*WELDING*


What size arc gap is suggested with a 5/32" (4.0mm) diameter electrode?

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

Given:

Assuming the transition to turbulence for flow over a flat plate happens at a Reynolds number of 5x105, determine the following for air at 300 K and engine oil at 380 K. Assume the free stream velocity is 3 m/s.

To Find:

a. The distance from the leading edge at which the transition will occur.

b. Expressions for the momentum and thermal boundary layer thicknesses as a function of x for a laminar boundary layer

c. Which fluid has a higher heat transfer

Calculation:

The transition from the lamina to turbulent begins when the critical Reynolds

number reaches [tex]5\times 10^5[/tex]

[tex](a). \;\text{Rex}_{cr}=5 \times 10^5\\\\\frac{\rho\;vx}{\mu}=5 \times 10^5\\\text{density of of air at}\;300K=1.16 \frac{kg}{m\cdot s}\\\text{viscosity of of air at}\;300K=1.846 \times 10^{-5} \frac{kg}{m\cdot s} \\v=3m/s\\\Rightarrow x=\frac{5\times 10^5 \times 1.846 \times 10^{-5} }{1.16 \times 3} =2.652 \;m \;\text{for air}\\(\text{similarly for engine oil at 380 K for given}\; \rho \;\text{and} \;\mu)\\[/tex]

[tex](b).\; \text{For the lamina boundary layer momentum boundary layer thickness is given by}:\\\frac{\delta}{x} =\frac{5}{\sqrt{R_e}}\;\;\;\;\quad\text{for}\; R_e <5 \times 10^5\\\\\text{for thermal boundary layer}\\\delta _t=\frac{\delta}{{P_r}^{\frac{1}{3}}}\quad\quad \text{where} \;P_r=\frac{C_p\mu}{K}\\\Rightarrow \delta_t=\frac{5x}{\sqrt{R_e}{P_r}^{\frac{1}{3}}}[/tex][tex](c). \frac{\delta}{\delta_t}={P_r}^{\frac{r}{3}}\\\text{For air} \;P_r \;\text{equivalent 1 hence both momentum and heat dissipate with the same rate for oil}\; \\P_r >>1 \text{heat diffuse very slowly}\\\text{So heat transfer rate will be high for air.}\\\text{Convective heat transfer coefficient will be high for engine oil.}[/tex]

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]

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.

Explanation:

Thermodynamics is the branch of science that deals with the relation between radiation, energy, physical properties of matter.

There are 4 laws of thermodynamics namely as follows :

Zeroth Law, First Law, Second Law and Third Law.

There are 4 branches of thermodynamics:

Classical thermodynamics, Statistical mechanics, Chemical thermodynamics and Equilibrium thermodynamics.

Hence, this is the required solution.

Answer:

The movement of heat

Explanation:

Answer: See explanation

Explanation:

Emerging technologies are simply referred to as the technical innovations which are done within a particular sector and typically brings about growth to such field or sector.

Some of the new and emerging technologies that are used in the electronics and telecommunications industry include:

1. Digital scent technology - This is a technology that is used for the sensing and the transmission of digital media such as video games, music etc that are event enabled.

2. Electronic nose - This refers to an electronic sensing device which is used in the detection of flavors.

3. Ambient intelligence - It is an electronic environments which shows sensitivity when there are people.

Answer:

mH275 units

Explanation: