The human eye, as well as the light-sensitive chemicals on color photographic film, respond differently to light sources with different spectral distributions. Daylight lighting corresponds to the spectral distribution of the solar disk, which may be approximated as a blackbody at 5800K. Incandescent lighting from the usual household bulb corresponds approximately to the spectral distribution of a black body at 2900K. Calculate the band emission fractions for the visible region, 0.47 mu m to 0.65 mum, for each of the lighting sources. Calculate the wavelength corresponding to the maximum spectral intensity for each of the light sources
Answers
Answer:
a) at T = 5800 k
band emission = 0.2261
at T = 2900 k
band emission = 0.0442
b) daylight (d) = 0.50 μm
Incandescent ( i ) = 1 μm
Explanation:
To Calculate the band emission fractions we will apply the Wien's displacement Law
The ban emission fraction in spectral range λ1 to λ2 at a blackbody temperature T can be expressed as
F ( λ1 - λ2, T ) = F( 0 ----> λ2,T) - F( 0 ----> λ1,T )
Values are gotten from the table named: blackbody radiation functions
a) Calculate the band emission fractions for the visible region
at T = 5800 k
band emission = 0.2261
at T = 2900 k
band emission = 0.0442
attached below is a detailed solution to the problem
b)calculate wavelength corresponding to the maximum spectral intensity
For daylight ( d ) = 2898 μm *k / 5800 k = 0.50 μm
For Incandescent ( i ) = 2898 μm *k / 2900 k = 1 μm
Related Questions
Please help me answer this engineering question
Answers
calculate force and moment reactions at bolted base O of overhead traffic signal assembly. each traffic signal has a mass 36kg, while the masses of member OC and AC are 50Kg and 55kg, respectively. The mass center of mmber AC at G.
Answers
Answer:
The free body diagram of the system is, 558 368 368 508 O ?? O, Consider the equilibrium of horizontal forces. F
Explanation:
I hope this helps you but I think and hope this is the right answer sorry if it’s wrong.
An FPC 4 m2 in area is tested during the night to measure the overall heat loss coefficient. Water at 60 C circulates through the collector at a flow rate of 0.06 l/s. The ambient temperature is 8 C and the exit temperature is 49 C. Determine the overall heat loss coefficient.
Answers
Answer:
- 14.943 W/m^2K ( negative sign indicates cooling )
Explanation:
Given data:
Area of FPC = 4 m^2
temp of water = 60°C
flow rate = 0.06 l/s
ambient temperature = 8°C
exit temperature = 49°C
Calculate the overall heat loss coefficient
Note : heat lost by water = heat loss through convection
m*Cp*dT = h*A * ( T - To )
∴ dT / T - To = h*A / m*Cp ( integrate the relation )
In ( [tex]\frac{49-8}{60-8}[/tex] ) = h* 4 / ( 0.06 * 10^-3 * 1000 * 4180 )
In ( 41 / 52 ) = 0.0159*h
hence h = - 0.2376 / 0.0159
= - 14.943 W/m^2K ( heat loss coefficient )
A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 75 MPa (68.25 ksi). If the plate is exposed to a tensile stress of 361 MPa (52360 psi) during use, determine the minimum length of a surface crack that will lead to fracture. Assume a value of 1.03 for Y.
Answers
Answer:
Explanation:
From the given information:
Strain fracture toughness [tex]K_k[/tex]= 75 MPa[tex]\sqrt{m}[/tex]
Tensile stress [tex]\sigma[/tex] = 361 MPa
Value of Y = 1.03
Thus, the minimum length of the critical interior surface crack which will result to fracture can be determined by using the formula:
[tex]a_c = \dfrac{1}{\pi} ( \dfrac{k_k}{\sigma Y})^2 \\ \\ a_c = \dfrac{1}{\pi} \Big [ \dfrac{75 \times \sqrt{10^3}}{361 \times 1.03 } \Big]^2 \\ \\ a_c = \dfrac{1}{\pi} \Big [ 6.378474693\Big]^2 \\ \\ \mathbf{ a_c = 12.95 \ mm}[/tex]
If it took a 30m capacity tank mediated by 2cm waterproof water faucet for 10 hours, calculate the flow speed (exit) water from the
Answers
Complete question:
If it took a 30m³ capacity tank mediated by 2cm waterproof water faucet for 10 hours, calculate the flow speed (exit) of water from the tank.
