Answer:
yes it is certainly good ice cream
Consider a single crystal of nickel oriented such that a tensile stress is applied along a [001] direction. If slip occurs on a (111) plane and in a [101]direction and is initiated at an applied tensile stress of 13.9 MPa (2020 psi), compute the critical resolved shear stress
Answer:
[tex]\mathbf{\tau_c =5.675 \ MPa}[/tex]
Explanation:
Given that:
The direction of the applied tensile stress =[001]
direction of the slip plane = [[tex]\bar 1[/tex]01]
normal to the slip plane = [111]
Now, the first thing to do is to calculate the angle between the tensile stress and the slip by using the formula:
[tex]cos \lambda = \Big [\dfrac{d_1d_2+e_1e_2+f_1f_2}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_2^2+e_2^2+f_2^2) }} \Big][/tex]
where;
[tex][d_1\ e_1 \ f_1][/tex] = directional indices for tensile stress
[tex][d_2 \ e_2 \ f_2][/tex] = slip direction
replacing their values;
i.e [tex]d_1[/tex] = 0 ,[tex]e_1[/tex] = 0 [tex]f_1[/tex] = 1 & [tex]d_2[/tex] = -1 , [tex]e_2[/tex] = 0 , [tex]f_2[/tex] = 1
[tex]cos \lambda = \Big [\dfrac{(0\times -1)+(0\times 0) + (1\times 1) }{\sqrt{(0^2+0^2+1^2)+((-1)^2+0^2+1^2) }} \Big][/tex]
[tex]cos \ \lambda = \dfrac{1}{\sqrt{2}}[/tex]
Also, to find the angle [tex]\phi[/tex] between the stress [001] & normal slip plane [111]
Then;
[tex]cos \ \phi = \Big [\dfrac{d_1d_3+e_1e_3+f_1f_3}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_3^2+e_3^2+f_3^2) }} \Big][/tex]
replacing their values;
i.e [tex]d_1[/tex] = 0 ,[tex]e_1[/tex] = 0 [tex]f_1[/tex] = 1 & [tex]d_3[/tex] = 1 , [tex]e_3[/tex] = 1 , [tex]f_3[/tex] = 1
[tex]cos \ \phi= \Big [ \dfrac{ (0 \times 1)+(0 \times 1)+(1 \times 1)} {\sqrt {(0^2+0^2+1^2)+(1^2+1^2 +1^2)} } \Big][/tex]
[tex]cos \phi= \dfrac{1} {\sqrt{3} }[/tex]
However, the critical resolved SS(shear stress) [tex]\mathbf{\tau_c}[/tex] can be computed using the formula:
[tex]\tau_c = (\sigma )(cos \phi )(cos \lambda)[/tex]
where;
applied tensile stress [tex]\sigma =[/tex] 13.9 MPa
∴
[tex]\tau_c =13.9\times ( \dfrac{1}{\sqrt{2}} )( \dfrac{1}{\sqrt{3}})[/tex]
[tex]\mathbf{\tau_c =5.675 \ MPa}[/tex]
Every time I take a photo, that photo has to be stored in a file somewhere within "My Files" correct?
How would I be able to take that photo out of the file it was stored?
Cut that photo by
1. Left click your mouse on the photo
2. Click cut
Then enter the file where you want to transfer and press
1. ctrl+v
Answer:
you can go to your file and then select the phpto and hold on a little bit and choose the delete option
Conditions of special concern: i. Suggest two reasons each why distillation columns are run a.) above or b.) below ambient pressure. Be sure to state clearly which explanation is for above and which is for below ambient pressure. ii. Suggest two reasons each why reactors are run at a.) elevated pressures and/or b.) elevated temperatures. Be sure to state clearly which explanation is for elevated pressure and which is for elevated temperature
Solution :
Methods for selling pressure of a distillation column :
a). Set, [tex]\text{based on the pressure required to condensed}[/tex] the overhead stream using cooling water.
(minimum of approximate 45°C condenser temperature)
b). Set, [tex]\text{based on highest temperature}[/tex] of bottom product that avoids decomposition or reaction.
c). Set, [tex]\text{based on available highest }[/tex] not utility for reboiler.
