Answer: Technology is the application of scientific knowledge for practical purposes, especially in industry.
Mechanical.
Electronic.
Industrial and manufacturing.
Medical.
Communications.
Explanation:
Answer:
Technology is something like a phone or like a tablet or like an electrace car
Explanation:
In the feature i belive that there is going to be more and higher andanved technology or like people say flying cars a robots. But i feel more advanced stuff like phones and computers or labtops or like tablets.
Due within next 5 minutes... Should scientists be allowed to research human cloning? Explain and defend your answer.
Answer:
science is hard but human cloning this is legendary but still its up to the government but me i would allow this because its something that science can upgrade to
Explanation:
Which type of design does not need special care for the placement of dimensions?
Answer:
cubes
Explanation:
Design for software engineering does not need special care for the placement of dimensions. Software engineering generally deals with arranging lines of coding.
What is Software Engineering Design?In order to achieve a software's goal by using lower-level, machine-understandable algorithms and a set of parameters based on project requirements or environmental constraints, software design is the process of outlining solutions, planning, and specifying a method of implementation based on the needs of the user.
Software engineering is the application of engineering principles, which are often employed in the design, development, testing, deployment, and management of physical systems to the design.
Therefore, the main focus of software engineering is organizing coded lines that are kept in a device's or computer's memory rather than in actual physical space.
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Whats wrong with the wind turbine?
Answer:
The wind turbine has little impact on the environment compared to conventional power plants. Meanwhile the Wind turbine can affect the wildlife meaning birds are being harmed by the blades.
Explanation: Hope that helps:)
Match each titration term with its definition.
Process of slowly adding a solution to react with another solution and determine the Choose... concentration of one of the solutions tased on the reaction between them
Solution of an unknown concentration that has another solution slowly added to it Choose...
When the required amount of one solution has been added to the second solution to Choose... complete the reaction
Solution of known concentration that is slowly added to a solution of unknown Choose... concentration
Glassware that allows a solution to be precisely and slowly added to another solution Choose...
A reagent added to the analyte solution that changes color when the reaction is Choose... complete
Answer:
1) titration
2) titrand
3) equivalence point
4) titrant
5) Burette
6) Indicator
Explanation:
The process in which a known volume of a standard solution is added to another solution so that the standard solution can react with the solution of unknown concentration such that its concentration is determined can be referred to as titration.
The solution which is added to another solution is called the titrant. The titrand is the solution of unknown concentration
A burette is a glassware used to slowly add a known volume of the titrant to the titrand.
The indicator used signals the point when the reaction is complete by a color change. At this point, a stoichiometric amount of titrant has been added to the titrand. This is also referred to as the equivalence point.
What are the two most important things to remember when at the end of your interview?
1. When the interviewer asked if you have any questions at the end of the interview don't say no. You should always say yes your interviewer is expecting you to ask a few good questions before ending the interview.
2. Always thank the interviewer for their time and effort to interview you. This would look very good for you and its a nice way to help wrap up the interview.
Tech A says that most electric fuel pumps are now mounted inside of the fuel tank. Tech B says that fuel flows through the center of the electric pump and is used to cool the pump. Who is correct?
Answer:
both
Explanation:
Tech A says that most electric fuel pumps are now mounted inside of the fuel tank. Tech B says that fuel flows through the center of the electric pump and is used to cool the pump. Who is correct? quizlet said so
In this case, both Technicians are right.
The electric fuel pump is a device that powers a vehicle's engine system by providing fuel.There are two classes of electric fuel pumps: external fuel pumps and in-tank fuel pumps.In modern vehicles (also in motorcycles) the electric fuel pump is now located inside the fuel tank.In conclusion, both Technicians are right.
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Which color of white light bends the most when it is refracted by a prism? red green yellow orange
The color of white light bends the most when it is refracted by a prism is green after the violet.
Why green bents towards the light?The shorter the wavelength of the mild, the greater its miles refracted. As a result, crimson mild is refracted the least, and violet mild is refracted the most - inflicting the colored mild to unfold out to shape a spectrum.
The hues of white mild withinside the order of lowering wavelength are crimson orange-yellow inexperienced blue indigo violet. Fro the listing we are given withinside the alternatives inexperienced has the shortest wavelength than others. Therefore inexperienced will bend the most.
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Read the following paragraph as quickly as you can, and see if you encounter any diculties.
