Answer:
An algotherum is a finite set of sequential instructions to accomplish a task where instructions are written in a simple English language
Fill in the truth table for output A.
A = (x+y)(x'+z')(x'+z')
Answer:
1+1×1 multiplay then you get the answer
Identify the transformation. Note: you can only submit this question once for marking. translation rotation shear projection none of the above
Answer:
The answer is "shear".
Explanation:
Every transformation could be shown by some kind of conventional matrix.
Its standard matrices of shape k here could be any real value enabling shear transformation to parallel to the y axis.
[tex]\left[\begin{array}{cc}1&0\\k&1\\\end{array}\right][/tex]
This coefficient matrix A is the standard matrix during transformation. With the use of a the transformation could be entered into T(x)=Ax
And standard transfer function is standard.
[tex]\left[\begin{array}{cc}1&0\\6&1\\\end{array}\right][/tex]
The matrix in the form of the shear matrix.
Air at 40C flows over a 2 m long flat plate with a free stream velocity of 7 m/s. Assume the width of the plate (into the paper) is 0.5 m. If the plate is at a constant temperature of 100C, find:
Complete Question
Air at 40C flows over a 2 m long flat plate with a free stream velocity of 7m/s. Assume the width of the plate (into the paper) is 0.5 m. If the plate is at a co temperature of 100C,find:
The total heat transfer rate from the plate to the air
Answer:
[tex]q=1.7845[/tex]
Explanation:
From the question we are told that:
Air Temperature [tex]T_1=40c[/tex]
Length [tex]l=2m[/tex]
Velocity [tex]v=7m/s[/tex]
Width [tex]w=0.5[/tex]
Constant temperature [tex]T_t= 100C[/tex]
Generally the equation for Total heat Transfer is mathematically given by
[tex]q=hA(T_s-T_\infty)[/tex]
Where
h=Convective heat transfer coefficient
[tex]h=29.9075w/m^2k[/tex]
Therefore
[tex]q=h(L*B)(T_s-T_\infty)[/tex]
[tex]q=29.9075*(2*0.5)(100+273-(40+273))[/tex]
[tex]q=1794.45w[/tex]
[tex]q=1.7845[/tex]
What is code in Arduino to turn led on and off
here's your answer..
A 2400-lb rear-wheel drive tractor carrying 900 lb of gravel starts from rest and accelerates forward at 3ft/s2. Determine the reaction at each of the two (a) rear wheels A, (b) front wheels B.
Answer:
Explanation:
The missing diagram attached to the question is shown in the attached file below:
The very first thing we need to do in other to solve this question is to determine the mass of both the tractor and the mass of the gravel
For tractor, the mass is:
[tex]m_1 = \dfrac{2400 \ lb }{32.2 \ ft/s^2}[/tex]
[tex]m_1 = 74.53 \ lb.s^2/ft[/tex]
For gravel, the mass is:
[tex]m_2 = \dfrac{900 \ lb}{32.2 \ft/s^2}[/tex]
[tex]m_2 = 27.95 \ lb.s^2/ft[/tex]
From the diagram, let's consider the force along the horizontal components and vertical components;
So,
[tex]\sum F_x = ma_x \\ \\ 2F = (m_1+m_2) a \\ \\ F = \dfrac{1}{2}(74.53 4 + 27.950)lb.s^2/ft(2 \ ft/s^2) \\ \\ F = 102.484 \ lb[/tex]
[tex]\sum F_y = 0 \\ \\ 2N_A+2N_B - 2400 -900 = 0 \\ \\ N_A +N_B = 1650 \ lb[/tex]
Consider the algebraic sum of moments in the plane of A, with counter-clockwise moments being positive.
