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sergeinik [125]
3 years ago
13

To learn to apply the concept of current density and microscopic Ohm's law. A "gauge 8" jumper cable has a diameter d of 0.326 c

m. The cable carries a current I of 30.0 A. The electric field E in the cable is 0.062 V/m. Below is a list of some common metals with their:____________
Physics
2 answers:
alukav5142 [94]3 years ago
8 0

Answer:

Resistivity = (1.726 × 10⁻⁸) Ωm

Comparing this answer with the resistivity of the listed materials in the question, this resistivity matches that of Copper the most.

Hence, the material is Copper.

Option A is correct. Copper.

Explanation:

Given,

I = 30.0 A

d = 0.326 cm = 0.00326 m

E = 0.062 V/m

The current density for the wire is given as

J = (I/A)

where J = current density = ?

I = current = 30.0 A

A = Cross sectional Area of the wire = (πd²/4) = [π×(0.00326²)/4] = 0.0000083503 m²

J = (30 ÷ 0.0000083503) = 3,592,685.3 A/m²

On a microscopic scale, Ohm's law can be stated as

E = Jρ

where E = Electric field = 0.062 V

J = Current density = 3,592,685.3 A/m²

ρ = Resistivity of the material = ?

ρ = (E/J)

ρ = 0.062 ÷ 3,592,685.3 = (1.726 × 10⁻⁸) Ωm

Comparing this answer with the resistivity of the listed materials in the question, this resistivity matches that of Copper the most.

Hence, the material is Copper.

Hope this Helps!!!

serious [3.7K]3 years ago
6 0

Complete question:

To learn to apply the concept of current density and microscopic Ohm's law.

A "gauge 8" jumper cable has a diameter of 0.326 centimeters. The cable carries a current of 30.0 amperes. The electric field in the cable is 0.062 newtons per coulomb.

What is the material of the cable?

A. Copper

B. Silver

C. Aluminum

D. Gold

E. Steel

Answer:

It is made of Copper

Explanation:

We are given:

•diameter d= 0.0326cm = 0.00326m

• Current I = 30A

• Electric field e = 0.062v/m

To find the material of cable, we'll need to find the resistivity of the material using the formula:

R = e/J

We already know e = 0.062v/m

To find current density J, we have:

J = \frac{I}{pi * r^2}

But r = d/2 = 0.00326 / 2 = 0.00163

= \frac{30}{pi * 0.00163 m^2} = 3.594 x 10^6 A/m^2

Hence resistivity will be,

R = e/J =

R = \frac{0.062}{3.594*10^6 A/m^2}

R = 1.725*10^-^8 ohms/m

From the resistivity of the material we can conclude that it is made of Copper

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Andreas93 [3]

The angle of incline of the hill above the horizontal is 8.81°.

Since the hill has a 15.5% grade is one that rises 15.5 m vertically for every 100.0 m of distance in the horizontal direction.

<h3>Tangent of the angle of the incline of the hill,</h3>

The tangent of the angle of the incline of the hill, Ф is

tanФ = vertical rise/horizontal distance = grade of hill

Now, the vertical rise = 15.5 m and the horizontal distance = 100.0 m

So, substituting the values of the variables into the equation, we have

tanФ = vertical rise/horizontal run

tanФ = 15.5 m/100.0 m

tanФ = 0.155

<h3>Angle of incline of the hill</h3>

Taking inverse tan of both sides, we have

Ф = tan⁻¹(0.155)

Ф = 8.81°

So, the angle of incline of the hill above the horizontal is 8.81°.

Learn more about angle of incline of a hill here:

brainly.com/question/10056962

6 0
2 years ago
620 J of heat is added to the cylinder of an engine, which causes the gas inside to expand. As a result, the piston of the engin
Maru [420]

Answer:

<u>400</u> J work is done BY the engine.

The internal energy of the gas is <u>620</u> J

Explanation:

The given information are;

The heat added to the cylinder = 620 J

The force applied by the piston of the engine = 8.0 kN = 8,000 N

The distance over which the force moves (the piston) = 5.0 cm = 0.05 m

The work done (by the engine) = Force × Distance = 8,000 N × 0.05 m = 400 J

The internal energy is the sum of the kinetic and potential energy of the system

Therefore, given that the internal energy, U, is the sum total of the energy in the system

∴ U = The heat supplied to the system = 620 J

Which gives;

<u>400</u> J work is done BY the engine.

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6 0
3 years ago
The unit meters corresponds to what variable
Sati [7]

Do you have any options? My guess would be distance but I could be wrong.

7 0
3 years ago
A 5.00-kg object is attached to one end of a horizontal spring that has a negligible mass and a spring constant of 280 N/m. The
lina2011 [118]

Answer:

1) v = 0.45 m/s

2) v = 0.65 m/s

3) v = 0.75 m/s  

Explanation:

1) We can find the speed of the object by conservation of energy:

E_{i} = E_{f}

\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}

Where:

k: is the spring constant = 280 N/m

v: is the speed of the object =?

m: is the mass of the object = 5.00 kg

x: is the displacement of the spring

\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.08 m)^{2} + \frac{1}{2}5.00 kgv^{2}                              

v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.08 m)^{2}}{5.00 kg}} = 0.45 m/s

2) When the object is 5.00 cm (0.050 m) from equilibrium, the speed of the object is:

\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}    

\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.05 m)^{2} + \frac{1}{2}5.00 kgv^{2}      

v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.05 m)^{2}}{5.00 kg}} = 0.65 m/s      

 

3) When the object is at the equilibrium position, the speed of the object is:

\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}    

\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0 m)^{2} + \frac{1}{2}5.00 kgv^{2}      

v = \sqrt{\frac{280N/m(0.10 m)^{2}}{5.00 kg}} = 0.75 m/s

I hope it helps you!                                                                                        

8 0
3 years ago
Suppose you want to operate an ideal refrigerator with a cold temperature of − 15.5 °C , and you would like it to have a coeffic
tensa zangetsu [6.8K]

Answer:

15.65 °C

Explanation:

cold temperature (Tc) = -15.5 degree C = 273.15 - 15.5 = 257.65 kelvin

minimum coefficient of performance (η) = 8.25

find the maximum hot reservoir temperature of such a generator (Th)

η = \frac{Tc}{Th-Tc}

Th = Tc x (\frac{1}{η} + 1)

Th = 257.65 x (\frac{1}{8.25} + 1)

Th = 288.8 K

Th = 288.8 - 273.15 = 15.65 °C

8 0
3 years ago
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