*WELDING*
What size arc gap is suggested with a 5/32" (4.0mm) diameter electrode?
Solution :
The correct size of the arc of a welding process depends upon the application and the electrode. As a rule, the arc length should not be more than a diameter of the core of the electrode.
As for the electrode of diameter size of 5/32" or 4 mm, the arc length should be more than its core diameter. Also for 5/32 " diameter electrode, the welding time for the one electrode must be one minute as well as the length of the weld be the same as the length of the electrode consumed.
Using the formula XC=1/(2πfC) in your answer, how would a capacitor influence a simple DC series circuit?
The capacitive reactance of a DC series circuit increases when its capacitance decreases and vice-versa.
What is a DC series circuit?A DC series circuit can be defined as a type of circuit in which all of its resistive components are connected end to end, so as to form a single path for the flow of current.
This ultimately implies that, the same amount of current flows through a direct current (DC) series circuit.
The capacitive reactance of a DC series circuit.Mathematically, the capacitive reactance of a DC series circuit is given by this formula:
[tex]X_C = \frac{1}{2\pi fC}[/tex]
Where:
is the capacitive reactance.f is the frequency.C is the capacitance.From the above formula, we can deduce that the capacitive reactance of a DC series circuit is inversely proportional to both frequency and capacitance. Thus, the capacitive reactance of a DC series circuit increases when its capacitance decreases and vice-versa.
In conclusion, a capacitor would influence a simple DC series circuit by blocking the flow of direct current (DC) through it.
Read more on capacitance here: https://brainly.com/question/22989451
Thin cuts called Sipes allow the tread pattern blocks to flex during stops and starts? Group of answer choices True False
Answer:
I think its True
it should be True
What flight patterns do groups of birds utilize and why?
Identify at least three new and emerging technologies that are used in the electronics and telecommunications industry.
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.
A fine-grained soil has a liquid limit of 200%, determined from the Casagrande cup method. The plastic limit was measured by rolling threads of soil to be 45%. The water content of the soil is determined to be 60% in the field by oven-drying a field sample. The clay content is 63%, determined via sieve and hydrometer testing and the plasticity chart. Determine: (a) Plasticity index, liquidity index, and the activity. (b) What state is the clay (liquid, plastic, semi-solid, etc.) and why
Answer:
a)
Plasticity index [tex]I_P[/tex] = 155%
Liquidity index [tex]I_L[/tex] = 0.09677
Activity A = 2.4603
b)
Consistency index of the clay = 0.9032
Since Liquidity index of soil( 0.09677) is between 0 and 0.25 and Consistency index (0.9032 ) is between 0.75 and 1; Then, The Clay is in Plastic State
Explanation:
Given that;
Liquid Limit [tex]W_{L}[/tex] = 200%
Plastic Limit [tex]W_{P}[/tex] = 45%
Natural Water Content [tex]W_N[/tex] = 60%
Clay Content [tex]C_C[/tex] = 63%;
(a) Plasticity index, liquidity index, and the activity.
Plasticity index [tex]I_P[/tex] = Liquid Limit [tex]W_{L}[/tex] - Plastic Limit [tex]W_{P}[/tex]
Plasticity index [tex]I_P[/tex] = 200% - 45%
Plasticity index [tex]I_P[/tex] = 155%
Liquidity index [tex]I_L[/tex] = [ (Natural Water Content [tex]W_N[/tex] - Plastic Limit [tex]W_{P}[/tex] = 45%) / Plasticity index [tex]I_P[/tex]]
so;
Liquidity index [tex]I_L[/tex] = ( 60 - 45)/155
Liquidity index [tex]I_L[/tex] = 15 / 155
Liquidity index [tex]I_L[/tex] = 0.09677
the activity A is;
A = Plasticity index [tex]I_P[/tex] / Clay Content [tex]C_C[/tex]
Activity A = 155 / 63
Activity A = 2.4603
b) What state is the clay;
To get the state in which is in, we use the consistency index of Soil. which is given as follows;
[tex]I_C[/tex] = [( Liquid Limit [tex]W_{L}[/tex] - Natural Water Content [tex]W_N[/tex] ) / Plasticity index [tex]I_P[/tex]]
we substitute
[tex]I_C[/tex] = ( 200 - 60) / 155
[tex]I_C[/tex] = 140 / 155
[tex]I_C[/tex] = 0.9032
Consistency index of the clay = 0.9032
Since Liquidity index of soil( 0.09677) is between 0 and 0.25 and Consistency index (0.9032 ) is between 0.75 and 1; Then, The Clay is in Plastic State
What is integrated circuit package
Answer:
Explanation:
, integrated circuit packaging is the final stage of semiconductor device fabrication, in which the block of semiconductor material is encapsulated in a supporting case that prevents physical damage and corrosion.