[tex]\sum M_A = I_o \alpha + \sum ma (d) \\ \\ = -2400 (20) + 2N_B (60) -900(110) = 0 - (74.534)(2)(20) - (27.950)(2)(40)[/tex]
[tex]=-48000 + 2N_B (60) -99000 = -2981.36-2236 \\ \\ = + 2N_B (60) = -2981.36-2236+48000+99000 \\ \\ = + 2N_B (60) = 141782.64 \\ \\ N_B = \dfrac{141782.64}{120} \\ \\ N_B = 1181.522 \ lb[/tex]
Replacing the value of 1181.522 lb for [tex]N_B[/tex] in equation (1)
[tex]N_A[/tex] + 1181.522 lb = 1650 lb
[tex]N_A[/tex] = (1650 - 1181.522)lb
[tex]N_A[/tex] = 468.478 lb
The net reaction on each of the rear wheels now is:)
[tex]F_R = \sqrt{N_A^2 +F^2}[/tex]
[tex]F_R = \sqrt{(468.478)^2 + (102.484)^2}[/tex]
[tex]\mathbf{F_R =479.6 \ lb}[/tex]
Now, we can determine the angle at the end of the rear wheels at which the resultant reaction force is being made in line with the horizontal
[tex]\theta = tan ^{-1}( \dfrac{468.478 }{102.484})[/tex]
[tex]\theta = 77.7^0[/tex]
Finally, the net reaction on each of the front wheels is:
[tex]F_B = N_B[/tex]
[tex]F_B =[/tex] 1182 lb
A main cable in a large bridge is designed for a tensile force of 2,600,000 lb. The cable consists of 1470 parallel wires, each 0.16 in. in diameter. The wires are cold-drawn steel with an average ultimate strength of 230,000 psi. What factor of safety was used in the design of the cable
Answer:
the factor of safety was used in the design of the cable is 2.6146
Explanation:
Given the data in the question;
Load on the main capable [tex]P_{initial[/tex] = 2600000 lb
number of parallel wires n = 1470
Diameter d = 0.16 in
average ultimate strength [tex]S_{ultimate[/tex] = 230000 psi
First we calculate the Load acting on each cable;
[tex]P_{initial[/tex] = P × n
P = [tex]P_{initial[/tex] / n
we substitute
P = 2600000 lb / 1470
P = 1768.70748 lb
Next we determine the working stress acting in a member;
[tex]S_{working[/tex] = P/A
{ Area A = [tex]\frac{\pi }{4}[/tex]d² }
[tex]S_{working[/tex] = P / [tex]\frac{\pi }{4}[/tex]d²
we substitute
[tex]S_{working[/tex] = 1768.70748 / [tex]\frac{\pi }{4}[/tex](0.16)²
[tex]S_{working[/tex] = 1768.70748 / 0.02010619298
[tex]S_{working[/tex] = 87968.29 psi
Now we calculate the factor of safety F.S
F.S = [tex]S_{ultimate[/tex] / [tex]S_{working[/tex]
we substitute
F.S = 230000 psi / 87968.29 psi
F.S = 2.6145785 ≈ 2.6146
Therefore, the factor of safety was used in the design of the cable is 2.6146
Tech A says that some relays are equipped with a suppression diode in parallel with the winding. Tech B says that some relays are equipped with a resistor in parallel with the winding. Who is correct
Answer:
Both are correct.
Explanation:
Both the technician A and B are correct. Some relays are equipped with a suppression diode in parallel windings while some relays are equipped with resistors. This is due to different requirements of the electromagnetic objects. Some require resistors to stop the flow of current towards the magnet.
