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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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Misha Larkins [42]

The correct answer is option B, representational

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3 years ago
Read 2 more answers
A car is traveling north with a velocity of 18.1 m/s. Find the velocity of the car after 7.50 seconds if the acceleration is 2.4
Vedmedyk [2.9K]
Hello!

Vx = V0x + Ax*t
Vx = 18.1 + 2.4t

Let’s take time as 7.50 seconds:
Vx = 18.1 + 2.4*7.50
Vx = 18.1 + 18 = 36.1 m/s

Then, the final velocity of the car is 36.1 m/s.
3 0
3 years ago
Paul lifts a sack weighing 245 newtons vertically from the ground and places it on a platform at a height of 0.7 meters. If he t
d1i1m1o1n [39]

Answer:

B. 17.15 watts

Explanation:

Given that

Time = 10 seconds

height = distance = 0.7 meters

weight of sack = mg = F = 245 newtons

Power = work done/ time taken

Where work done = force × distance

Substituting the given parameters into the formula

Work done = 245 newton × 0.7 meters

Work done = 171.5 J

Recall,

Power = work done/time

Power = 171.5 J ÷ 10

Power = 17.15 watts

Hence the power expended is B. 17.15 watts

6 0
3 years ago
Susan, driving north at 53 mphmph , and Shawn, driving east at 63 mphmph , are approaching an intersection. Part A What is Shawn
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Answer:

Shawn's speed relative to Susan's speed = 10 mph

Resultant velocity = 82.32 mph

Explanation:

The given data :-

i) Susan driving in north and speed of Susan is ( v₁ ) = 53 mph.

ii) Shawn driving in east and speed of Shawn is ( v₂ ) = 63 mph.

iii) The speed of both Susan and Shawn is relative to earth.

iv) The angle between Susan in north and Shawn in east is 90°.

We have to find Shawn's speed relative to Susan's speed.

v₂₁ = v₂ - v₁   = 63 - 53 = 10 mph

Resultant velocity,

v = \sqrt{v_{2} ^{2}+ v_{1} ^{2}  }  =\sqrt{63^{2} +53^{2} }

v = 82.32 mph

5 0
3 years ago
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