Running the distillation column above the ambient pressure because :
The components to be distilled have very high vapor pressures and the temperature at which they can be condensed at or below the ambient pressure.
Run the reactor at an evaluated temperature because :
a). The rate of reaction is taster. This results in a small reactor or high phase conversion.
b). The reaction is endothermic and equilibrium limited increasing the temperature shifts the equilibrium to the right.
Run the reaction at an evaluated pressure because :
The reaction is gas phase and the concentration and hence the rate is increased as the pressure is increased. This results in a smaller reactor and /or higher reactor conversion.
The reaction is equilibrium limited and there are few products moles than react moles. As increase in pressure shifts the equilibrium to the right.
The inputs of two registers R0 and R1 are controlled by a 2-to-1 multiplexer. The multiplexer select line and the register load enable inputs are controlled by inputs C0 and C1. Only one of the control inputs may be equal to 1 at a time. The required transfers are:
Answer: Hello your question is incomplete attached below is the complete question
answer :
Attached below
Explanation:
Given data:
The inputs of two registers are controlled by a 2-to-1 multiplexer.
The multiplexer select line and the register load enable inputs are controlled by inputs Co, C1, and C2.
Using the required transfers in the question to complete the detailed logic diagrams ( attached below )
lời mở đầu cho môn lí thuyết tài chính tiền tệ
Answer:
No puedo entenderte solo en español
A 120-volt fluorescent ballast has an input current of 0.34 ampere and an input power rating of 22 watts. The power factor of the ballast is ____.
Answer:
PF= .54
Explanation:
Power Factor equals working/real power (W) over apparent power (VA). 1.0 PF is an efficient equipment. PF= 22/(120*.34)
The power factor of the ballast is 0.54 which is the ratio of working power to apparent power.
What is the power factor?The power factor is a measure of energy efficiency. It is typically expressed as a percentage, with a lower percentage indicating inefficient power usage.
The power factor (PF) is the ratio of working power (in kW) to apparent power (in kilovolt amperes) (kVA).
Given that a 120-volt fluorescent ballast has an input current of 0.34 ampere and an input power rating of 22 watts.
PF = (True power)/(Apparent power)
PF = W/VA
Here W = 22 watts, V = 120-volt, and A = 0.34 ampere
PF = 22 / (120 × 0.34)
PF = 22 / 40.8
PF = 0.5392
PF = 0.54
Thus, the power factor of the ballast is 0.54.
Learn more about the power factor here:
https://brainly.com/question/19567608
#SPJ2
The number of telephone calls that arrive at a call center is often modeled as a poisson random variable. Assume that on the average there are 10 calls per hour
If the operators of the call center take a 30 minutes break for lunch what is probability that they don’t miss any call ?
with the aid of a labbled diagram describe the operation of a core type single phase transformer
Answer:
A simple single-phase transformer has each winding being wound cylindrically on a soft iron limb separately to provide a necessary magnetic circuit, which is commonly referred to as “transformer core”. It offers a path for the flow of the magnetic field to induce voltage between two windings.
A series circuit has 4 identical lamps. The potential difference of the energy source is 60V. The total resistance of the lamps is 20 Ω. Calculate the current through each lamp.
Answer:
[tex]I=3A[/tex]
Explanation:
From the question we are told that:
Number of lamps [tex]N=4[/tex]
Potential difference [tex]V=60v[/tex]
Total Resistance of the lamp is [tex]R= 20ohms[/tex]
Generally the equation for Current I is mathematically given by
[tex]I=\frac{V}{R}[/tex]
[tex]I=\frac{60}{20}[/tex]
[tex]I=3A[/tex]
Butters Furnishings makes hand crafted furniture for sale in its retail stores. The furniture maker has recently installed a new assembly process, including a new sander and polisher. With this new system, production has increased to 70 pieces of furniture per day from the previous 60 pieces of furniture per day. The number of defective items produced has dropped from 10 pieces per day to 4 per day. The production facility operates strictly 7 hours per day. Five people work daily in the plant. What is the growth rate in productivity for Aztec using the new assembly process?