In a tension test of steel, the ultimate load was 13,100 lb and the elongation was 0.52 in. The original diameter of the specimen was 0.50 in. and the gage length was 2.00 in. Calculate (a) the ultimate tensile stress (b) the ductility of the material in terms of percent elongation
Answer:
a) the ultimate tensile stress is 66717.8 psi
b) the ductility of the material in terms of percent elongation is 26%
Explanation:
Given the data in the question;
ultimate load P = 13,100 lb
elongation δl = 0.52 in
diameter of specimen d = 0.50 in
gage length l = 2.00 inch
First we determine the cross-sectional area of the specimen
A = [tex]\frac{\pi }{4}[/tex] × d²
we substitute
A = [tex]\frac{\pi }{4}[/tex] × ( 0.50 )²
A = 0.1963495 in²
a) the ultimate tensile stress σ[tex]_u[/tex]
tensile stress σ[tex]_u[/tex] = P / A
we substitute
tensile stress σ[tex]_u[/tex] = 13,100 / 0.1963495
tensile stress σ[tex]_u[/tex] = 66717.766 ≈ 66717.8 psi
Therefore, the ultimate tensile stress is 66717.8 psi
b) ductility of the material in terms of percent elongation;
percentage elongation of specimen = [change in length / original length]100
% = [ δl / l ]100
we substitute
% = [ 0.52 in / 2.00 in ]100
= [ 0.26 ]100
= 26
Therefore, the ductility of the material in terms of percent elongation is 26%
Para conseguir jugo de naranja concentrada, se parte de un extracto con 7% en peso de sólidos el cual se mete a un evaporador al vacío. En el evaporador se elimina el agua necesaria para que el jugo salga con una concentración del 60% de peso de sólidos. Si se introducen al proceso 1000 kg/h de jugo diluido, calcule la cantidad de agua evaporada y de jugo concentrado saliente.
Answer:
Se obtendrán 116.66 litros de jugo concentrado, y el agua evaporada será por un total de 883.33 litros.
Explanation:
Dado que para conseguir jugo de naranja concentrada, se parte de un extracto con 7% en peso de sólidos el cual se mete a un evaporador al vacío, y en el evaporador se elimina el agua necesaria para que el jugo salga con una concentración del 60% de peso de sólidos, si se introducen al proceso 1000 kg/h de jugo diluido, para calcular la cantidad de agua evaporada y de jugo concentrado saliente se debe realizar el siguiente cálculo;
1000 x 0.07 = 70
60 = 70
100 = X
100 x 70 / 60 = X
7000 / 60 = X
116.66 = X
Por lo tanto, se obtendrán 116.66 litros de jugo concentrado, y el agua evaporada será por un total de 883.33 litros.
Find the equation of the output voltage as a function of time assuming the switch closes at t = 0 and the capacitor is fully discharged for t < 0.
Answer: Hello your question is incomplete attached below is the complete question
answer : V(out) (t) = 1 - e^-100t
Explanation:
The equation of the output voltage as a function of time assuming at t = 0 switch closes and capacitor will be discharged when t < 0
V(out) (t) = 1 - e^-100t
attached below is the step by step explanation
Words and numbers can be
printed using many different
or type styles.
Peter B. is admitted to a local hospital emergency room (ER) suffering from an anxiety attack. He tells the ER physician that he is anxious about a job promotion for which he is being considered. Peter’s secretary is worried about him and asks her father, Dr. K., who is on the medical staff at the hospital, to go to the ER and see how Peter is doing. Dr. K., who is often in the ER, knows all the staff and they willingly give him Peter’s chart when he asks for it. Dr. K. calls his daughter to tell her that Peter is being treated for anxiety with an anti-depressant drug and will probably be discharged. She relays this encouraging message to Peter’s boss. Peter does not receive the promotion.
a. Will it be an easy matter for Peter to prove that the ER staff caused Peter to lose his promotion? Explain your answer.
b. What precautions can be taken to avoid giving confidential information to medical personnel who have no need to see it?
c. In your opinion, should a diagnosis of anxiety be a concern for an employer? Why or why not?
a.No it will be not an easy matter for Peter to prove that the Er staff caused peter to lose his promotion
Here's why?
1)The boss may think he might not able to stretch himself in emergency situations
2)He can think he can't handle normal pressure
b.What precautions can be taken to avoid giving confidential information to medical personnel who have no need to see it?
1)Peter should not let the medical personnel about his personal life
2)He might have not said extra information about his job promotion
3)Will it be easy matter for Peter to prove that the ER staff caused Peter to lose his promotion? Explain your answer
No, it will not be easy, Peter has to prove in one or in another situation that he can face the difficulties and be able to handle anxiety attacks in the future.
