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ikadub [295]
3 years ago
13

When electric current is flowing in a metal, the electrons are moving.A. at nearly the speed of lightB. at the speed of lightC.

at the speed of sound in the metalD. at the speed of sound in airE. at none of the above speeds
Physics
2 answers:
larisa86 [58]3 years ago
4 0

Answer:

E. at none of the above speeds

Explanation:

When current flowing through the metal then the speed of electrons in metal is not very high.

This speed of all electrons inside metal is opposite the the electric field which is due to the applied potential difference on the metal by external battery.

As we know that

\Delta V = i R

here for the current flowing in the metal the all the free electrons will move at drift speed which is given as

i = neAv_d

here speed of electrons will be

v_d = drift speed

n = number density of electrons

A = crossectional area

e = charge of an electron

in general this speed is very small and approximately of order cm/s

irina [24]3 years ago
3 0

I believe its E.

Traveling through metal, electrons can not go the speed of light

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In a magnet, the domains all point in the same direction; in an ordinary piece of metal, they're all jumbled up.

Explanation:

In a magnet, the domains all point toward the north pole; in an ordinary piece of metal, they all point to the south pole.



Side note:
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A team of dogs accelerates a 290kg dogsled from 0 to 6.0m/s in 3.0 s. Assume that the acceleration is constant.Part AWhat is the
Mademuasel [1]

Answer:

(a) a=2m/sec^2

(b) 5220 j

(c) 1740 watt

(d) 3446.66 watt

Explanation:

We have given mass m = 290 kg

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Final velocity v = 6 m/sec

Time t = 3 sec

From first equation of motion

v = u+at

So a=\frac{v-u}{t}=\frac{6-0}{3}=2m/sec^2

(a) We know that force is given by

F = ma

So force will be F=290\times 2=580N

(b) From second equation of motion we know that

s=ut+\frac{1}{2}at^2=0\times 3+\frac{1}{2}\times 2\times 3^2=9m

We know that work done is given by

W = F s = 580×9 =5220 j

(c) Time is given as t = 3 sec

We know that power is given as

P=\frac{W}{t}=\frac{5220}{3}=1740Watt

(d) Time t = 1.5 sec

So P=\frac{W}{t}=\frac{5220}{1.5}=3466.66Watt

5 0
3 years ago
Calculate the acceleration of a 270000-kg jumbo jet just before takeoff when the thrust on the aircraft is 160000 N .
Radda [10]

Answer:

<h3>The answer is 0.59 m/s²</h3>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

f is the force

m is the mass

From the question we have

a =  \frac{160000}{270000}  =  \frac{16}{27}  \\  = 0.592592...

We have the final answer as

<h3>0.59 m/s²</h3>

Hope this helps you

7 0
3 years ago
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