Answer:
0.320
Explanation:
Productivity is measured as the total output divided by total input.
From the given information, the output is taken into consideration when we finish subtracting the defective materials.
The labor hours usually regarded as the input = 7 hours per day × 5 = 35
For the prior productivity, we have:
[tex]\mathbf{Prior \ Productivity = \dfrac{60 - 10}{35}}[/tex]
[tex]\mathbf{Prior \ Productivity = \dfrac{50}{35}}[/tex]
[tex]\mathbf{Prior \ Productivity = 1.43}[/tex]
For the current productivity:
[tex]\mathbf{current \ productivity= \dfrac{70-4}{35}}[/tex]
[tex]\mathbf{current \ productivity= \dfrac{66}{35}}[/tex]
[tex]\mathbf{current \ productivity= 1.89}[/tex]
Thus, the growth rate i.e. the increased change in productivity is:
[tex]= \dfrac{1.89 - 1.43}{1.43}[/tex]
= 0.320
You are hired as the investigators to identify the root cause and describe what should have occurred based on the following information. The mass of jet fuel required to travel from Toronto to Edmonton is 22,300 kg. The fuel gage correctly indicated that the plane already had 7,682 L of jet fuel in the tank. The specific gravity of the jet fuel is 0.803. Using this information, the crew added 4,916 L of fuel and took off, only to run out of fuel and crash a short while later. Use your knowledge of dimensions and units to work out what went wrong.
Answer:
The mass of fuel added, which is 10,166.2 kg is less than 22,300 kg which is the mass of fuel required to travel from Toronto to Edmonton, the plane therefore crashed.
Explanation:
Since density ρ = m/v where m = mass of fuel and v = volume of fuel, we need to find the mass of each volume of fuel.
So, m = ρv now ρ = specific gravity × density of water = 0.803 × 1000 kg/m³ = 803 kg/m³.
To find the mass of the 7,682 L of fuel, its volume is 7,682 dm³ = 7,682 dm³ × 1 m³/1000 dm³ = 7.682 m³.
It's mass, m = 803 kg/m³ × 7.682 m³ = 6168.646 kg
To find the mass of the extra 4,916 L of fuel added, we have
m' = ρv' where v' = 4,916 L = 4,916 dm³ = 4916 dm³ × 1 m³/1000 dm³ = 4.916 m³
m' = 803 kg/m³ × 4.916 m³ = 3947.548 kg
So, the total mass of the fuel is m" = m + m' = 6168.646 kg + 3947.548 kg = 10116.194 kg ≅ 10,166.2 kg
Since this mass of fuel added, which is 10,166.2 kg is less than 22,300 kg which is the mass of fuel required to travel from Toronto to Edmonton, the plane therefore crashed.
A 6-hp pump is used to raise water to an elevation of 12 meters. If the mechanical efficiency of the pump is 85 percent, determine the maximum volume flow rate of water in gallons per hour.
Answer:
29481.60 gallons/hour
Explanation:
Pump ( in ) = 6 - hp
Elevation ( h ) = 12 meters
mechanical efficiency of pump ( η pump ) = 85% = 0.85
Calculate for the maximum volume flow rate of water
1 - hp = 745.7 w
η pump = mgh / W * in
mgh = η pump * W * in
i.e. pVgh = η pump * W * in
∴ V = η pump * W * in / pgh
= 0.82 * ( 745.7 * 6 ) / 1000 * 9.81 * 12
= 3668.844 / 117720
= 0.031 m^3 /s = 29481.601 gallons/hour
Question 3
Marked out of 1.00
Question text
When you sell a car, what do you need to do to tell the state that you are not liable for that car anymore, including parking tickets?
Select one:
a. Call the DMV and tell them you sold the car
b. On the "Title" of the car, fill out the "Release of Liability" form and mail it in as instructed on the form.
c. The person you are selling the car to will take care of it.
d. There is nothing to do; if you do not own the car anymore, it is not your responsibility.