Hence above answers are suitable for the given situation
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Consider the following example: The 28-day compressive strength should be 4,000 psi. The slump should be between 3 and 4 in. and the maximum aggregate size should not exceed 1 in. The coarse and fine aggregates in the storage bins are wet. The properties of the materials are as follows:________.
Cement : Type I, specific gravity = 3.15
Coarse Aggregate: Bulk specific gravity (SSD) = 2.70; absorption
capacity = 1.1%; dry-rodded unit weight = 105 lb./ft.3
surface moisture = 1%
Fine Aggregate: Bulk specific gravity (SSD) = 2.67; absorption
capacity = 1.3%; fineness modulus = 2.70;
surface moisture = 1.5%
What is the Code requirement for receptacles connected to circuits of different voltages and installed in one building
Answer:
National Electrical Code
Explanation:
The National Electrical Code (NEC) is the foundational code that establishes the minimum requirements for electrical installation work. The NEC protects individuals and properties from being discriminated against dangers associated with the use of electricity. For electrical installation work, local and state electrical standards and codes should be consulted in addition to the NEC.
Section 406.4 of the NEC refers to On the same premises, receptacles linked to circuits with distinct voltages frequencies, or kinds of current (AC or DE) must be designed in such a way that the attachment plugs utilized on these circuits cannot be interchanged.
A lamp and a coffee maker are connected in parallel to the same 120-V source. Together, they use a total of 140 W of power. The resistance of the coffee maker is 300 Ohm. Find the resistance of the lamp.
Answer:
Resistance of the lamp (r) = 156.52 ohm (Approx.)
Explanation:
Given:
Total power p = 140 W
Resistance of the coffee maker = 300 Ohm
Voltage v = 120 V
Find:
Resistance of the lamp (r)
Computation:
We know that
p = v² / R
SO,
Total power p = [Voltage²/Resistance of the lamp (r)] + [Voltage²/Resistance of the coffee maker]
140 = [120² / r] + [120²/300]
140 = 120²[1/r + 1/300]
140 = 14,400 [1/r + 1/300]
Resistance of the lamp (r) = 156.52 ohm (Approx.)
explain "columnar transposition" for the transposition ciphers using the following plaintext as an example:
abcd
efgh
ijkl
mnop
Answer:
wa 4tefaw gfawe f
Explanation:
so you do e Fse gaewr gware gfawe agfawe gwar g
Analyze the rate of heat transfer through a wall of an industrial furnace which is constructed from 0.15-m-thick fireclay brick having a thermal conductivity of 1.7 W/m.K. Measurements made during steady state operation reveal temperatures of 1400 and 1150 K at the inner and outer surfaces, respectively. The wall dimension is 0.5 m by 1.2 m by side.
Answer:
1700 W
Explanation:
The heat transfer rate P = kA(T - T')/d where k = thermal conductivity of wall = 1.7 W/m-K, A = area of wall = 0.5 m × 1.2 m = 0.6 m², T = temperature of inner surface = 1400 K, T = temperature of outer surface = 1150 K and d = thickness of wall = 0.15 m
So, P = kA(T - T')/d
substituting the values of the variables into the equation, we have
P = 1.7 W/m-K × 0.6 m²(1400 K - 1150 K)/0.15 m
P = 1.7 W/m-K × 0.6 m² × 250 K/0.15 m
P = 255 Wm/0.15 m
P = 1700 W
So, the heat transfer rate through the wall is 1700 W
True or false all workers who do class 1 asbestos work must be part of a medical surveillance program
Answer:
Yes
Explanation:
Answer:
true
Explanation:
hehehe
How would you increase the size of the base unit of length in the metric system
Estimate the theoretical fracture strength of a brittle material if it is known that fracture occurs by the propagation of an elliptically shaped surface crack of length 0.25 mm (0.01 in) and tip radius of curvature of 1.2 x 10-3 mm (4.7 x 10-5 in.) when a stress of 1200 MPa (174,000 psi) is applied.