Aoccdrnig to rscheearch at an Elingsh uinervtisy, it deosn't mttaer in waht oredr the ltteers in a wrod are, the olny iprmoetnt tihng is taht the frist and lsat ltteer is at the rghit pclae. The rset can be a toatl mses and you can sitll raed it wouthit a porbelm. Tihs is bcuseae we do not raed ervey lteter by itslef but the wrod as a wlohe.
This has been presented as an example of a principle of human reading compre-hension. If you keep the rst letter and the last letter of a word in their correct positions, then scramble the letters in between, the word is still quite readable in the context of an accompanying paragraph. However, it seems that this is a bit of a myth and not truly based on solid research.1 In short, for longer words the task is much more dicult. Nonetheless, we are going to imitate the process on some English text.
The task will be to read in a paragraph from a le, scramble the internal letters of each word, and then write the result to a le. Handling punctuation is tricky. You are required to deal with punctuation that comes at the end of a word (period, question mark, exclamation, etc.)|that is, punctuation is left untouched and does not count as the nal unscrambled letter. Optionally, one can deal with the more dicult task of handling all punctuation, such as apostrophes for possessives or hyphenated words. Truly randomizing the order of letters is a task for later in the text, but we can do some reasonable approximations now. Attacking this problem in a divide-and-conquer way should begin by writing code to scramble the letters in a word.
Create a dierent solution by dening scrambling functions for each of the following approaches: (Each approach counts as a dierent problem)
(a) For each letter choose a random number and rotate that letter by the random amount. Import random and use the random.randint(a,b) function where `a' and `b' dene the range of random numbers returned.
(b) Implement a dierent method to scrambling the letters. You must clearly describe in the comments the process to scramble the letters in place.It must be dierent than the process from part a.
Answer:
a
Explanation:
just took the quiz
Consider a W21x93. Determine the moment capacity of the beam. Assume the compression flange is not laterally braced and that the unbraced length is 15 feet. Assume Cb
Answer:
The answer is "828.75"
Explanation:
Please find the correct question:
For W21x93 BEAM,
[tex]Z_x = 221.00 in^3 \\\\\to \frac{b_t}{2t_f} =4.53\\\\\to \frac{h}{t_w}=32.3[/tex]
For A992 STREL,
[tex]F_y= 50\ ks[/tex]
Check for complete section:
[tex]\to \frac{b_t}{2t_f} =4.53 < \frac{65}{\sqrt{f_y = 9.19}}\\\\\to \frac{h}{t_w} =32.3 < \frac{640}{\sqrt{f_y = 90.5}}[/tex]
Design the strength of beam =[tex]\phi_b Z_x F_y\\\\[/tex]
[tex]=0.9 \times 221 \times 50\\\\=9945 \ in \ \ kips\\\\=\frac{9945}{12}\\\\= 828.75 \ft \ kips \\[/tex]
Assume that 20 percent of the dynamic count of the instructions executed for a program are branch instructions. Delayed branching is used, with one delay slot. Assume that there are no stalls caused by other factors. First, derive an expression for the execution time in cycles if all delay slots are filled with NOP instructions. Then, derive another expression that reflects the execution time with 70 percent of delay slots filled with useful instructions by the optimizing compiler. From these expressions, determine the compiler’s contribution to the increase in performance, expressed as a speedup percentage.