Answer:
the flow rate of the water from the tank is 0.05 m³/min
Explanation:
Given;
volume of water in the tank, v = 30 m³
length of the waterproof faucet, L = 2cm = 0.02 m
duration of water flow through the tank, t = 10 hours
The flow rate of the water from the tank is calculated as;
[tex]Q = \frac{V}{t} = \frac{30 \ m^3}{10\ h \ \times \ 60 \min} = 0.05 \ m^3/ \min[/tex]
Therefore, the flow rate of the water from the tank is 0.05 m³/min
. Briefly describe the simplest heat treatment procedure that would be used in converting a 0.76 wt% C steel from one microstructure to the other, as follows: (a) Martensite to spheroidite (b) Spheroidite to martensite (c) Bainite to pearlite (d) Pearlite to bainite (e) Spheroidite to pearlite (f) Pearlite to spheroidite (g) Tempered martensite to martensite (h) Bainite to spheroidite
Answers
Answer:
a) converting Martensite to spheroidite
The heat treatment procedure for converting Martensite to spheroidite involves heating Martensite of 0.76 wt% C steel for approximately 24 hours at a temperature of 700°C
b) Converting Spheroidite to martensite
The heat treatment entails heating Spheroidite of 0.76 wt% C steel to a temperature of 760°C to austenization then it will be quenched at temperature > 140°C
c) Converting Bainite to Pearlite
The heat treatment involves heating Bainite of 0.76 wt% C steel to a temperature of 720°C until austenization then it will be quenched at temperature < 35°C
d) Converting Pearlite to Bainite
The heat treatment involves heating Pearlite of 0.76 wt% C steel to a temperature of 720°C until austenization then it will be cooled at a temperature range 220°C to 540°C. the temperature is maintained at this range until the complete formation of Bainite
e) Converting Spheroidite to perlite
The heat treatment involves heating Spheroidite of 0.76 wt% C steel to a temperature of 720°C until austenization then it will be quenched at temperature < 35°C
f) Perlite to Spheroidite
The heat treatment procedure for converting Perlite to spheroidite involves heating Perlite of 0.76 wt% C steel for approximately 24 hours at a temperature of 700°C
g) Tempered martensite to martensite
The heat treatment entails heating Tempered martensite of 0.76 wt% C steel to a temperature of 760°C until austenization then it will be quenched at temperature > 140°C
h) Bainite to spheroidite
The heat treatment procedure for converting Perlite to spheroidite involves heating Bainite of 0.76 wt% C steel for approximately 24 hours at a temperature of 700°C
Explanation:
The heat treatment procedure is simply the heating of a metal to a high temperature and cooling the metal back. during this process the metal will undergo certain mechanical changes
a) converting Martensite to spheroidite
The heat treatment procedure for converting Martensite to spheroidite involves heating Martensite of 0.76 wt% C steel for approximately 24 hours at a temperature of 700°C
b) Converting Spheroidite to martensite
The heat treatment entails heating Spheroidite of 0.76 wt% C steel to a temperature of 760°C to austenization then it will be quenched at temperature > 140°C
c) Converting Bainite to Pearlite
The heat treatment involves heating Bainite of 0.76 wt% C steel to a temperature of 720°C until austenization then it will be quenched at temperature < 35°C
d) Converting Pearlite to Bainite
The heat treatment involves heating Pearlite of 0.76 wt% C steel to a temperature of 720°C until austenization then it will be cooled at a temperature range 220°C to 540°C. the temperature is maintained at this range until the complete formation of Bainite
e) Converting Spheroidite to perlite
The heat treatment involves heating Spheroidite of 0.76 wt% C steel to a temperature of 720°C until austenization then it will be quenched at temperature < 35°C
f) Perlite to Spheroidite
The heat treatment procedure for converting Perlite to spheroidite involves heating Perlite of 0.76 wt% C steel for approximately 24 hours at a temperature of 700°C
g) Tempered martensite to martensite
The heat treatment entails heating Tempered martensite of 0.76 wt% C steel to a temperature of 760°C until austenization then it will be quenched at temperature > 140°C
h) Bainite to spheroidite
The heat treatment procedure for converting Perlite to spheroidite involves heating Bainite of 0.76 wt% C steel for approximately 24 hours at a temperature of 700°C
Q1) Assuming that in a room full of 13 students born in 2000 and 2004 only, calculate the probability that that two persons or more have the same birthday (same day, month and year) knowing that 2000 contains 366 days and 2004 contains 366 days.