Answer:
D
Explanation:
Your answer is D.There is nothing to do; if you do not own the car anymore, it is not your responsibility
A project has an initial cost of $100,000 and uniform annual benefits of $12,500. At the end of its 8-year useful life, its salvage value is $30,000. At a 10% interest rate, the net present worth of the project is approximately:__________
1- $-19,318
2- $0
3- $+30,000
4- $+100,000
Answer:
[tex]X=-\$19318[/tex]
Explanation:
From the question we are told that:
Initial cost [tex]P= $100,000[/tex]
Annual benefits [tex]A= $12,500[/tex]
Salvage value [tex]S= $30,000[/tex]
Interest rate [tex]I=10\%=>0.10[/tex]
Time [tex]t=8years[/tex]
Generally the equation for Net Project worth X is mathematically given by
[tex]X=-P+A+S[/tex]
Where
Present worth of Annual benefits A is
[tex]A'=A(P/a,0.10,8)[/tex]
[tex]A'=12500*5.3349[/tex]
[tex]A'=\$66686.25[/tex]
Present worth of Salvage Price S is
[tex]S'=S(P/a,0.10,8)[/tex]
[tex]S'=30000*0.46651[/tex]
[tex]S'=\$13995.3[/tex]
Therefore
[tex]X=-P+A'+S'[/tex]
[tex]X=-100000+66686.25+\$13995.3[/tex]
[tex]X=-\$19318[/tex]
Find at the terminals of the circuit
The fracture toughness of a stainless steel is 137 MPa*m12. What is the tensile impact load sustainable before fracture that a rod can withstand with a pre-existing surface crack of 2 mm, given a square cross-section of 4.5 mm on each side, in kiloNewtons
Answer:
7.7 kN
Explanation:
The capacity of a material having a crack to withstand fracture is referred to as fracture toughness.
It can be expressed by using the formula:
[tex]K = \sigma Y \sqrt{\pi a}[/tex]
where;
fracture toughness K = 137 MPa[tex]m^{1/2}[/tex]
geometry factor Y = 1
applied stress [tex]\sigma[/tex] = ???
crack length a = 2mm = 0.002
∴
[tex]137 =\sigma \times 1 \sqrt{ \pi \times 0.002 }[/tex]
[tex]137 =\sigma \times 0.07926[/tex]
[tex]\dfrac{137}{0.07926} =\sigma[/tex]
[tex]\sigma = 1728.489 MPa[/tex]
Now, the tensile impact obtained is:
[tex]\sigma = \dfrac{P}{A}[/tex]
P = A × σ
P = 1728.289 × 4.5
P = 7777.30 N
P = 7.7 kN
diffrerentiate y=cos^{4} (3x+1)
Answer:
-6sin(6x+2)cos²(3x+1)dx.
Explanation:
[tex]dy=4*cos^3(3x+1)*3*(-sin(3x+1))dx=-6sin(6x+2)cos^2(3x+1)dx.[/tex]
The answer to the question is what the
Which of the following is not a correct statement about a probability select one
It must have a value between 0 and 1
It can be reported as a decimal or a fraction
A value near means that the event is not likely to occur/happens
It is the collection of several experiments
Answer:
but how does that make sense
Answer:
option b is the wrong statement
Step-by-step explanation:
it can be reported as decimal and fraction. probability must have a value between 0 and 1.probabilty zero means it is an impossible event, which is not likely to occur or happen.
Find at the terminals of the circuit
Answer:
ok i will on day
Explanation:
technician A says that incandescent bulbs resist vibration well. Technician B says that HID headlamps require up to approximately 25,000 volts to start. Who is correct
Answer:
Both a and b.
Explanation:
HID headlamps require high voltage ignition to start just like street lamps. It requires almost 25,000 volts to start HID head lamps but require only 80 to 90 volts to keep it operating.
Derive the expression for electrical-loading nonlinearity error (percentage) in a rotatory potentiometer in terms of the angular displacement, maximum displacement (stroke), potentiometer element resistance, and load resistance. Plot the percentage error as a function of the fractional displacement for the three cases: RL/RC = 0.1, 1.0, and 10.0
Answer:
The plot for percentage error as a function of fractional displacement ( [tex]\frac{R_{L} }{R_{C} }[/tex]) for the values of 0.1,1.0,10.0 is shown in image attached below.
