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Serga [27]
3 years ago
15

Residential building codes typically require the use of 12-gauge copper wire (diameter 0.205 cm) for wiring receptacles. Such ci

rcuits carry currents as large as 20.0 A. If a wire of smaller diameter (with a higher gauge number) carried that much current, the wire could rise to a high temperature and cause a fire. Calculate the rate at which internal energy is produced in 1.00 m of 12-gauge copper wire carrying 20.0 A.
Physics
1 answer:
Alecsey [184]3 years ago
8 0

Given Information:  

Current = I = 20 A

Diameter = d = 0.205 cm = 0.00205 m

Length of wire = L = 1 m

Required Information:  

Energy produced = P = ?

Answer:  

P = 2.03 J/s

Explanation:  

We know that power required in a wire is

P = I²R

and R = ρL/A

Where ρ is the resistivity of the copper wire 1.68x10⁻⁸ Ω.m

L is the length of the wire and A is the area of the cross-section and is given by

A = πr²

A = π(d/2)²

A = π(0.00205/2)²

A = 3.3x10⁻⁶ m²

R = ρL/A

R = 1.68x10⁻⁸*(1)/3.3x10⁻⁶

R = 5.09x10⁻³ Ω

P = I²R

P = (20)²*5.09x10⁻³

P = 2.03 Watts or P = 2.03 J/s

Therefore, 2.03 J/s of energy is produced in 1.00 m of 12-gauge copper wire carrying a current of 20 A

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Answer:

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6 0
3 years ago
What is the car's speed at the bottom of the dip?The passengers in a roller coaster car feel 50% heavier thantheir true weight a
Rashid [163]

Answer:

v = 14 m/s

Explanation:

given,

radius of dip = 40 m

The passengers in a roller coaster car feel 50% heavier than their true weight.

Apparent weight

A = W + \dfrac{W}{2}

A =\dfrac{3W}{2}

A =\dfrac{3mg}{2}

When the car is at the bottom,  the weight will be acting downwards and the centripetal force will also be acting downward where as Normal force which is apparent weight will be acting in upward direction.

now,

N = m g + \dfrac{mv^2}{r}

\dfrac{3mg}{2} = m g + \dfrac{mv^2}{r}

\dfrac{mg}{2} = \dfrac{mv^2}{r}

v = \sqrt{\dfrac{rg}{2}}

v = \sqrt{\dfrac{40\times 9.8}{2}}

v = 14 m/s

8 0
2 years ago
The forces acting on this object are _______ and the net force is ________ Balanced, 0 N Balanced, 10 N Unbalanced, 10 N Unbalan
Amanda [17]
1st Blank Answer: Balanced   
2nd Blank Answer: 0n 

I took the test and got it right
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5 0
2 years ago
Read 2 more answers
The electric potential at the origin of an xy-coordinate system is 40 V. A -8.0-μC charge is brought from x = +∞ to that point.
vredina [299]

Answer:

-320 μJ.

Explanation:

Consider a point with an electrical charge of q. Assume that V is the electrical potential at the position of that charge. The electrical potential of that point charge will be equal to:

\text{Potential Energy} = q \cdot V.

Keep in mind that since both q and V might not be positive, the size of the electrical potential energy might not be positive, either.

For this point charge,

  • q = \rm -8.0\; \mu C; (that's -8.0 microjoules, which equals to \rm -8.0\times 10^{-6}\; J)
  • V = \rm 40\; V.

Hence its electrical potential energy:

\text{Potential Energy} = q\cdot V = \rm (-8.0\; \mu C) \times 40\; V = -320\; \mu J.

Why is this value negative? The electrical potential energy of a charge is equal to the work needed to bring that charge from infinitely far away all the way to its current position. Also, negative charges are attracted towards regions of high electrical potential. Bringing this \rm -8.0\; \mu C negative charge to the origin will not require any external work. Instead, this process will release 320 μJ of energy. As a result, the electrical potential energy is a negative value.

7 0
3 years ago
An ideal spring hangs from the ceiling. A 1.25-kg mass is hung from the spring. After all vibrations have died away, the spring
ch4aika [34]

The kinetic energy of the mass at the instant it passes back through its equilibrium position is about 1.20 J

\texttt{ }

<h3>Further explanation</h3>

Let's recall Elastic Potential Energy formula as follows:

\boxed{E_p = \frac{1}{2}k x^2}

where:

<em>Ep = elastic potential energy ( J )</em>

<em>k = spring constant ( N/m )</em>

<em>x = spring extension ( compression ) ( m )</em>

Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

mass of object = m = 1.25 kg

initial extension = x = 0.0275 m

final extension = x' = 0.0735 - 0.0275 = 0.0460 m

<u>Asked:</u>

kinetic energy = Ek = ?

<u>Solution:</u>

<em>Firstly , we will calculate the spring constant by using </em><em>Hooke's Law</em><em> as follows:</em>

F = k x

mg = k x

k = mg \div x

k = 1.25(9.8) \div 0.0275

k = 445 \frac{5}{11} \texttt{ N/m}

\texttt{ }

<em>Next , we will use </em><em>Conservation of Energy</em><em> formula to solve this problem:</em>

Ep_1 + Ek_1 = Ep_2 + Ek_2

\frac{1}{2}k (x')^2 + mgh + 0 = \frac{1}{2}k x^2 + Ek

Ek = \frac{1}{2}k (x')^2 + mgh - \frac{1}{2}k x^2

Ek = \frac{1}{2}k ( (x')^2 - x^2 ) + mgh

Ek = \frac{1}{2}(445 \frac{5}{11}) ( 0.0460^2 - 0.0275^2 ) + 1.25(9.8)(0.0735)

\boxed {Ek \approx 1.20 \texttt{ J}}

\texttt{ }

<h3>Learn more</h3>
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302
  • Young Modulus : brainly.com/question/9202964
  • Simple Harmonic Motion : brainly.com/question/12069840

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Elasticity

8 0
3 years ago
Read 2 more answers
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