Answers
Answer:
D
Explanation:
Got it wrong so i could answer
1) Each of the following would be considered company-confidential except
A) a contract bid B) employee salaries C) your company's strategic plan D) your company's address
Answers
An air conditioning system is to be filled from a rigid container that initially contains 5 kg of saturated liquid at 24° Celsius the valve connecting this container to the air conditioning system is not open until the mass in this container is .25 Cal and the quality is going 506 at which time the valve is closed during this time only saturated liquid R134a flows from the container presuming that the process is isothermal wild the valve is open.
Required:
Determine the final quality of the R-134a in the container and the total heat transfer.
Answers
Water exiting the condenser of a power plant at 45 Centers a cooling tower with a mas flow rate of 15,000 kg/s. A stream of cooled water is returned to the condenser at the same flowrate. Makeup water is added in a separate stream at 20 C. Atmosphericair enters the cooling tower at 30 C, with a wet bulb temperature of 20 C. The volumetric flow rate of moist air into the cooling tower is 8000 m3/s. Moist air exits the tower at 40C and 90% RH. Assume atmospheric pressure is at 101.3 kPa. Determine: a.T
Answers
Answer: hello your question is incomplete below is the missing part
question :Determine the temperature of the cooled water exiting the cooling tower
answer : T = 43.477° C
Explanation:
Temp of water at exit = 45°C
mass flow rate of cooling tower = 15,000 kg/s
Temp of makeup water = 20°C
Assuming an atmospheric pressure of = 101.3 kPa
Determine temperature of the cooled water exiting the cooling tower
Water entering cooling tower at 45°C
Given that Latent heat of water at 45°C = 43.13 KJ/mol
Cp(wet air) = 1.005+ 1.884(y1)
where: y1 - Inlet mole ratio = (0.01257) / (1 - 0.01257) = 0.01273
Hence : Cp(wet air) = 29.145 + (0.01273) (33.94) = 29.577 KJ/kmol°C
First step : calculate the value of Q
Q = m*Cp*(ΔT) + W(latent heat)
Q = 321.6968 (29.577) (40-30) + 43.13 (18.26089)
Q = 95935.8547 KJ/s
Given that mass rate of water = 15000 kg/s
Hence the temperature of the cooled water can be calculated using the equation below
Q = m*Cp*∆T
Cp(water) = 4.2 KJ/Kg°C
95935.8547 = (15000)*(4.2)*(45 - T)
( 45 - T ) = 95935.8547/ 63000. ∴ T = 43.477° C
An acid-base indicator is usually a weak acid with a characteristic color in the protonated (acid) and deprotonated (conjugate base) forms. In this assignment, you will monitor the color of an acetic acid solution containing Bromocresol Green as an indicator, as the pH is changed and then you will use your data to calculate the ionization constant, Ka, for the bromocresol green indicator and compare to an accepted value.
1. Start Virtual Chemlab, select Acid-Base Chemistry, and then select lonization Constants of Weak Acids from the list of assignments. The lab will open in the Titration laboratory. Bottles of 0.1104 M NaOH and 0.1031 M HAC (acetic acid) will be on the lab bench. The buret will be filled with the NaOH solution and a beaker containing 10.00 mL of the HAc solution will be on the stir plate. The stir plate will be on, Bromocresol Green indicator will have been added to the beaker, and a calibrated pH probe will also be in the beaker so the pH of the solution can be monitored.