Answer:
the theoretical fracture strength of the brittle material is 5.02 × 10⁶ psi
Explanation:
Given the data in the question;
Length of surface crack α = 0.25 mm
tip radius ρ[tex]_t[/tex] = 1.2 × 10⁻³ mm
applied stress σ₀ = 1200 MPa
the theoretical fracture strength of a brittle material = ?
To determine the the theoretical fracture strength or maximum stress at crack tip, we use the following formula;
σ[tex]_m[/tex] = 2σ₀[tex]([/tex] α / ρ[tex]_t[/tex] [tex])^{\frac{1}{2}[/tex]
where α is the Length of surface crack,
ρ[tex]_t[/tex] is the tip radius,
and σ₀ is the applied stress.
so we substitute
σ[tex]_m[/tex] = (2 × 1200 MPa)[tex]([/tex] 0.25 mm / ( 1.2 × 10⁻³ mm ) [tex])^{\frac{1}{2}[/tex]
σ[tex]_m[/tex] = 2400 MPa × [tex]([/tex] 208.3333 [tex])^{\frac{1}{2}[/tex]
σ[tex]_m[/tex] = 2400 MPa × 14.43375
σ[tex]_m[/tex] = 34641 MPa
σ[tex]_m[/tex] = ( 34641 × 145 )psi
σ[tex]_m[/tex] = 5.02 × 10⁶ psi
Therefore, the theoretical fracture strength of the brittle material is 5.02 × 10⁶ psi
A rapid sand filter has a loading rate of 8.00 m/h, surface dimensions of , an effective filtration rate of 7.70 m/h, and a production efficiency of 96 percent. A complete filter cycle duration is 52 h and the filter is rinsed for 20 minutes at the start of each cycle
Required:
a. What flow rate (m3/s) does the filter handle during production?
b. What volume of water is needed for backwashing plus rinsing the filter in each filter cycle?
Answer: hello there is a Missing information below is the missing information
surface dimensions of 10m * 8m
answer :
a) 640 m^3/hr
b) 1334.66 m^3
Explanation:
a) Determine the flow rate ( m^3/s ) of the filter handle
Va = Q /A[tex]_{f}[/tex]
where ; Va = filtration rate ( 8.00 m/h ) , Q = flow rate of filter handle , A[tex]_{f}[/tex] = surface area ( 10 m * 8 m )
Q = 8 * ( 10m * 8m ) = 640 m^3 / hr
b) Determine the volume of water needed for backwashing plus rinsing the filter in each filter cycle
Лf = 0.96 ( production efficiency )
Vb + Vr = 0.04 Vf
∴ Vf = ( Vb + Vr ) / 0.04 ------ ( 1 )
next step ; determine the volume of filtered water making use of effective filtration rate
Ref = ( Vf - Vb - Vr ) / A[tex]_{f}[/tex]T[tex]_{c}[/tex]
therefore : Vb + Vr = Vf - ( 80 * 52 * 7.7 ) ---- ( 2 )
Input equation 1 into 2
Vb + Vr = ( ( Vb + Vr ) / 0.04 ) - 32032 ---- ( 3 )
Resolve equation 3
hence the Volume of water needed for Backwashing and rinsing the filter in each filter cycle
= 1334.66 m^3
Unit of rate of heat transfer
Answer:
The units on the rate of heat transfer are Joule/second, also known as a Watt.
Explanation:
Heat flow is calculated using the rock thermal conductivity multiplied by the temperature gradient. The standard units are mW/m2 = milli Watts per meter squared. Thus, think of a flat plane 1 meter by 1 meter and how much energy is transferred through that plane is the amount of heat flow.
hope it helps .
stay safe healthy and happy..The rate of heat transfer is measured in Joules per second, also known as Watts.
What is heat transfer?Heat transfer is a thermal engineering discipline that deals with the generation, use, conversion, and exchange of thermal energy between physical systems.
Heat transfer mechanisms include thermal conduction, thermal convection, thermal radiation, and energy transfer via phase changes.