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. 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 the higher heat transfer
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]
A timing light checks the ignition timing in relation to the ____ position.
Answer:
The timing light is connected to the ignition circuit and used to illuminate the timing marks on the engine's crankshaft pulley or flywheel, with the engine running. The apparent position of the marks, frozen by the stroboscopic effect, indicates the current timing of the spark in relation to piston position.
Explanation:
:)
Can we modify the soil’s composition?
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.
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
(CO 3) A nonrecursive filter may best be described as _____. Group of answer choices a filter whose current output depends on past and present inputs and outputs a filter whose current output depends on past and present inputs a filter whose current output depends on past and present outputs a filter whose current output depends on past and present inputs and only the current output
Answer: a filter whose current output depends on past and present inputs.
Explanation:
A nonrecursive filter may best be described as a filter whose current output depends on past and present inputs.
It should be noted that in this case, the output is typically stable. For the non recursive filter, the current output which is denoted as (yn) will have to be calculated from the input values which are for both previous and current values.
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]
When measuring a Brake Drum, the Brake Micrometer is set to a Base Drum Diameter of 10 Inches plus four notches, and the dial reads 22. What is the diameter of this drum
diameter of this drum?
Answer:
10.5
Explanation:
Measuring a Brake Drum, the Brake Micrometer is set to a Base Drum Diameter of 10 Inches plus four notches, and the dial reads 22. Diameter of the drum is 10.5.
What is diameter?
Any straight line segment that cuts through the center of a circle and has ends that are on the circle is considered a circle's diameter in geometry. It is also known as the circle's longest chord.
The diameter of a sphere can be defined using either of the two methods. In more recent usage, the diameter can also refer to a diameter's length, or displaystyle d.
As opposed to a diameter, which refers to the line segment itself, one uses the term "diameter" in this context since all diameters of a circle or sphere have the same length, which is equal to twice the radius "displaystyle r.
Therefore, Measuring a Brake Drum, the Brake Micrometer is set to a Base Drum Diameter of 10 Inches plus four notches, and the dial reads 22. Diameter of the drum is 10.5.
To learn more about diameter, refer to the link:
https://brainly.com/question/5501950
#SPJ5
what's the best way to plan an organized
Answer:
Get ready and comfortable.
List all of the tasks you need to accomplish over the next week. .
Next schedule everything.
Get a planner/calender.
Cut those tasks that do not fit into your
. Air entrainment is a process of entrapping tiny air bubbles in concrete mix in order to increase the durability of the hardened concrete in freeze-thaw climates. After 35 days, the breaking stress [in psi] was measured for concrete samples with and without air entrainment. Based on the data, does the air entrainment process increase the breaking stress of the concrete
This question is incomplete, the complete question is;
Air entrainment is a process of entrapping tiny air bubbles in concrete mix in order to increase the durability of the hardened concrete in freeze-thaw climates. After 35 days, the breaking stress [in psi] was measured for concrete samples with and without air entrainment. Based on the data, does the air entrainment process increase the breaking stress of the concrete. { ∝ = 0.05 }, assuming population variances are equal.
Air Entrainment No Air Entrainment
4479 4118
4436 4531
4358 4315
4724 4237
4414 3888
4358 4279
4487 4311
3984
4197
4327
Answer:
Since p-value ( 0.088) > significance level ( 0.05)
hence, Failed to reject Null hypothesis
It is then concluded that the null hypothesis H₀ is NOT REJECTED.
Therefore, there is no sufficient evidence to claim that population mean μ1 is greater than μ2 at 0.05 significance level.
We conclude that Air entrainment process can't increase the breaking stress of the concrete.