2. What is the color and pH of the solution?
3. On the buret, the horizontal position of the orange handle is off for the stopcock. Open the stopcock by pulling down on the orange handle. The vertical position delivers solution the fastest with three intermediate rates in between (slow dropwise, fast dropwise, and slow stream). Turn the stopcock to the second position or fast drop-wise addition. Observe the color of the solution and close the stopcock when you observe the change in color by double clicking on the center of the stopcock.
4. What is the color and pH of the solution now?
5. Continue to add NaOH as before or at a faster rate. What is the final color and pH of the solution after all of the NaOH is added?
6. An acid-base indicator is usually a weak acid with a characteristic color in the protonated and deprotonated forms. Because bromocresol green is an acid, it is convenient to represent its rather complex formula as HBOG. HBOG ionizes in water according to the following equation:
HBOG + H2O = BCG + H3O+
(yellow) (blue)
The K. (the equilibrium constant for the acid) expression is:
Ka = [BCG-][H3O+]/[HBCG)
When [BCG-[ = [HBCG), then Ka = [H3O+). If you know the pH of the solution, then the [H3O+] and Ka can be determined.
What would be the color of the solution if there were equal concentrations of HBCG and BCG-?
What is the pH at the first appearance of this color?
What is an estimate for the Ka for bromocresol green?
Answers
A line of students are arranged in odd and even positions. Now the students in the odd positions are to be sorted in the descending order and the students in the even position are to be sorted in ascending order, given a 1 D array. The maximum length of the line is 20. Display – ‘Invalid Size', if the input specified is zero or negative. Write an algorithm to implement the above scenario.
Answers
Answer:
The algorithm is as follows:
0. Start
1. MyArray = []
2. Input n
3. if n <1 or n > 20:
3.1print("Invalid Size")
4. else:
4.1 for [tex]i = 0[/tex] to n - 1:
4.1.1 Input MyArray[i]
4.2 even = []; odd = []
4.3 enum = 0; onum = 0
4.4 for i = 0 to n - 1:
4.4.1 if i%2 == 0:
4.4.1.1 even[enum]=list[i]
4.4.1.2 enum = enum + 1
4.4.2 else:
4.4.2.1 odd[onum]=list[i]
4.4.2.2 onum= onum + 1
4.5 MyArray.clear()
4.6 enum=0
4.7 while even:
4.7.1 minm = even[0]
4.7.2 for x in even:
4.7.2.1 if x < minm:
4.7.2.1.1 minm = x
4.7.3 MMyArray[enum] = minm
4.7.4 even.remove(minm)
4.7.5 enum = enum + 1
4.8 while odd:
4.8.1 maxm = odd[0]
4.8.2 for x in odd:
4.8.2.1 if x > maxm:
4.8.2.1.1 maxm = x
4.8.3 MMyArray[enum] = maxm
4.8.4 odd.remove(maxm)
4.8.5 enum = enum + 1
4.9 for i = 0 to n - 1:
4.9.1 print MyArray[i]
5. Stop
Explanation:
This starts the algorithm
0. Start
This creates an empty array
1. MyArray = []
This gets input for n (the length of the array)
2. Input n
If n is is less than 1 or greater than 20, then the input is invalid
3. if n <1 or n > 20:
3.1print("Invalid Size")
For valid values of n, we have:
4. else:
The italicized gets input into the array
4.1 for [tex]i = 0[/tex] to n - 1:
4.1.1 Input MyArray[i]
This creates empty arrays for even index and for odd index
4.2 even = []; odd = []
This initializes even index and odd index to 0
4.3 enum = 0; onum = 0
This iterates through the array indices
4.4 for i = 0 to n - 1:
If index is even, add array element to even
4.4.1 if i%2 == 0:
4.4.1.1 even[enum]=list[i]
4.4.1.2 enum = enum + 1
If otherwise, add array element to odd
4.4.2 else:
4.4.2.1 odd[onum]=list[i]
4.4.2.2 onum= onum + 1
Clear elements of MyArray
4.5 MyArray.clear()
Set index to 0
4.6 enum=0
Iterate through the even array
4.7 while even:
Set minimum to the first index
4.7.1 minm = even[0]
Sort the array in ascending order
4.7.2 for x in even:
4.7.2.1 if x < minm:
4.7.2.1.1 minm = x
Add the sorted array into MyArray
4.7.3 MMyArray[enum] = minm
Delete the elements of even array
