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-Dominant- [34]

3 years ago

5

A straight section of silver wire has a cross sectional area A=2.0 mm2. A total of 9.4×1018 electrons pass through the wire in 3

.0s. The conduction electron density in silver is 5.8×1028 electrons/m3. How far along the straight section do the electrons in the wire move, on average, during the 3.0s in question?

1 answer:

Sergio [31]3 years ago

3 0

Answer:

8.08 x 10^-5 m

Explanation:

A = 2 mm^2 = 2 x 10^-6 m^2

Total number of electrons, N = 9.4 x 10^18

time, t = 3 s

n = 5.8 x 10^28 electrons/ m^3

Current, i = Q / t = N x e / t = (9.4 x 10^18 x 1.6 x 10^-19) / 3 = 0.5 A

Let vd be the drift velocity.

i = n e A vd

0.5 = 5.8 x 10^28 x 1.6 x 10^-19 x 2 x 10^-6 x vd

vd = 2.7 x 10^-5 m/s

Distance traveled by the electrons = velocity x time

= vd x t = 2.7 x 10^-5 x 3 = 8.08 x 10^-5 m

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A light plane must reach a speed of 33 m/s per take off. How long a runway is needed if the constant acceleration is 3.0 m/s^2.

aleksley [76]

Given the final velocity (Vf) and the acceleration (a), the distance that should be traveled by the plane is calculated through the equation,
d = (Vf² - Vi²) / 2a
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d = ((33 m/s)² - 0)) / 2(3 m/s²)
The distance is therefore equal to 181.5 meters.

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2 years ago

An electron in a mercury atom changes from energy level b to a higher energy level when the

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2 years ago

series RC circuit is built with a 15 kΩ resistor and a parallel-plate capacitor with 18-cm-diameter electrodes. A 18 V, 36 kHz s

andre [41]

Answer:

$d=1.84\ mm$

Explanation:

<u>Capacitance</u>

A two parallel-plate capacitor has a capacitance of

$\displaystyle C=\frac{\epsilon_o A}{d}$

where

$\epsilon_o=8.85\cdot 10^{-12}\ F/m$

A = area of the plates = $\pi r^2$

d = separation of the plates

$\displaystyle d=\frac{\epsilon_o A}{C}=\frac{\epsilon_o \pi r^2}{C}$

We need to compute C. We'll use the circuit parameters for that. The reactance of a capacitor is given by

$\displaystyle X_c=\frac{1}{wC}$

where w is the angular frequency

$w=2\pi f=2\pi \cdot 36000=226194.67\ rad/s$

Solving for C

$\displaystyle C=\frac{1}{wX_c}$

The reactance can be found knowing the total impedance of the circuit:

$Z^2=R^2+X_c^2$

Where R is the resistance, $R=15 K\Omega=15000\Omega$ . Solving for Xc

$X_c^2=Z^2-R^2$

The magnitude of the impedance is computed as the ratio of the rms voltage and rms current

$\displaystyle Z=\frac{V}{I}$

The rms current is the peak current Ip divided by $\sqrt{2}$ , thus

$\displaystyle Z=\frac{\sqrt{2}V}{I_p}$

$I_p=0.65\ mA/1000=0.00065\ A$

Now collect formulas

$\displaystyle X_c^2=Z^2-R^2=\left(\frac{\sqrt{2}V}{I_p}\right)^2-R^2$

Or, equivalently

$\displaystyle X_c=\sqrt{\frac{2V^2}{I_p^2}-R^2}$

$\displaystyle X_c=\sqrt{\frac{2\cdot 18^2}{0.00065^2}-15000^2}$

$X_c=36176.34\ \Omega$

The capacitance is now

$\displaystyle C=\frac{1}{226194.67\cdot 36176.34}=1.22\cdot 10^{-10}\ F$

The radius of the plates is

$r=18\ cm/2=9 \ cm = 0.09 \ m$

The separation between the plates is

$\displaystyle d=\frac{8.85\cdot 10^{-12} \cdot \pi\cdot 0.09^2}{1.22\cdot 10^{-10}}$

$d=0.00184\ m$

$\boxed{d=1.84\ mm}$

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