Answer:
Explanation:
From the given information;
The first objective is to derive an expression for the execution time in cycles
For an execution time T for the non-pipelined processor;
[tex]T = \dfrac{N \times S}{R}[/tex]
where
N = instruction count
S = average number of clock cycles to fetch & execute an instruction
R = clock rate
From the average number of clock
[tex]S = 1 + \delta _{branch\_penality }[/tex]
where;
[tex]\delta _{branch\_penality }= {Dynamic \ count \times Delay \ Slots}[/tex]
Similarly, we are given that the dynamic count is 20%
∴
The execution time required for all the delayed slots that is filled with NOP instructions can be estimated as:
[tex]\delta _{branch\_penality }=0.20 \times 1.00[/tex]
[tex]\delta _{branch\_penality }=0.20[/tex]
Therefore;
S = 1 + 0.20
S = 1.2
Recall that:
For an execution time T for the non-pipelined processor;
[tex]T = \dfrac{N \times S}{R}[/tex]
[tex]T = \dfrac{N \times 1.2}{R}[/tex]
[tex]T = \dfrac{ 1.2 \ N}{R}[/tex]
To derive another expression that reflects the execution time of 70% delay; we have:
[tex]\delta_{branch\_penality} = 0.20 \times ( 1-0.7)[/tex]
[tex]\delta_{branch\_penality} = 0.20 \times ( 0.3)[/tex]
[tex]\delta_{branch\_penality} = 0.06[/tex]
S = 1 + 0.06
S = 1.06
For an execution time T for the non-pipelined processor;
[tex]T = \dfrac{N \times S}{R}[/tex]
[tex]T = \dfrac{N \times 1.06}{R}[/tex]
[tex]T = \dfrac{ 1.06 \ N}{R}[/tex]
Finally, the compiler's contribution to the increase in speed up percentage for the above two cases is:
[tex]=\begin {pmatrix}\dfrac{T_{all \ delay \ slots }}{T_{70\% \ delay \ slots } } -1 \end {pmatrix} \times 100[/tex]
[tex]= \begin {pmatrix}\dfrac{1.2 }{1.06 } -1 \end {pmatrix} \times 100[/tex]
= (1.1320 - 1) × 100
= 0.1320 × 100
= 13.20%
Therefore, the compiler's contribution to increasing performance as expressed as speed up percentage is 13.20%
A steam pipe passes through a chemical plant, where wind passes in cross-flow over the outside of the pipe. The steam is saturated at 17.90 bar and you can assume that the resistances to heat transfer inside the pipe and of the pipe itself are minimal, so that the outside surface temperature of the pipe is equal to the saturation temperature of the steam. The pipe is stainless steel and has an outside diameter of 6.75 cm and a length of 34.7 m. The air flows over the pipe at 7.6 m/s and has a bulk fluid temperature of 27 °C.
A. What is the rate of heat transfer from the pipe to the air?
B. would your answer change if the air flow direction changes to parallel flow? If so, calculate that q as well.
Answer:
a) the rate of heat transfer from the pipe to the air is 23.866 watts
b) YES, the rate of heat transfer changes to 3518.61 watt
Explanation:
Given that:
steam is saturated at 17.90 bar.
the pipe is stainless steel and has an outside diameter of 6.75 cm
length = 34.7 m
Air flows over the pipe at 7.6 m/s
Bulk fluid temperature of 27°C
we know that
hD/k = 0.028 (Re)^0.8 (Pr)^0.33
Outside diameter of pipe = 6.75 cm
length of the pipe = 34.7 m
velocity of air = 7.6 m/s
Cp of air = 1.005 kJ/Kgk
viscosity of air = 1.81 × 10⁻⁵ kg/(m.sec)
thermal conductivity of air = 2.624 × 10⁻⁵ kw/m.k
so as
hD/k = 0.028 (Re)^0.8 (Pr)^0.33
hD/k = 0.028 (Dvp / u)^0.8 (Cpu / k)^0.33
(h × 0.0675 / 2.624 × 10⁻⁵) = (0.028 ([0.0675 × 7.6 × 1.225] / [1.81 ×10⁻⁵])^0.8) (((1.005 × 1.81 × 10⁻⁵) / (2.624 × 10⁻⁵))^0.33))
h = 0.0414 w/m².k
a)
Now to find the rate of heat transfer Q
Q = hAΔT
Q = 0.0414 × (2π × 0.03375 × 34.7) × (105.383 - 27)
Q = 23.866 watts
therefore the rate of heat transfer from the pipe to the air is 23.866 watts
b)
Now the flow direction changes to parallel flow, then
(h × 0.0675 / 2.624 × 10⁻⁵) = (0.028 ([34.7 × 7.6 × 1.225] / [1.81 ×10⁻⁵])^0.8) (((1.005 × 1.81 × 10⁻⁵) / (2.624 × 10⁻⁵))^0.33))
h = 6.1036 w/m².k
so from the steam table, saturated steam at 17.70 bar, temperature of steam will be 105.383°C
so to find the rate of heat transfer Q
Q = hAΔT
Q = 6.1036 × (2π × 0.03375 × 34.7) × (105.383 - 27)
Q = 3518.61 watt
Therefore the rate of heat transfer changes to 3518.61 watt
The point at 0°, 0° is
Suppose an underground storage tank has been leaking for many years, contaminating a groundwater and causing a contaminant concentration directly beneath the site of 0.30 mg/L. The contamination is flowing at the rate of 0.5 ft/day toward a public drinking water well 1 mile away. The contaminant degrades with a rate constant of 1.94 x 10^-4 1/day. Draw a picture of the system. Estimate the steady-state pollutant concentration expected at the well. If the slope factor is 0.02 (mg/kg-day)^-1.
