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DochEvi
2 months ago
12

At room temperature what is the strength of the electric field in a 12-gauge copper wire (diameter 2.05 mm that is needed to cau

se a 2.75 a current to flow?
Physics
1 answer:
Maru [3.3K]2 months ago
6 0
Electric field strength can be calculated as resistivity of copper multiplied by the current density.
where
p= 1.72 x 10^-8 <span>ohm meter
diameter = 2.05mm=.00205 m
current = 2.75 A
</span>First, we determine the current density:
current density = current / cross-sectional area
Calculating the cross-sectional area
cross-sectional area = pi.(d/2)^2;
cross-sectional area = 3.3 006x10-6 m^2
Substituting values gives us
current density = 2.75A/3.3006x 10-6m^2
current density=35.55 x10^2 A/m^2
Thus, electric field strength =1.72 x 10^-8 ohm meter x 35.55 x10^2 A/m^2
Resulting in electric field strength of 46.415 Volts/m.


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A rear window defroster consists of a long, flat wire bonded to the inside surface of the window. When current passes through th
ValentinkaMS [3465]

Answer: 2.6*10^-8 Ωm

Explanation:

the length of the wire, l = 12.2 m

the width of the wire, w = 1.8 mm

the thickness of the wire, T = 0.11 mm

the potential difference of the battery, v = 12 V

the current in the battery, I = 7.5 A

Recall, Ohm’s law states, V = IR.

Hence, R = V/I

R = 12/7.5 = 1.6 Ω

Resistivity of a material is given by

ρ = RA/l

ρ = [1.6 * 1.8*10^-3 * 0.11*10^-3] / 12.2

ρ = (3.168*10^-7) / 12.2

ρ = 2.596*10^-8

Thus, the resistivity of the wire is 2.60*10^-8 Ωm

7 0
1 month ago
Read 2 more answers
A rod 150 cm long and of diameter 2.0 cm is subjected to an axial pull of 20 kn. if the modulus of elasticity of the material of
Sav [3153]
Given:
a rod with a circular cross section is experiencing uniaxial tension.
Length, L=1500 mm
radius, r = 10 mm
E=2*10^5 N/mm^2
Force, F=20 kN = 20,000 N
[note: newton (unit) in abbreviation is written in upper case, as in N ]

From the details provided, the cross-section area = &pi; r^2 = 100 &pi; =314 mm^2
(i) Stress,
&sigma;
=F/A
= 20000 N / 314 mm^2
= 6366.2 N/mm^2
= 6370 N/mm^2 (to 3 significant figures)

(ii) Strain
&epsilon;
= ratio of extension / original length
= &sigma; / E
= 6366.2 /(2*10^5)
= 0.03183 
= 0.0318 (to three significant figures)

(iii) elongation
= &epsilon; * L
= 0.03183*1500 mm
= 47.746 mm
= 47.7 mm  (to three significant figures)

5 0
1 month ago
A particular brand of gasoline has a density of 0.737 g/mLg/mL at 25 ∘C∘C. How many grams of this gasoline would fill a 14.9 gal
Yuliya22 [3333]

Answer:

The answer to your inquiry is Mass = 41230.7 g or 41.23 kg.

Explanation:

Data

Density = 0.737 g/ml

Mass = ?

Volume = 14.9 gal

1 gal = 3.78 l

Process

1.- Convert gallons to liters

1 gal ---------------- 3.78 l

14.9 gal ------------- x

x = 56.44 l

2.- Convert liters to milliliters

1 l ------------------- 1000 ml

56.44 l --------------- x

x = (56.44 x 1000) / 1

x = 56444 ml

3.- Calculate the mass

Formula

Density = \frac{mass}{volume}

Solving for mass

Mass = density x volume

Substituting values

Mass = 0.737 x 56444

Result

Mass = 41230.7 g or 41.23 kg.

3 0
2 months ago
Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of
inna [3103]

Answer:

\int\limits^2_0 {} \, \int\limits^{5\pi}_0 {F(S(u,v)).N} \,dvdu ≈ 3077.34

Explanation:

To calculate the flux of F (vector field) across surface S, where

F(x,y,z) = y i − x j + z^{2} k

and S(u,v) = u cos v i + u sin v j + v k, 0 ≤ u ≤ 2, 0 ≤ v ≤ 5π

We will evaluate the following integral:

\int\limits^2_0 {} \, \int\limits^{5\pi}_0 {F(S(u,v)).N} \,dvdu

Substituting for the surface

x = u cos v

y = u sin v

z = v

Then

F(S(u,v)) = u sin v i - u cos v j + v^{2}k

The normal vector N is computed as

N = S_{u}XS_{v}

Where:

S_{u} = =

S_{v} = =

N = < cos v, sin v, 0 > X <- u sin v, u cos v, 2v

N = < 2v sin v, -2v cos v, u >

F(S(u,v)).N = < u sin v, -u cos v, v^{2}>. < 2v sin v, -2v cos v, u >

F(S(u,v)).N = 2uv + uv^{2}

Thus

\int\limits^2_0 {} \, \int\limits^{5\pi}_0 {2uv}+u v^{2}\,dvdu ≈ 3077.34

8 0
1 month ago
A person weighing 55 kg walks by applying 160 N of force on the ground, while pushing a 10-kg object. If the person accelerates
Sav [3153]

Answer:

I'm uncertain

Explanation:

since I didn't provide a correct answer, continue with my inquiries and you can use 'I'm uncertain' for 100 points.

6 0
2 months ago
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