Technician A says that acid core solder should be used whenever aluminum wires are to be soldered.
Technician B says that solderless connectors should not be used if a weather-resistant connection is desired.
Who is correct?
a. A only
b. B only
c. Both A and B
O d. Neither Anor B

Answer:

mH275 units

Explanation:

This question is incomplete, the complete question is;

Given the complex numbers A₁ = 6∠30 and A₂ = 4 + j5;

(a) convert A₁ to rectangular form

(b) convert A₂ to polar and exponential form

(c) calculate A₃ = (A₁ + A₂), giving your answer in polar form

(d) calculate A₄ = A₁A₂, giving your answer in rectangular form

(e) calculate A₅ = A₁/([tex]A^{*}[/tex]₂), giving your answer in exponential form.

Answer:

a) A₁ in rectangular form is 5.196 + j3

b) value of A₃  in polar form is 12.19∠41.02°

The polar form of A₂ is 6.403 ∠51.34°, exponential form of A₂ = 6.403[tex]e^{j51.34 }[/tex]

c) value of A₃  in polar form is 12.19∠41.02°

d) A₄ in rectangular form is 5.784 + j37.98

e) A₅ in exponential form is 0.937[tex]e^{j81.34 }[/tex]

Explanation:

Given data in the question;

a) A₁ = 6∠30

we convert A₁ to rectangular form

so

A₁ = 6(cos30° + jsin30°)

= 6cos30° + j6cos30°

= (6 × 0.866) + ( j × 6 × 0.5)

A₁  =  5.196 + j3

Therefore, A₁ in rectangular form is 5.196 + j3

b) A₂ = 4 + j5

we convert to polar and exponential form;

first we convert to polar form

A₂ = √((4)² + (5)²) ∠tan⁻¹( [tex]\frac{5}{4}[/tex] )

= √(16 + 25) ∠tan⁻¹( 1.25 )

= √41 ∠ 51.34°

A₂ = 6.403 ∠51.34°

The polar form of A₂ is 6.403 ∠51.34°

next we convert to exponential form;

A∠β can be written as A[tex]e^{j\beta }[/tex]

so, A₂  in exponential form will be;

A₂ = 6.403[tex]e^{j51.34 }[/tex]

exponential form of A₂ = 6.403[tex]e^{j51.34 }[/tex]

c) A₃ = (A₁ + A₂)

giving your answer in polar form

so, A₁ = 6∠30 = 5.196 + j3 and A₂ = 4 + j5

we substitute

A₃ = (5.196 + j3) + ( 4 + j5)

= 9.196 + J8

next we convert to polar

A₃ = √((9.196)² + (8)²) ∠tan⁻¹( [tex]\frac{8 }{9.196}[/tex] )

A₃ = √(84.566416 + 64) ∠tan⁻¹( 0.8699)

A₃ = √148.566416 ∠41.02°    

A₃ = 12.19∠41.02°

Therefore, value of A₃  in polar form is 12.19∠41.02°

d) A₄ = A₁A₂

giving your answer in rectangular form

we substitute

A₄ = (5.196 + j3) ( 4 + j5)

= 5.196( 4 + j5) + j3( 4 + j5)

= 20.784 + j25.98 + j12 - 15

A₄ = 5.784 + j37.98

Therefore, A₄ in rectangular form is 5.784 + j37.98

e) A₅ = A₁/([tex]A^{*}[/tex]₂)

giving your answer in exponential form

we know that [tex]A^{*}[/tex]₂ is the complex  conjugate of A₂

so

[tex]A^{*}[/tex]₂ = (6.403 ∠51.34° )*

= 6.403 ∠-51.34°

we convert to exponential form

A∠β can be written as A[tex]e^{j\beta }[/tex]

[tex]A^{*}[/tex]₂  = 6.403[tex]e^{-j51.34 }[/tex]

also

A₁ = 6∠30

we convert to polar form

A₁ = 6[tex]e^{j30 }[/tex]

so A₅ = A₁/([tex]A^{*}[/tex]₂)

A₅ = 6[tex]e^{j30 }[/tex] / 6.403[tex]e^{-j51.34 }[/tex]

A₅  = (6/6.403) [tex]e^{j(30+51.34) }[/tex]

A₅  = 0.937[tex]e^{j81.34 }[/tex]

Therefore A₅ in exponential form is 0.937[tex]e^{j81.34 }[/tex]

Answer:

Resistance of copper = 1.54 * 10^18 Ohms

Explanation:

Given the following data;

Length of copper, L = 3 kilometers to meters = 3 * 1000 = 3000 m

Resistivity, P = 1.7 * 10^8 Ωm

Diameter = 0.65 millimeters to meters = 0.65/1000 = 0.00065 m

[tex] Radius, r = \frac {diameter}{2} [/tex]

[tex] Radius = \frac {0.00065}{2} [/tex]

Radius = 0.000325 m

To find the resistance;

Mathematically, resistance is given by the formula;

[tex] Resistance = P \frac {L}{A} [/tex]

Where;

First of all, we would find the cross-sectional area of copper.

Area of circle = πr²

Substituting into the equation, we have;

Area  = 3.142 * (0.000325)²

Area = 3.142 * 1.05625 × 10^-7

Area = 3.32 × 10^-7 m²

Now, to find the resistance of copper;

[tex] Resistance = 1.7 * 10^{8} \frac {3000}{3.32 * 10^{-7}} [/tex]

[tex] Resistance = 1.7 * 10^{8} * 903614.46 [/tex]

Resistance = 1.54 * 10^18 Ohms