Required:
Estimate the cancer risk for an adult male drinking the water for 10 years.
Answer: the cancer risk for an adult male drinking the water for 10 years is 2.88 × 10⁻⁶
Explanation:
firstly, we find the time t required to travel for the contaminant to the well;
Given that, contamination flowing rate = 0.5 ft/day
Distance of well from the site = 1 mile = 5280 ft
so t = 5280 / 0.5 = 10560 days
k is given as 1.94 x 10⁻⁴ 1/day
next we find the Pollutant concentration Ct in the well
Ct = C₀ × e^-( 1.94 x 10⁻⁴ × 10560)
Ct = 0.3 x e^-(kt)
Ct= 0.0386 mg/L
next we determine the chronic daily intake, CDI
CDI = (C x CR x EF x ED) / (BW x AT)
where C is average concentration of the contaminant(0.0368mg/L), CR is contact rate (2L/day), EF is exposure frequency (350days/Year), ED is exposure duration (10 years), BW is average body weight (70kg).
now we substitute
CDI = (0.0368 x 2 x 350 x 10) / ((70x 365) x 70)
= 257.7 / 1788500
= 0.000144 mg/Kg.day
CDI = 1.44 x 10⁻⁴ mg/kg.day
Finally we calculate the cancer risk, R
Slope factor SF is given as 0.02 Kg.day/mg
Risk, R = I x SF
= 1.44 x 10⁻⁴ mg/kg.day x 0.02Kg.day/mg
R = 2.88 × 10⁻⁶
therefore the cancer risk for an adult male drinking the water for 10 years is 2.88 × 10⁻⁶
(a) At a simple interest rate of 12% per year, determine how long it will take $5000 to increase to twice as much. (b) Compare the time it will take to double if the rate is 20% per year simple interest.
Explanation:
10000=5000(1.12^x)
2=1.12^x
(log_1.12)(2)=x
x= about 6.1163
10000=5000(1.2^x)
2=1.2^x
(log_1.2)(2)=x
x= about 3.8019
compare them by saying like 20% will be 6.12/3.8 times faster
Determine the greatest pressure drop allowed over the 10-m-long pipe caused by viscous friction, so the flow remains laminar.
This question is incomplete, the complete question is;
Determine the greatest pressure drop allowed over the 10-m-long pipe caused by viscous friction, so the flow remains laminar.
The 125-mm-diameter smooth pipe is transporting SAE 10W-30 oil with ρ=920 kg/m3 and µ=0.2 N.s/m2 .
Answer: the greatest pressure drop allowed is 14247 Pascals
Explanation:
LAMINAR FLOW INF PIPE
greatest pressure drop
(P1 - P2) = pressure drop
(P1 - P2) = 32uvl / D²
so (P1 - P2) ∝ V
greatest pressure drop is only at very high velocity
now Re (Reynold Number = svd / u
Re ∝ V
so velocity is high when Re is high
Re = 2000 ( Highest for LAMINAR)
Re = 2000 = svd / u
now in the question we were given that
s = 920 kg/m³, u = 0.2 NS/m², d = 125mm = 0.125m, l = 10m
so we substitute;
2000 = (920 × v × 0.125) / 0.2
v = 3.48 m/s
now pressure drop = (P1 - P2) = 32UVL / d²
we substitute
(P1 - P2) = (32(0.2) (3.48)(10)) / (0.125)²
(P1 - P2) = 14247 Pascals
therefore the greatest pressure drop allowed is 14247 Pascals
A spherical ball has 3cm internal diameter and its inner surface temperature is 250˚C and the outer surface temperature is 30˚C. Calculate the outer diameter of the ball if the heat loss from the ball is 1600W and has 2.75W/m.˚C thermal conductivity.
Answer:
The outer diameter of the ball is 6.2138 cm
Explanation:
The formula to apply is ;
Heat loss ,
[tex]Q/t=kA*\frac{( T_1-T_2)}{d}[/tex]
where ;
Q/t=total heat loss from the ball = 1600 w
k=coefficient of heat transmission through the ball= 2.75 W/m.˚C
A=area in m² of the ball with the coefficient of heat transmission
T₁=Hot temperature
T₂=Cold temperatures
d=thickness of the ball
Area of spherical ball using internal diameter, 3cm= 0.03 m will be
Radius = half the diameter = 0.03/2 = 0.015
Area = 4 *π*r²
Area = 4*π*0.015² = 0.002827 m²
Apply the formula for heat loss to get the thickness as:
1600 = {2.75 * 0.002827 *(250-30 ) }/d
1600 =1.711/d
1600d = 1.711
d=1.711/1600 = 0.001069 m
d= 0.1069
Using internal radius and the thickness to get outer radius as;
3 + 0.1069 = 3.1069 cm
Outer diameter will be twice the outer radius
2*3.1069 = 6.2138 cm
Which of the following is considered a bated standard
Answer:
is it multiple choice or....