4.7.4 even.remove(minm)
4.7.5 enum = enum + 1
Iterate through the odd array
4.8 while odd:
Set maximum to the first index
4.8.1 maxm = odd[0]
Sort the array in descending order
4.8.2 for x in odd:
4.8.2.1 if x > maxm:
4.8.2.1.1 maxm = x
Add the sorted array into MyArray
4.8.3 MMyArray[enum] = maxm
Delete the elements of odd array
4.8.4 odd.remove(maxm)
4.8.5 enum = enum + 1
Iterate through the indices of the double sorted MyArray
4.9 for i = 0 to n - 1:
Print each array element
4.9.1 print MyArray[i]
End algorithm
5. Stop
See attachment for the program implemented in Python
Methane (CH4) at 298 K, 1 atm enters a furnace operating at steady state and burns completely with 140% of theoretical air entering at 400 K, 1 atm. The products of combustion exit at 500 K, 1 atm. The flow rate of the methane is 1.4 kg/min. Kinetic and potential energy effects are negligible and air can be modeled as 21% O2 and 79% N2 on a molar basis.
Required:
Determine the dew point temperature of the products, in K.
Answers
Explain two reason why it is important for DG08 Engineering to refer to electronic component pin configuration specifications when designing and building printed circuit boards?
Answers
Answer:
refer to aja
Explanation:
Help, quick please. I need help with my engineering word problem
Answers
Answer:
a
Explanation:
Can someone put each letter by the correct word for my automotive class !
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Answer:
L = spindle
M = lower ball joint
part without the letter showing = steering knuckle
Explanation:
ihjpr2 ywjegnak'evsinawhe2'qwmasnh ngl,;snhy
Answers
Answer:
ummm ok?
Explanation:
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A Class III two-lane highway is on level terrain, has a measured free-flow speed of 45 mi/h, and has 100% no-passing zones. During the peak hour, the analysis direction flow rate is 150 veh/h, the opposing direction flow rate is 100 veh/h, and the PHF-0.95. There are 5% large trucks and 10% recreational vehicles. Determine the level of service.
Answers
Answer:
LOS = A
Explanation:
Given all the parameters the level of service as seen from the attached graph
is LOS = A
To determine the LOS from the attached graph
calculate the trial value of Vp
Vp = V / PHF
= (100 + 150) / 0.95 = 263 pc/h
since the trial value of Vp = ( 0 to 600 ) pc/h . hence E.T = 1.7 , ER = 1
next we will calculate the flow rate
flow rate = 1 / [ ( 1 + PT(ET - 1 ) + PR ( ER - 1 ) ]
Fhr = 1 / 1.035 = 0.966 ≈ 1
next calculate the real value of Vp
Vp = V / ( PHF * N * Fhr * Fp )
= ( 100 + 150 ) / ( 0.95 * 2 * 1 * 1 )
Vp ≈ 126 pc/h/In
Next calculate the density
D = Vp / S = 126 / ( 45 * 1.61 ) = 1.74 pc/km/In
Determine the slopes and deflections at points B and C for the beam shown below by the moment-area method. E=constant=70Gpa I=500 (10^6)mm^4
Answers
Answer:
hello your question is incomplete attached below is the complete question
answer :
Slopes : B = 180 mm , C = 373 mm
Deflection: B = 0.0514 rad , C = 0.077 rad
Explanation:
Given data :
I = 500(10^6) mm^4
E = 70 GPa
The M / EI diagram is attached below
Deflection angle at B
∅B = ∅BA = [ 150 (6) + 1/2 (300)*6 ] / EI
= 1800 / ( 500 * 70 ) = 0.0514 rad
slope at B
ΔB = ΔBA = [ 150(6)*3 + 1/2 (300)*6*4 ] / EI
= 6300 / ( 500 * 70 ) = 0.18 m = 180 mm
Deflection angle at C
∅C = ∅CA = [ 1800 + 300*3 ] / EI
= 2700 / ( 500 * 70 )
= 2700 / 35000 = 0.077 rad
Slope at C
ΔC = [ 150 * 6 * 6 + 1/2 (800)*6*7 + 300(3) *1.5 ]
= 13050 / 35000 = 373 mm
The fracture strength of glass may be increased by etching away a thin surface layer. It is believed that the etching may alter the surface crack geometry (i.e. reduce crack length and increase tip radius). Calculate the ratio of the etched and original crack tip radii if the fracture strength is increased by a factor of 6 when 16.0% of the crack length is removed.
