I need help with this problem.

Copper and aluminum are being considered for a high-voltage transmission line that must carry a current of 60.7 A. The resistanc

lisov135 [29]

Answer:

a) The magnitude JJ of the current density for a copper cable is 5.91 × 10⁵A.m⁻²

b)The mass per unit length \lambdaλ for a copper cable is 0.757kg/m

c)The magnitude J of the current density for an aluminum cable is 3.5 × 10⁵A/m²

d)The mass per unit length \lambdaλ for an aluminum cable is 0.380kg/m

Explanation:

The expression for electric field of conductor is,

$E = \frac{V}{L}$

The general equation of voltage is,

V = iR

The expression for current density in term of electric field is,

$J = \frac{E}{p}$

Substitute (V/L) for E in the above equation of current density.

$J = \frac{V}{pL} ------(1)$

Substitute iR for V in equation (1)

$J = \frac{iR}{pL} ------(2)$

Substitute 1.69 × 10⁸ Ω .m for p

50A for i

0.200Ω.km⁻¹ for (R/L) in eqn (2)

$J = \frac{(50) (0.200\times 10^-^3) }{1.69 \times 10^-^8 } \\\\= 5.91 \times 10^5A.m^-^2$

The magnitude JJ of the current density for a copper cable is 5.91 × 10⁵A.m⁻²

b) The expression for resistivity of the conductor is,

$p = \frac{RA}{L}$

$A = \frac{pL}{R}$

The expression for mass density of copper is,

m = dV

where, V is the density of the copper.

Substitute AL for V in equation of the mass density of copper.

m=d(AL)

m/L = dA

λ is use for (m/L)

substitute,

pL/R for A and λ is use for (m/L) in the eqn above

$\lambda = d\frac{p}{\frac{R}{L} } ------(3)$

Substitute 0.200Ω.km⁻¹ for (R/L)

8960kgm⁻³ for d and 1.69 × 10⁸ Ω .m

$\lambda = (8960) \frac{(1.69 \times 10^-^8 }{0.200\times 10^-^3} \\\\= 0.757kg.m^-^1$

c) Using the equation (2) current density for aluminum cable is,

$J = \frac{iR}{pL}$

p is the resistivity of the aluminum cable.

Substitute 2.82 × 10⁻⁸Ω.m for p ,

50A for i and 0.200Ω.km⁻¹ for (R/L)

$J = \frac{(50)(0.200\times10^-^3) }{2.89\times 10^-^8} \\\\= 3.5 \times10^5A/m^2$

The magnitude J of the current density for an aluminum cable is 3.5 × 10⁵A/m²

d) Using the equation (3) mass per unit length for aluminum cable is,

$\lambda = d\frac{p}{\frac{R}{L} }$

p is the resistivity and is the density of the aluminum cable.

Substitute 0.200Ω.km⁻¹ for (R/L), 2700 for d and 2.82 × 10⁻⁸Ω.m for p

$\lambda = (2700) \frac{(2.82 \times 10^-^8) }{(0.200 \times 10^-^3) } \\\\= 0.380kg/m$

The mass per unit length \lambdaλ for an aluminum cable is 0.380kg/m

7 0

3 years ago

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What is energy and types of energy<br>

pishuonlain [190]

Energy- the ability to do work/how things can change and move

Types
Potential Energy
Kinetic Energy
Nuclear Energy
Mechanical Energy
Sound Energy
Heat

8 0

2 years ago

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A train brakes from 25 m/s to rest in 30 sec. What is its deceleration?

givi [52]

Answer:

a= -0.83m\s^2

Explanation:

a = v \ t

a = -25 \ 30 = -0.833 m\s^2

the object is slowing down 0.83 meter every second

8 0

2 years ago

A bird flies at a speed of 2.3 m/s if it has 14 j of kinetic energy what is the mass

Sidana [21]

Kinetic Energy = 1/2 * mv²

Kinetic Energy = 14 J, v = 2.3 m/s , m = ?

14 = 1/2 * m* 2.3²

14 = 0.5*m*2.3*2.3

m = 14 / (0.5*2.3*2.3)

m = 5.29 kg.

Mass = 5.29 kg.

7 0

3 years ago

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A child in a boat throws a 5.80-kg package out horizontally with a speed of 10.0 m/s. The mass of the child is 24.6kg and the ma

Sergio [31]

Answer:

-0.912 m/s

Explanation:

When the package is thrown out, momentum is conserved. The total momentum after is the same as the total momentum before, which is 0, since the boat was initially at rest.

$(m_c + m_b)v_b + m_pv_p = 0$

where $m_c = 24.6 kg, m_b = 39 kg, m_p = 5.8 kg$ are the mass of the child, the boat and the package, respectively. $, v_p = 10m/s, v_b$ are the velocity of the package and the boat after throwing.

$(24.6 + 39)v_b + 5.8*10 = 0$

$63.6v_b + 58 = 0$

$v_b = -58/63.6 = -0.912 m/s$

3 0

3 years ago