Set up the following characteristic equations in the form suited to Evanss root-locus method. Give L(s), a(s), and b(s) and the parameter, K, in terms of the original parameters in each case. Be sure to select K so that a(s) and b(s) are monic in each case and the degree of b(s) is not greater than that of a(s).
a) s + (1/τ) = 0 versus parameter τ
b) s2 + cs + c + 1 = 0 versus parameter c
c) (s + c)3 + A(Ts + 1) = 0
i. versus parameter A
ii. versus parameter T
iii. versus the parameter c, if possible. Say why you can or can not. Can a plot of the roots be drawn versus c for given constant values of A and T by any means at all.
d) 1 + (kp + k1/s + kDs/Ts + 1)G(s) = 0. Assume that G(s) = A c(s)/d(s), where c(s) and d(s) are monic polynomials with the degree of d(s) greater than that of c(s).
i. versus kp
ii. versus kI
iii. versus kD
iv. versus τ
Answer:
attached below is the detailed solution and answers
Explanation:
Attached below is the detailed solution
C(iii) : versus the parameter C
The parameter C is centered in a nonlinear equation, therefore the standard locus will not apply hence when you use a polynomial solver the roots gotten would be plotted against C
Answer: a(s) is of degree n and b(s) is of degree n-1.
a)Characteristic equation in form suitable for Evans's root-locus method is given by a(s) + b(s)K = 0, where a(s) and b(s) are polynomials of s with real coefficients. Now, given equation can be represented as: s + (1/τ) = 0 => s + (1/τ) = 0s + (1/τ)K = 0=> K = -τs/τ + 0Thus, L(s) = s, a(s) = 1, b(s) = 1 and K = -τ.
b)The characteristic equation is given by:s^2 + cs + c + 1 = 0For the root-locus method, we have to write the characteristic equation in the form a(s) + b(s)K = 0. Since the degree of a(s) is 2, we select K such that the degree of b(s) is also 2, and so b(s) will be monic.s^2 + cs + c + 1 = 0=> (s + c/2)^2 + 1/4 - c^2/4 + c = 0=> s^2 + (2c)s + (1 + c - c^2/4) = 0Now, taking a(s) = s^2 + (2c)s + (1 + c - c^2/4) and b(s) = 1 with K = -1/c^2, we have:a(s) + b(s)K = s^2 + (2c)s + (1 + c - c^2/4) - 1/c^2 = 0
c) The characteristic equation is given by:(s + c)3 + A(Ts + 1) = 0
i. versus parameter A:s = -c is a repeated root of multiplicity 3 when A = 0For s = -c, the characteristic equation becomes:-A(Tc + 1) = 0If A = 0, the characteristic equation will be (s + c)^3 = 0 and will have a repeated root at s = -c with a multiplicity of 3.
ii. versus parameter T:s = -c is a repeated root of multiplicity 3 when T = 0.For s = -c, the characteristic equation becomes:3c^2s + A = 0If A = 0, the characteristic equation will be (s + c)^3 = 0 and will have a repeated root at s = -c with a multiplicity of 3.
iii. versus the parameter c:We cannot draw the root locus of (s + c)3 + A(Ts + 1) = 0 with respect to the parameter c because there is no c term in the characteristic equation; it only contains c cubed.
d) The characteristic equation is given by:1 + (kp + k1/s + kDs/Ts + 1)G(s) = 0Assuming G(s) = Ac(s)/d(s), where c(s) and d(s) are monic polynomials with the degree of d(s) greater than that of c(s), the characteristic equation becomes:
d(s) + kp d(s) + k1 d(s)/s + kDs c(s)/T = 0.
Thus, a(s) = d(s) + k1 d(s)/s and b(s) = kDc(s)/T + kp d(s) with K = -1.
Therefore, a(s) is of degree n and b(s) is of degree n-1.
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Describe the blade design for a wind turbine that you would choose. Why did you choose this design?
Answer: aerofoil type blades
Explanation: they are more difficult to make but offer better performance and higher rotational speeds making them ideal for electrical energy generation.