Answers
Answer:
the ratio of the etched to the original crack tip radius is 30.24
Explanation:
Given the data in the question;
we determine the initial fracture stress using the following expression;
(σf)₁ = 2(σ₀)₁ [tex][[/tex] α₁/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex] ----- let this be equation 1
where; (σ₀)₁ is the initial fracture strength
([tex]p_t[/tex])₁ is the original crack tip radius
α₁ is the original crack length.
first, we determine the final crack length;
α₂ = α₁ - 16% of α₁
α₂ = α₁ - ( 0.16 × α₁)
α₂ = α₁ - 0.16α₁
α₂ = 0.84α₁
next, we calculate the final fracture stress;
the fracture strength is increased by a factor of 6;
(σ₀)₂ = 6( σ₀ )₁
Now, expression for the final fracture stress
(σf)₂ = 2(σ₀)₂ [tex][[/tex] α₂/([tex]p_t[/tex])₂ [tex]]^{1/2[/tex] ------- let this be equation 2
where ([tex]p_t[/tex])₂ is the etched crack tip radius
value of fracture stress of glass is constant
Now, we substitute 2(σ₀)₁ [tex][[/tex] α₁/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex] from equation for (σf)₂ in equation 2.
0.84α₁ for α₂.
6( σ₀ )₁ for (σ₀)₂.
∴
2(σ₀)₁ [tex][[/tex] α₁/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex] = 2(6( σ₀ )₁) [tex][[/tex] 0.84α₁/([tex]p_t[/tex])₂ [tex]]^{1/2[/tex]
divide both sides by 2(σ₀)₁
[tex][[/tex] α₁/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex] = 6 [tex][[/tex] 0.84α₁/([tex]p_t[/tex])₂ [tex]]^{1/2[/tex]
[tex][[/tex] 1/([tex]p_t[/tex])₁ [tex]]^{1/2[/tex] = 6 [tex][[/tex] 0.84/([tex]p_t[/tex])₂ [tex]]^{1/2[/tex]
[tex][[/tex] 1/([tex]p_t[/tex])₁ [tex]][/tex] = 36 [tex][[/tex] 0.84/([tex]p_t[/tex])₂ [tex]][/tex]
1 / ([tex]p_t[/tex])₁ = 30.24 / ([tex]p_t[/tex])₂
([tex]p_t[/tex])₂ = 30.24([tex]p_t[/tex])₁
([tex]p_t[/tex])₂/([tex]p_t[/tex])₁ = 30.24
Therefore, the ratio of the etched to the original crack tip radius is 30.24
Create a 6-bit full subtractor that uses the Borrow method to subtract two 6-bit binary numbers. You can use the proper basic sub-circuit.
