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 could be a great tool to plan and build a
website?
Calculate the compression resistance of a compound column consisting of UC-305x305x198 with one cover plate
of 350mm x 20mm on each flange and having a length of 5m. assume that bottom of column is fixed and top is
pinned in both x & y axis?
Answer:
350×305π{^ not sure my answer please follow me
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
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.
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]
What is thermodynamics
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:
Hi can you help me pls?
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
A 20-mm thick draw batch furnace front is subjected to uniform heat flux on the inside surface, while the outside surface is subjected to convection and radiation heat transfer. Assuming that (1) heat conduction is steady; (2) one dimensional heat conduction across the furnace front thickness; (3) Thermal properties are constant; (4) inside and outside surface temperatures are constant. Determine the surface temperature T0 and TL based on the known conditions provided in the drawing.
Answer:
hello your question is incomplete attached below is the complete question
answer :
To ( inside temperature ) = 598 K
TL ( outside temperature ) = 594 k
Explanation:
a) Determine the surface temperature To and TL based on the known conditions provided in the drawing
To ( inside temperature ) = 598 K
TL ( outside temperature ) = 594 k
attached below is the detailed solution
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.
If the reading of mercury manometer was 728 mmHg, what is the reading for another liquid such as water in mH20 units?
Answer:
mH275 units
Explanation:
that was trueIdentify 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.
*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.
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
. 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.
How do birds achieve take-off, gliding, dive, etc.? How do they modify their body shape?
What flight patterns do groups of birds utilize and why?