The signboard truss is designed to support a horizontal wind load of 800 lb. A separate analysis shows that 5 8 of this force is transmitted to the center connection at C and the rest is equally divided between D and B. Calculate the forces in members BE and BC.
Answer:
BE = 559lb T
BC = 300lb T
Explanation:
Base on the figure attached, let us first find the load on joint B, C and D
FC = 5/8 * 800 = 500lb
FD = 13/28 * 800 = 150lb
FB is also equal to 150lb
The angle alpha α at joint D will be:
Tan α = 6/12
α = tan ^-1 (0.5)
α = 26.6
by using resolution of forces, resolve the force at that point to horizontal and vertical component.
Horizontal component
Fd + DE sin α = 0
DE sinα = -Fd
DE = -Fd/sinα
DE = -150/26.6
DE = - 335.4lb
Vertical component
-CD - DEcosα = 0
CD = -DEcosα
CD = -(-335.4)cos26.6
CD = 300lb
So let us do the same at joint C and joint E
At joint C
Horizontal component
Fc + CE = 0
CE = -Fc
CE = -500lb
Vertical component
CD - BC = 0
BC = CD
BC = 300lb T
At joint E
The horizontal component will be:
-CE - DEsinα - BEsinα + EFsinα = 0
-(-500) - (-335.4)sin26.6 - BEsin26.6 + EFsin26.6 + EFsin 26.6
650 - BEsin26.6 + EFsin26.6 = 0
Make BE the subject of the formular
BEsin26.6 = 650 + EFsin26.6
BE = (650 - EFsin26.6)/ sin26.6
BE = 1453.2 + EF ................ (1)
The vertical component
DEcosα - BEcosα - EFcosα = 0
substitute all the parameters
- 335.4cos26.6 - BEcos26.6 - EFcos26.6 = 0
Substitute BE in the equation 1 into the equation above
-300 - (1453.2 + EF)cos26.6 - EFcos26.6 = 0
-1599.7 - 1.789EF = 0
EF = -1599.7 / 1.789
EF = -894.2lb
Substitute EF in equation 1
BE = 1453.2 - 894.2
BE = 559lb T
Therefore the forces in members BE and BC
BE = 559lb T
BC = 300lb T
Match each titration term with its definition.
Process of slowly adding a solution to react with another solution and determine the Choose... concentration of one of the solutions tased on the reaction between them
Solution of an unknown concentration that has another solution slowly added to it Choose...
When the required amount of one solution has been added to the second solution to Choose... complete the reaction
Solution of known concentration that is slowly added to a solution of unknown Choose... concentration
Glassware that allows a solution to be precisely and slowly added to another solution Choose...
A reagent added to the analyte solution that changes color when the reaction is Choose... complete
Answer:
1) titration
2) titrand
3) equivalence point
4) titrant
5) Burette
6) Indicator
Explanation:
The process of adding a known volume a standard solution to another solution to react with it in order to determine the concentration of the unknown solution is known as titration.
The solution to which another solution of known concentration is added is called the titrand while the solution of known concentration is called the titrant.
A burette is a glassware used to slowly add a known volume of the titrant to the titrand. An indicator shows the point when the reaction is complete by a color change. This is the point when the required amount of one solution has been added to the second solution. It is also called the equivalence point.
I. The process of slowly adding a solution to react with another solution and determine the concentration of one of the solutions based on the reaction between them: Titration.
II. A solution of an unknown concentration that has another solution slowly added to it: Analyte.
III. When the required amount of one solution has been added to the second solution to complete the reaction: Endpoint.
IV. A solution of known concentration that is slowly added to a solution of unknown concentration: Titrant.
V. A glassware that allows a solution to be precisely and slowly added to another solution: Burette.
VI. A reagent added to the analyte solution that changes color when the reaction is complete: Indicator.
Titration is also referred to as titrimetry or volumetric analysis and it can be defined as a chemical process in which a solution of known concentration is slowly added to a solution of unknown concentration, in order to determine the concentration or quantity of the analyte (solution of an unknown concentration)
For example, titration can be used to determine the concentration of a basic solution by titrating it with an acid solution of known concentration that is required to neutralize the basic solution.
In Chemistry, the following terms and apparatus are used for the titration of solutions:
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First Order Logic
Translate into First Order Logic (FOL) the following statements. Then write the negations of the FOL propositions found.
