moves with the drift velocity when the electric current is flowing through the conductor.

The drift velocity is due to the applied electric field across the conductor.

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Acid rain contributes to soil erosion. Acid rain can be caused by

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4 years ago

Read 2 more answers

Three long parallel wires each carry 2.0-A currents in the same direction. The wires are oriented vertically, and they pass thro

blagie [28]

Answer:

$21.2\times 10^{-6}$ T

Explanation:

$i$ = magnitude of current in each wire = 2.0 A

$a$ = length of the side of the square = 4 cm = 0.04 m

$r$ = length of the diagonal of the square = $\sqrt{2}$ a = $\sqrt{2}$ (0.04) = 0.057 m

$B$ = magnitude of magnetic field by wires at A and C

$B = \left ( \frac{\mu _{o}}{4\pi } \right )\left ( \frac{2i}{a} \right )$

$B = (10^{-7}) \left ( \frac{2(2)}{0.04} \right )$

$B = 10\times 10^{-6}$ T

$B'$ = magnitude of magnetic field by wire at B

$B' = \left ( \frac{\mu _{o}}{4\pi } \right )\left ( \frac{2i}{r} \right )$

$B' = (10^{-7}) \left ( \frac{2(2)}{0.057} \right )$

$B' = 7.02\times 10^{-6}$ T

Net magnitude of the magnetic field at D is given as

$B_{net} = \sqrt{B^{2}+B^{2}} + B'$

$B_{net} = \sqrt{2} B + B'$

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3 years ago

A 180 g model airplane charged to 18 mC and traveling at 2.2 m/s passes within 8.6 cm of a wire, nearly parallel to its path, ca

viva [34]

Answer:

$a=0.2*10^{-5}g$

Explanation:

From the question we are told that:

Mass $M=180=>0.18kg$

Charge $Q=18mC=18*10^-^3C$

Velocity $v=2.2m/s$

Length of Wire $L=8.6cm=>0.086$

Current $I=30A$

Generally the equation for Magnetic Field of Wire B is mathematically given by

$B=\frac{\mu_0*I}{2\pi*l}$

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Generally the equation for Force on the plane F is mathematically given by

$F=qvB$

Therefore

$ma=qvB$

$a=\frac{qvB}{m}$

$a=\frac{18*10^{-5}83.4*6.978*10^{-5}}{0.18kg}$

$a=2.37*10^{-5}$

Therefore in Terms of g's

$a=\frac{2.37*10^{-5}}{9.8}$

$a=0.2*10^{-5}g$

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3 years ago

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steposvetlana [31]

c) only from warmer areas to colder areas.

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3 years ago