Explanation:
Given the data in the question;
mean x" = (4479 + 4436 + 4358 + 4724 + 4414 + 4358 + 4487 + 3984 + 4197 + 4327) / 10
mean x"1 = 43764 / 10 = 4376.4
x ( x - x" ) ( x - x" )²
4479 102.6 10526.76
4436 59.6 3552.16
4358 -18.4 338.56
4724 347.6 120825.76
4414 37.6 1413.76
4358 -18.4 338.56
4487 110.6 12232.36
3984 -382.4 153977.76
4197 -179.4 32184.36
4327 -49.4 2440.36
∑ 337830.4
Standard deviation s1 = √( (∑( x - x" )²) / n -1
Standard deviation s1 = √( 337830.4 / (10 - 1 ))
Standard deviation s1 = 193.74
x2 ( x2 - x"2 ) ( x2 - x"2 )²
4118 -121.9 14859.61
4531 291.1 84739.21
4315 75.1 5640.01
4237 -2.9 8.41
3888 -351.9 123833.61
4279 39.1 1528.81
4311 71.1 5055.21
∑ 235664.87
mean x"2 = (4118 + 4531 + 4315 + 4237 + 3888 + 4279 + 4311) / 7
mean x"2 = 29679 / 7 = 4239.9
Standard deviation s2 = √( (∑( x2 - x" )²) / n2 - 1
Standard deviation s1 = √( 337830.4 / (7 - 1 ))
Standard deviation s1 = 198.19
so
Mean x"1 = 4376.4, S.D1 = 193.74, n1 = 10
Mean x"2 = 4239.9, S.D2 = 198.19, n2 = 7
so;
Null Hypothesis H₀ : μ1 = μ2
Alternative Hypothesis H₁ : μ1 > μ2
Lets determine our rejection region;
based on the data provided. the significance level ∝ = 0.005
with degree of freedom DF = n1 + n2 - 2 = 10 + 7 - 2 = 15
so, Critical Value = 1.753
The rejection region for this right -tailed is R = t:t > 1.753
Test statistics
since it is assumed that the population variances are equal, so we calculate pooled standard deviation;
Sp = √{ [ (n1 -1)S.D1² + (n2 - 1)S.D2²] / [ n1 + n2 -2 ]
we substitute
Sp = √{ [ (10 -1)(193.74)² + (7 - 1)(198.19)²] / [ 10 + 7 -2 ]
Sp = √ [ 573492.345 / 15 ]
Sp = 195.53
so the Test statistics will be;
t = (x"1 - x"2) / Sp√([tex]\frac{1}{n1}[/tex] + [tex]\frac{1}{n2}[/tex] )
t = (4376.4 - 4239.9) / 195.53√([tex]\frac{1}{10}[/tex] + [tex]\frac{1}{7}[/tex] )
t = 136.5 / 96.36
t = 1.42
so
P-value = 0.088
Since p-value ( 0.088) > significance level ( 0.05)
hence, Failed to reject Null hypothesis
It is then concluded that the null hypothesis H₀ is NOT REJECTED.
Therefore, there is no sufficient evidence to claim that population mean μ1 is greater than μ2 at 0.05 significance level.
We conclude that Air entrainment process can't increase the breaking stress of the concrete.
Suppose that for the general population, 1 in 5000 people carries the human immunodeficiency virus (HIV). A test for the presence of HIV yields either a positive (+) or negative (-) response. Suppose the test gives the correct answer 99% of the time. What is P[-|H], the conditional probability that a person tests negative given that the person does have the HIV virus? What is P[H|+], the conditional probability that a randomly chosen person has the HIV virus given that the person tests positive?
Answer:
The answer is "0.019".
Explanation:
For HIV positive: [tex]P[HIV{+}] =\frac{1}{5000}[/tex]
For HIV negative: [tex]P[HIV{-}] =\frac{4999}{5000}[/tex]
calculating the proability for test gives right reslut: [tex]P[TV] =\frac{99}{100}[/tex]
calculating the proability for test gives wrong reslut: [tex]P[TX] =\frac{1}{100}[/tex]
For HIV negative: [tex]P[HIV{-}] = p[TX]= \frac{1}{100}[/tex]
calculating proability to have HIV:
[tex]P[HIV{+}] = \frac{P[HIV{-} \times HIV{-}]}{P{+}}[/tex]
[tex]= \frac{\frac{1}{5000} \times \frac{99}{100}}{\frac{1}{5000} \times \frac{99}{100}+\frac{4999}{5000} \times\frac{1}{100}}\\\\= \frac{\frac{1}{5000} \times 0.99}{\frac{1}{5000} \times 0.99+\frac{4999}{5000} \times0.01}\\\\= \frac{\frac{0.99}{5000} }{\frac{0.99}{5000} +\frac{49.99}{5000}}\\\\= \frac{0.000198}{0.000198 +0.009998}\\\\= \frac{0.000198}{0.010196}\\\\=0.019[/tex]