1. All tigers are fast. Domain: animals.
2. Some tigers are fierce and dangerous. Domain: animals.
3. Every prime number is odd. Domain: positive integers.
4. All prime numbers except two are odd. Domain: positive integers.
5. All fruits are either yellow or red. Domain: produce.
6. For every integer number, there exist a bigger integer. Domain: integers.
Answer:
1) ∀x [ Tiger(x) → Fast(x) ]
2) эx [ Tiger (x) ∧ Fierce (x) ∧ Dangerous(x) ]
3) ∀x [ Prime(x) → Odd(x) ]
4) ∀x [ prime (x) ∧ ~Two(x) → Odd (x) ]
5) ∀x [ Fruits(x) → ( yellow(x) ∨ Red(x) ]
6) ∀xэy [ I(x) → greater (y, x) ]
Explanation:
Translating the statements into first Order Logic and their negations
1) All tigers are fast. Domain: animals.
∀x [ Tiger(x) → Fast(x) ]
2) Some tigers are fierce and dangerous. Domain: animals
эx [ Tiger (x) ∧ Fierce (x) ∧ Dangerous(x) ]
3) Every prime number is odd. Domain: positive integers
∀x [ Prime(x) → Odd(x) ]
4) All prime numbers except two are odd. Domain: positive integers
∀x [ prime (x) ∧ ~Two(x) → Odd (x) ]
5) All fruits are either yellow or red. Domain: produce.
∀x [ Fruits(x) → ( yellow(x) ∨ Red(x) ]
6) For every integer number, there exist a bigger integer. Domain: integers.
∀xэy [ I(x) → greater (y, x) ]
Answer:
Answer:
1) ∀x [ Tiger(x) → Fast(x) ]
2) эx [ Tiger (x) ∧ Fierce (x) ∧ Dangerous(x) ]
3) ∀x [ Prime(x) → Odd(x) ]
4) ∀x [ prime (x) ∧ ~Two(x) → Odd (x) ]
5) ∀x [ Fruits(x) → ( yellow(x) ∨ Red(x) ]
6) ∀xэy [ I(x) → greater (y, x) ]
Explanation:
Translating the statements into first Order Logic and their negations
1) All tigers are fast. Domain: animals.
∀x [ Tiger(x) → Fast(x) ]
2) Some tigers are fierce and dangerous. Domain: animals
эx [ Tiger (x) ∧ Fierce (x) ∧ Dangerous(x) ]
3) Every prime number is odd. Domain: positive integers
∀x [ Prime(x) → Odd(x) ]
4) All prime numbers except two are odd. Domain: positive integers
∀x [ prime (x) ∧ ~Two(x) → Odd (x) ]
5) All fruits are either yellow or red. Domain: produce.
∀x [ Fruits(x) → ( yellow(x) ∨ Red(x) ]
6) For every integer number, there exist a bigger integer. Domain: integers.
∀xэy [ I(x) → greater (y, x) ]
Explanation:
A closed system of mass 20 kg undergoes a process in which there is a heat transfer of 1000 Q6: ki from the system to the surroundings. The work done on the system is 200 kl. If the initial 5 specific internal energy of the system is (250+R:) kl/kg, what is the final specific internal energy, in kj/kg? Neglect changes in kinetic and potential energy:
The final specific internal energy : 190 kJ/kg
Further explanationThe laws of thermodynamics 1 state that: energy can be changed but cannot be destroyed or created
The equation is:
[tex]\tt E_{in}-E_{out}=\Delta E~system\\\\\Delta E=\Delta U+\Delta KE+\Delta PE\\\\\Delta U=m(U_2-U_1)\\\\Q-W=\Delta U+\Delta KE+\Delta PE[/tex]
Energy owned by the system is expressed as internal energy (U)
This internal energy can change if it absorbs heat Q (U> 0), or releases heat (U <0). Or the internal energy can change if the system does work or accepts work (W)
The sign rules for heat and work are set as follows:
• The system receives heat, Q +
• The system releases heat, Q -
• The system does work, W -
• the system accepts work, W +
A closed system of mass 20 kg⇒m=20 kg
Heat transfer of 1000 kJ from the system to the surroundings⇒Q=-1000 kJ
The work done on the system is 200 kJ⇒W=+200 kJ
The initial specific internal energy of the system is 250 kJ /kg⇒U₁ = 250 kj/kg
Neglect changes in kinetic and potential energy⇒ΔKE+ΔPE=0, so
Q-W = ΔU
Input in equation
[tex]\tt -1000-200=20(U_2-250)\\\\-1200=20U_2-5000\\\\3800=20U_2\\\\U_2=190~kJ/kg[/tex]
Which of the following tasks might a civil engineer perform during a project’s construction phase in a design-build approach? (Select all that apply.)
