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3 years ago

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The drift speed in a copper wire is 4.08 × 10-5 m/s for a typical electron current. Calculate the magnitude of the electric fiel

d inside the copper wire. The mobility of mobile electrons in copper is 4.5 × 10-3 (m/s)/(N/C). (Note that though the electric field in the wire is very small, it is adequate to push a sizable electron current through the copper wire.)

1 answer:

jeka943 years ago

6 0

Answer:

9.07 x 10^-3 N/C

Explanation:

The mobility of electrons is defined as the ratio of drift velocity to the applied electric field.

Vd = 4.08 x 10^-5 m/s

Mobility = 4.5 x 10^-3 m/s

Let the electric field is E.

Mobility = Vd / E

E = Vd / mobility

E = 4.08 x 10^-5 / (4.5 x 10^-3)

E = 9.07 x 10^-3 N/C

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A uniform electric field exists everywhere in the x, y plane. This electric field has a magnitude of 4500 N/C and is directed in

Anton [14]

Answer with Explanation:

We are given that

$E=4500 N/C$

$q=-5.5\times 10^{-9} C$

a.x=-0.2 m

$E'=\frac{Kq}{r^2}=\frac{9\times 10^9\times 5.5\times 10^{-9}}{(0.2)^2}=1237.5 N$

Where $k=9\times 10^9$

Net electric field=E+E'=4500+1237.5=5737.5 N/C

b.x=0.20 m

Net electric field=E-E'=4500-1237.5=3262.5 N/C

c.y=0.20 m

E=1237.5 N/C

Net electric field= $\sqrt{E^2+E'^2}=\sqrt{(4500)^2+(1237.5)^2}=4667.05 N/C$

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3 years ago

Another name for heat energy

scoundrel [369]

<span>Another name for heat energy is Thermal energy , hope this helps</span>

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4 years ago

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Scientists experimenting with two charged objects of 2.56 x 10-6 C and 3.34 x 10-7 C displayed a repulsion of 2.26 x 10-3 N. How

Pachacha [2.7K]

Answer:

The objects were 1.8m apart.

Explanation:

We will start stating the Coulomb's Law. It says that:

$F_e=\frac{Kq_1q_2}{r^{2}}$

Where F_e is the electric force between the objects, q_1 and q_2 are the magnitude of the charge of the objects, r is the distance between them and K is the Coulomb's constant ( $K=8.9*10^{9} \frac{Nm^{2} }{C^{2} }$ in vacuum). Solving for the distance r we have:

$r=\sqrt{\frac{Kq_1q_2}{F_e} }$

Plugging the given values into this equation, we obtain:

$r=\sqrt{\frac{(8.9*10^{9}\frac{Nm^{2} }{C^{2} })(2.56*10^{-6}C)(3.34*10^{-7}C)}{2.26*10^{-3}N}}=1.8m$

In words, the two charged objects were 1.8m apart.

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3 years ago

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On land, coal is transported primarily by train. A typical large coal train may be about 1.5 km long and may consist of 120 cars

WINSTONCH [101]

Answer:

equivalent power = 6.60 × $10^{12}$ W

Explanation:

given data

coal train length = 1.5 km = 1500 m

contain = 120 cars

holding= 110 tonnes of coal each

train traveling speed = 100 km/h = 27.78 m/s

to find out

equivalent power in watts

solution

we take here energy produce by coal is 27 × $10^{6}$ J/kg

and here total coal = 120 × 110 × 10³ kg

sop here energy produce by total coal is

energy produce by total coal = 27 × $10^{6}$ × 120 × 110 × 10³

and time required to cross that distance is

time = $\frac{distance}{speed}$

time = $\frac{1500}{27.78}$

time = 53.99 sec

so here equivalent power will be

equivalent power = $\frac{energy}{time}$

equivalent power = $\frac{27*10^6*120*110*10^3}{53.99}$

equivalent power = 6.60 × $10^{12}$ W

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3 years ago

If y = 0.02 sin (30x – 200t) (SI units), the frequency of the wave is

leva [86]

Answer:

31.831 Hz.

Explanation:

<u>Given:</u>

$\rm y = 0.02\sin(30x-200 t).$

The vertical displacement of a wave is given in generalized form as

$\rm y = A\sin(kx -\omega t).$

<em>where</em>,

A = amplitude of the displacement of the wave.
k = wave number of the wave = $\dfrac{2\pi }{\lambda}.$
$\lambda$ = wavelength of the wave.
x = horizontal displacement of the wave.
$\omega$ = angular frequency of the wave = $\rm 2\pi f$ .
f = frequency of the wave.
t = time at which the displacement is calculated.

On comparing the generalized equation with the given equation of the displacement of the wave, we get,

$\rm A=0.02.\\k=30.\\\omega =200.\\$

therefore,

$\rm 2\pi f=200\\\\\Rightarrow f = \dfrac{200}{2\pi}=31.831\ Hz.$

It is the required frequency of the wave.

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