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slava [35]
3 years ago
12

A copper wire of 1-m length and 1-mm radius has a resistance of 5.5 × 10 − 3 [Ω] and carries a current of 4 [A]. The number of f

ree electrons per unit volume in copper is 8.43 × 10 28 [m − 3 ]. Find (a) the current density, (b) the conductivity of the copper, (c) the power dissipated in the wire, (d) the (average) drift velocity of free electrons, and (e) the mobility of free electrons.
Physics
1 answer:
SOVA2 [1]3 years ago
8 0

Answer: a)1.27 *10^6 A/m^2; b) 57.9*10^6 Ω*m; c) 94.16*10^-6 m/s;

d) 4.29*10^-3 m^2/V*s

Explanation: In order to explain this problem we have to take into account the following expressions:

The current density is equal:

J=I/A  where A is the area of the wire.

J=4A/(π*1*10^-6m^2)=1.27 *10^6 A/m^2

The resistence of the wire is given by:

R=L/(σ*A) where σ and L are the conductivity and the length of the wire, respectively.

Then we have:

σ= L/(R*A)=1/(5.5*10^-3*π*1*10^-6)=57.9 *10^6 Ω*m

The drift velocity of free electrons is given by:

Vd=i/(n*A*e) where n is the numer of electrons per volume. e is the electron charge.

then we have:

Vd=J/(n*e)= 1.27 *10^6/(8.43*10^28*1.6*10^-19)=94.16 *10^-6 m/s

Finally, the mobility of free electrons is equal to:

μ=Vd/E  ;  E=J/σ

μ=E*σ/J= 94.16*10^-6*57.9 *10^6/1.27 *10^6=4.29*10^-3 m^2/V*s

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3 years ago
A person jogs eight complete laps around a quarter-mile track in a total time of 12.5 min. Calculate (a) the average speed and (
Margarita [4]

\large\displaystyle\text{$\begin{gathered}\sf \huge \bf{\underline{Data:}} \end{gathered}$}

  • \large\displaystyle\text{$\begin{gathered}\sf 1\ mile = 1609.34 \ m \end{gathered}$}
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                           \large\displaystyle\text{$\begin{gathered}\sf 12.5 \not{min}*\frac{60 \ s}{1\not{min}}=750 \ s \end{gathered}$}

                   \large\displaystyle\text{$\begin{gathered}\sf \bf{A) \ Calculate \ the \ average \ speed: } \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf 402.33 \ m*8 \ laps = 3218.64 \ m \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf d=3218.64 \ m \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf V=\frac{d}{t} \ \ \ \ \ \  V= \frac{3218.64 \ m }{750 \ s} \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V=4.29 \ m/s \end{gathered}$}

                  \large\displaystyle\text{$\begin{gathered}\sf \bf{B) \ Calculate \ the \ average \ speed \  in \ m/s} \end{gathered}$}

                          \large\displaystyle\text{$\begin{gathered}\sf V=402.33 \ m \end{gathered}$}  

                          \large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}

                          \large\displaystyle\text{$\begin{gathered}\sf V=\frac{D}{T} \ \ \ \ \ V=\frac{402.33 \ m}{750 \ s}   \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V= 0.53 \ m/s \end{gathered}$}

4 0
2 years ago
A wheel free to rotate about its axis that is not frictionless is initially at rest. A constant external torque of +46 N·m is ap
Rom4ik [11]

Answer:

Explanation:

Let Torque due to friction be

F  

Net torque

= 46 - F

Angular impulse = change in angular momentum

=(  46 - F ) x 17  = I X 580

When external torque is removed , only friction creates torque reducing its speed to zero in 120 s so

Angular impulse = change in angular momentum

F  x 120 = I X 580

(  46 - F ) x 17 = F  x 120

137 F = 46 x 17

F = 5.7 Nm

b )

Putting this value in first equation

5.7 x 120 = I x 580

I = 1.18 kg m²

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Calculate the magnitude of the acceleration of the box if you push on the box with a constant force 170.0 N that is parallel to
Degger [83]

The acceleration of the  box up the ramp is 9.65 m/s².

<h3>What is the magnitude of acceleration of the box?</h3>

The magnitude of the acceleration of the box is calculated by applying Newton's second law of motion as shown below;

F(net) = ma

where;

  • m is the mass of the box
  • a is the acceleration of the box

The net force on the box is calculated as follows;

F(net) = F - Ff

F(net) = F - μmgcosθ

where;

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  • μ is coefficient of friction

F(net) = 170 -  (0.3 x 15 x 9.8 x cos55)

F(net) = 144.7

The acceleration of the box is calculated as;

a = F(net) / m

a = (144.7) / (15)

a = 9.65 m/s²

Thus, the acceleration of the  box up the ramp is 9.65 m/s².

Learn more about acceleration here: brainly.com/question/14344386

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8 0
1 year ago
A car travels at an average speed of 60km/h for 15 minutes. How far does the car travel in this time?
jok3333 [9.3K]

Answer:

In 15 minutes the car travels a distance of 15 km.

Explanation:

4 0
3 years ago
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