Work with construction management to solve specific problems as they arise.
Draw up the contracts for construction management.
Ensure compliance with safety standards.
Provide information on design changes.
Answer:
Draw up the contract for construction management.
Answer:
Work with construction management to solve specific problems as they arise.
Ensure compliance with safety standards.
Explanation:
Timken rates its bearings for 3000 hours at 500 rev/min. Determine the catalog rating for a ball bearing running for 10000 hours at 1800 rev/min with a load of 2.75 kN with a reliability of 90 percent.
Answer:
C₁₀ = 6.3 KN
Explanation:
The catalog rating of a bearing can be found by using the following formula:
C₁₀ = F [Ln/L₀n₀]^1/3
where,
C₁₀ = Catalog Rating = ?
F = Design Load = 2.75 KN
L = Design Life = 1800 rev/min
n = No. of Hours Desired = 10000 h
L₀ = Rating Life = 500 rev/min
n₀ = No. of Hours Rated = 3000 h
Therefore,
C₁₀ = [2.75 KN][(1800 rev/min)(10000 h)/(500 rev/min)(3000 h)]^1/3
C₁₀ = (2.75 KN)(2.289)
C₁₀ = 6.3 KN
A transmitter has an output power of 0.1mW while the fiber has coupling loss of 12dB, attenuation of
6dB/km for the length of 500m. The link contains two connectors of 2dB average loss. The receiver has a
minimum acceptable power ( sensitivity) of -35dBm. The designer has allowed a 4dB margin. Based on
the given information,
a) Calculate total link loss
b) Determine the receiver's sensitivity
c) Identify this system is good practice or not
Experimental Design Application Production engineers wish to find the optimal process for etching circuit boards quickly. They create a single replicate 24experiment to test the effect of the following factors on time to etch a circuit board: (A) Concentration of nitric acid in the etchant, (B) temperature of the etchant, (C) stirring rate in the etching tank, (D) surface area of the board. Running this experiment, they obtain the data in HW_EDA_137.csv.Preview the document Which factors effects and interaction effects are significant?
Answer:
Hello your question is incomplete attached below is the missing part and answer
options :
Effect A
Effect B
Effect C
Effect D
Effect AB
Effect AC
Effect AD
Effect BC
Effect BD
Effect CD
Answer :
A = significant
B = significant
C = Non-significant
D = Non-significant
AB = Non-significant
AC = significant
AD = Non-significant
BC = Non-significant
BD = Non-significant
CD = Non-significant
Explanation:
The dependent variable here is Time
Effect of A = significant
Effect of B = significant
Effect of C = Non-significant
Effect of D = Non-significant
Effect of AB = Non-significant
Effect of AC = significant
Effect of AD = Non-significant
Effect of BC = Non-significant
Effect of BD = Non-significant
Effect of CD = Non-significant
Help please all of the numbers b4 the equal sign are wrong
Answer:
3/5" = 12'1 3/4" = 35'1 1/4" = 25'9/10" = 18'2 13/20" = 53'Explanation:
One number is wrong; they all lack units.
The basic ratio is 1" = 20', so you can divide feet by 20 to find inches.
3/5" = 12'1 3/4" = 35'1 1/4" = 25'9/10" = 18'2 13/20" = 53'Perhaps you want decimal inches:
0.60" = 12'1.75" = 35'1.25" = 25'0.90" = 18'2.65" = 53'Assume you have a steel beam that is 20 feet long and carries a concentrated load of 500 pounds at a distance of 15 feet from Reaction (R ). Reaction (R ) is pinned 5 feet from the load. What are the values in lbs for R and R ? A. 475 and 25 B. 500 and 00 C. 375 and 125 D. 125 and 375
Answer:
The answer is "Option D".
Explanation:
[tex]\to \epsilon_T= 0\\\\\to R+R_1=500 \ lb \\\\\to \epsilon \ M_{pin} > 0\\\\\to -R \times 20 +(500 \times 5) =0\\\\\to 2500 =20R\\\\\to 20R= 2500\\\\\to R= \frac{2500}{20}\\\\\to R= \frac{250}{2}\\\\\to R= 125 \ lb \\\\\to R_1 =500-125\\[/tex]
[tex]=375 \ lb \\[/tex]
Don't break or crush mercury-containing lamps because mercury powder may be released.
A) TrueB) False