The rate of heat transfer through a unit thickness of material per unit area per unit temperature difference is defined as thermal conductivity. Thermal conductivity varies with temperature and is measured experimentally.
Heat is typically transferred in a combination of these three types and occurs at random. Heat transfer rate is measured in Joules per second, also known as Watts.
Thus, Joules per second or watts is the unit of rate of heat transfer.
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A cylinder is internally pressurized to a pressure of 100 MPa. This causes tangential and axial stresses in the outer surface of 400 and 200 MPa, respectively. Make a Mohr circle representation of the stresses in the outer surface. What maximum normal and shear stresses are experienced by the outer surface?
Answer:
[tex]\mu_{max}=200Mpa[/tex]
Explanation:
From the question we are told that:
Internally pressurized [tex]P_i=100MPa[/tex]
Tangential Stress [tex]P_t=400mpa[/tex]
Axial stress [tex]P_a=200mpa[/tex]
Generally the equation for maximum normal and shear stresses are experienced by the outer surface is mathematically given by
[tex]\mu_{max}=|\frac{P_t-P_a}{2}|,|\frac{P_t}{2}|,|\frac{P_t}{2}|[/tex]
Therefore
[tex]\mu_{max}=|\frac{400-200}{2}|,|\frac{400}{2}|,|\frac{200}{2}|[/tex]
[tex]\mu_{max}=200Mpa[/tex]
Consider a single crystal of silver oriented such that tensile stress is applied along a (0 0 1) direction. If slip occurs on a (1 1 1) plane and in a (1 0 1) direction and is initiated at an applied tensile stress of 1.1 MPa (160 psi), compute the critical resolved shear stress.
Answer:
i don't know
Explanation:
sorry
(8 pts.) Air in an Otto cycle engine is compressed to a temperature and pressure of 450 °C and 2.5 MPa. After the power stroke, the conditions are 600 °C and 0.45 MPa. Find the peak cycle temperature (°C), heat addition (kJ/kg), and efficiency
Answer:
a) [tex]Tb=1845.05K[/tex]
b) [tex]Q=1000.25KJ[/tex]
c) [tex]\mu=0.59[/tex]
Explanation:
From the question we are told that:
Temperature x [tex]Tx=450c=>723K[/tex]
Pressure x [tex]Px=2.5MPa[/tex]
Temperature y [tex]Ty=600c=>873K[/tex]
Pressure y [tex]Py=0.45MPa[/tex]
Let
Air atmospheric temperature be [tex]25c[/tex]
Therefore
Temperature [tex]Ta=25+273=298k[/tex]
Generally the equation for Otto cycle is mathematically given by
[tex]\frac{Tb}{Tx}=\frac{Ty}{Ta}[/tex]
[tex]Tb=\frac{873*723}{298}[/tex]
[tex]Tb=2118.05[/tex]
Therefore the peak cycle temperature (°C)
[tex]Tb=2118.05k[/tex]
[tex]Tb=2118.05-273[/tex]
[tex]Tb=1845.05K[/tex]
Generally the equation for Heat addition is mathematically given by
[tex]Q=Cv(Tb-Tx)[/tex]
[tex]Q=Cv(2118.05-723)[/tex]
[tex]Q=1000.25KJ[/tex]
Generally the equation for Thermal efficiency is mathematically given by
[tex]\mu=1-\frac{Ta}{Tx}[/tex]
[tex]\mu=1-\frac{298}{723}[/tex]
[tex]\mu=0.59[/tex]
An article gave a scatter plot along with the least squares line of x = rainfall volume (m3) and y = runoff volume (m3) for a particular location. The accompanying values were read from the plot.
c) Calculate a point estimate of the true average runoff volume when rainfall volume is 51. (Round your answer to four decimal places.)
(d) Calculate a point estimate of the standard deviation . (Round your answer to two decimal places.)