Answers
You are using a Jupyter Notebook to explore data in a DataFrame named productDF. You want to write some inline SQL by using the following code, and visualize the results as a scatter plot: %%sql SELECT cost, price FROM product What should you do before running a cell with the %%sql magic? a. Create a new DataFrame named product from productDF.select("cost", "price") b. Persist the productDF DataFrame using productDF.createOrReplaceTempView("product") c. Filter the productDF dataframe using productDF.filter("cost == price") d. Rename the columns in the productDF DataFrame using productDF.withColumnRenamed("cost", "price")
Answers
Identify the following formulas:
1. Slope Formula
2. Slope-Intercept Form
3. Standard Form
4. Point-Slope Formula
Answers
2. y=mx+b
3. Ax+By=C (I think)
4. y-y1=m(x-x1)
Problem 9.11 A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain fracture toughness of 92 MPa m (klein.)) and a yield strength of 900 MPa (65270 psi). The flaw size resolution limit of the flaw detection apparatus is 3 mm (0.1181 in.). (a) If the design stress is one-half of the yield strength and the value of Y is 1.15, what is the critical flaw length
Answers
Answer:
the critical flaw length is 10.06 mm
Explanation:
Given the data in the question;
plane strain fracture toughness [tex]K_{tc[/tex] = 92 Mpa√m
yield strength σ[tex]_y[/tex] = 900 Mpa
design stress is one-half of the yield strength ( 900 Mpa / 2 ) 450 Mpa
Y = 1.15
we know that;
Critical crack length [tex]a_c[/tex] = 1/π( [tex]K_{tc[/tex] / Yσ )²
we substitute
[tex]a_c[/tex] = 1/π( 92 Mpa√m / (1.15 × 450 Mpa )²
[tex]a_c[/tex] = 1/π( 92 Mpa√m / (517.5 Mpa )²
[tex]a_c[/tex] = 1/π( 0.177777 )²
[tex]a_c[/tex] = 1/π( 0.03160466 )
[tex]a_c[/tex] = 0.01006 m = 10.06 mm
Therefore, the critical flaw length is 10.06 mm
{ [tex]a_c[/tex] = ( 10.06 mm ) > 3 mm
The critical flow is subject to detection
Al ejercer una fuerza de 50N sobre un resorte elastico esto se alarga desde los 15 cm hasta los 60cm¿cual es la constante elastica del resorte?
Answers
Answer:
Constante de resorte = 1.1 N/m
Explanation:
Dados los siguientes datos;
Fuerza = 50N
Extensión = 60cm - 15cm = 45cm
Para encontrar la constante del resorte;
Matemáticamente, la fuerza ejercida para estirar un resorte viene dada por la fórmula;
Fuerza = constante de resorte * extensión
Sustituyendo en la fórmula, tenemos;
50 = constante de resorte * 45
Constante de resorte = 50/45
Constante de resorte = 1.1 N/m
) A flow is divided into two branches, with the pipe diameter and length the same for each branch. A 1/4-open gate valve is installed in line A, and a 1/3-closed ball valve is installed in line B. The head loss due to friction in each branch is negligible compared with the head loss across the valves. Find the ratio of the velocity in line A to that in line B (include elbow losses for threaded pipe fittings).
Answers
Answer:
Va / Vb = 0.5934
Explanation:
First step is to determine total head losses at each pipe
at Pipe A
For 1/4 open gate valve head loss = 17 *Va^2 / 2g
elbow loss = 0.75 Va^2 / 2g
at Pipe B
For 1/3 closed ball valve head loss = 5.5 *Vb^2 / 2g
elbow loss = 0.75 * Vb^2 / 2g
Given that both pipes are parallel
17 *Va^2/2g + 0.75*Va^2 / 2g = 5.5 *Vb^2 / 2g + 0.75 * Vb^2 / 2g
∴ Va / Vb = 0.5934
.) If the charges attracting each other in the problem above have equal magnitude, what is the magnitude of each charge?
Answers
Answer:
Not seeing any other information, the best answer I can give is 2m.
Explanation:
M = magnitude
You see, if they have an equal charge, and you add them, it'd be 2 * m, or 2m.
what is the most common type of suspensions system used on body over frame vehicles?
Answers
Answer:
Engine
Explanation:
Semi-independent suspension is the most common type of suspension system used on body over frame vehicles.
What is a Semi-independent suspension?Semi-independent suspension give the front wheels some individual movement.
This suspension only used in rear wheels.
Thus, the correct option is Semi-independent suspension
Learn more about Semi-independent suspension
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Which of the following identifies the beginning phase of the engineering design process?
structural analysis
visual analysis
recognizing specifications and limitations
brainstorming possible designs
Answers
Answer:
<❤)structural analysi(❤>
Explanation:
(♨️)BRAINLEIST PLEASE(♨️)
Answer:
Structural analysis
Explanation:
A source current of 10 mA is supplied to a parallel circuit consisting of the following resistors three resistors, a 2200 a 500 and a 1KO. What is the source voltage required to
supply the current