While changing a soil's basic texture is very difficult, you can improve its structure–making clay more porous, sand more water retentive–by adding amendments. The best amendment for soil of any texture is organic matter, the decaying remains of plants and animals.
Given the complex numbers A1 5 6/30 and A2 5 4 1 j5, (a) convert A1 to rectangular form; (b) convert A2 to polar and exponential form; (c) calculate A3 5 (A1 1A2), giving your answer in polar form; (d) calculate A4 5 A1A2, giving your answer in rectangular form; (e) calculate A5 5 A1ysA* 2d, giving your answer in exponential form.
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]
1. A thin-walled cylindrical pressure vessel is capped at the end and is subjected to an internal pressure (p). The inside diameter of the vessel is 6 ft and the wall thickness is 1.5 inch. The vessel is made of steel with tensile yield strength and compressive yield strength of 36 ksi. Determine the internal pressure required to initiate yielding according to (a) The maximum-shear-stress theory of failure, and (b) The maximum-distortion-energy theory of failure, if a factor of safety (FS) of 1.5 is desired.
Determine the resistance of 3km of copper having a diameter of 0,65mm if the resistivity of copper is 1,7x10^8
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;
P is the resistivity of the material. L is the length of the material.A is the cross-sectional area of the material.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
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