(e) What proportion of the observed variation in runoff volume can be attributed to the simple linear regression relationship between runoff and rainfall? (Round your answer to four decimal places.)
x 6 12 14 16 23 30 40 52 55 67 72 81 96 112 127
y 4 10 13 14 15 25 27 48 38 46 53 72 82 99 100
Answer:
y = 0.834X - 1.58015
Slope = 0.8340 ; Intercept = - 1.5802
y = 40.9539
19.93
0.9765
Explanation:
X: Rainfall volume
6
12
14
16
23
30
40
52
55
67
72
81
96
112
127
Y : Runoff
4
10
13
14
15
25
27
48
38
46
53
72
82
99
100
The scatterplot shows a reasonable linear trend between the Rainfall volume and run off.
The estimated regression equation obtained using a linear regression calculator is :
y = 0.834X - 1.58015
y = Runoff ; x = Rainfall volume
Slope = 0.8340 ; Intercept = - 1.5802
Point estimate for Runoff, when, x = 51
y = 0.834X - 1.58015
y = 0.834(51) - 1.58015
y = 40.95385
y = 40.9539
d.)
Point estimate for standard deviation :
s = 5.145
σ = s * √n
σ = √15 * 5.145
= 19.93
e.)
r² = Coefficient of determination gives the proportion of explained variance in Runoff due to the regression line. From the model output, the r² value = 0.9765. Which means That about 97.65% Runoff is due to Rainfall volume.
tech a says that a slightly lean mixture offers good fuel economy and low exhaust emissions. Tech b says that a mixture that is too rich fouls spark plugs and causes incomplete burning. Who is correct
A steel bar with a diameter of .875 inches and a length of 15.0 ft is axially loaded with a force of 21.6 kip. The modulus of elasticity of the steel is 29 *106 psi. Determine
Answer:
35.92 kpsi
Explanation:
Given data:
diameter of the steel bar d = 0.875 in
Area A = πd^2/4 = π(0.875)^2/4
length L = 15.0 ft
Load P = 21.6 kip
Modulus of elesticity E = 29×10^6 Psi
Assume we are asked to determine axial stress in the bar which is given as
[tex]\sigma = Load, P/ Area, A[/tex]
[tex]\sigma = 4P/\pi d^2[/tex]
substitute the value
[tex]\sigma = \frac{4\times 21.6}{\pi \times (0.875)^2} \\=35.92\ kpsi[/tex]
A Si pin photodiode has an active light-receiving area of diameter 0.4 mm. When radiation of wavelength 700 nm (red light and intensity 0.1 mW cm^-2 is incident it generates a photocurrent of 56.6 nA. What is the responsivity (A W^-1) and quantum efficiency (QE) of the photodiode at 700 nm?
Answer:
Explanation:
here is your answer:
The responsivity will be 0.45 A / W. Then the quantum efficiency will be 80%.
How to calculate responsivity and quantum efficiency?A Si pin photodiode has an active light-receiving area of a diameter of 0.4 mm.
When radiation of wavelength 700 nm (red light and intensity 0.1 mW cm^-2 is incident it generates a photocurrent of 56.6 nA.
Diameter (d) = 0.4 mm = 0.04 cm
Then the area will be
Area (A) = π / 4 x d²
Area (A) = π / 4 x 0.04²
Area (A) = 1.26 x 10⁻³ square cm
Then the incident power will be
P₀ = intensity of light x area
P₀ = 1.26 x 10⁻³ x 0.1 x 10⁻³
P₀ = 0.126 x 10⁻⁶ μW
Then the responsivity will be
R = photocurrent / power
R = 56.6 x 10⁻⁹ / 0.126 x 10⁻⁶
R = 0.4492
R ≅ 0.45 A / W
Then the quantum efficiency will be
η = Rhc / qλ
h = plank constant
c = speed of light
q = charge of an electron
Then we have
η = 0.45 x 6.62 x 10⁻³⁴ x 3 x 10⁸ / 1.6 x 10⁻¹⁹ x 700 x 10⁻⁹
η = 0.7979
η = 0.8
η = 80%
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chemicals injected into shale rock stay there....true or false
Answer:
False
Explanation:
Flow back to surface