Explanation:
Electrical loading non linearity error (percentage) is shown below.
[tex]E=\frac{(\frac{v_{o} }{v_{r} }-\frac{Q}{Q_{max} } )}{\frac{Q}{Q_{max} } }[/tex]×[tex]100[/tex]
where Q= displacement of the slider arm
[tex]Q_{max}=[/tex] maximum displacement of a stroke
[tex]\frac{v_{o} }{ v_{r} } =[/tex][tex]\frac{(\frac{Q}{Q_{max} }(\frac{R_{L} }{R_{C} } ) )}{(\frac{R_{L} }{R_{C} } ) +(\frac{Q}{Q_{max} })-(\frac{Q}{Q_{max} })^{2} }[/tex]
here [tex]R_{L}=load resistance[/tex]
[tex]R_{C}=[/tex]total resistance of potentiometer.
Now the nonlinearity error in percentage is
[tex]E=\frac{(\frac{(\frac{Q}{Q_{max} }(\frac{R_{L} }{R_{C} } ) )}{(\frac{R_{L} }{R_{C} } ) +(\frac{Q}{Q_{max} })-(\frac{Q}{Q_{max} })^{2} }-\frac{Q}{Q_{max} } )}{\frac{Q}{Q_{max} } }[/tex]×[tex]100[/tex]
The following attached file shows nonlinear error in percentage as a function of [tex]\frac{R_{L} }{R_{C} }[/tex] displacement with given values 0.1, 1.0, 10.0. The plot is drawn using MATLAB.
The MATLAB code is given below.
clear all ;
clc ;
ratio=0.1 ;
i=0 ;
for zratio=0:0.01:1 ;
i=i+1 ;
tratioa (1,i)=zratio ;
E1(1,i)=((((zratio*ratio)/(ratio+zratio-zratio^2))-zratio)/zrtio)*100 ;
end
ratio=1.0 :
i=0 ;
for zratio=0:0.01:1 ;
i=i+1 ;
tratiob (1,i)=zratio ;
E2(1,i)=((((zratio*ratio)/(ratio+zratio-zratio^2))-zratio)/zratio)*100 ;
end
ratio=10.0 :
i=0 ;
for zratio=0:0.01:1 ;
i=i+1 ;
tratioc (1,i)=zratio ;
E3(1,i)=((((zratio*ratio)/(ratio+zratio-zratio^2))-zratio)/zrtio)*100 ;
end
k=plot(tratioa,E1,tratiob,E2,tratioc,E3)
grid
title({non linear error in % as a function of R_L/R_C})
k(1). line width = 2;
k(1).marker='*'
k(1).color='red'
k(2).linewidth=1;
k(2).marker='d';
k(2).color='m';
k(3).linewidth=0.5;
k(3).marker='h';
k(3).color='b'
legend ('location', 'south east')
legend('R_L/R_C=0.1','R_L/R_C=1.0','R_L/R_C=10.0')
Compute the first four central moments for the following data:
i xi
1 45
2 22
3 53
4 84
5 65
Answer:
Compute the first four central moments for the following data:
i xi
1 45
2 22
3 53
4 84Explanation:
Determine (a) the principal stresses and (b) the maximum in-plane shear stress and average normal stress at the point. Specify the orientation of the element in each case. 60 45 30
Answer:
a) 53 MPa, 14.87 degree
b) 60.5 MPa
Average shear = -7.5 MPa
Explanation:
Given
A = 45
B = -60
C = 30
a) stress P1 = (A+B)/2 + Sqrt ({(A-B)/2}^2 + C)
Substituting the given values, we get -
P1 = (45-60)/2 + Sqrt ({(45-(-60))/2}^2 + 30)
P1 = 53 MPa
Likewise P2 = (A+B)/2 - Sqrt ({(A-B)/2}^2 + C)
Substituting the given values, we get -
P1 = (45-60)/2 - Sqrt ({(45-(-60))/2}^2 + 30)
P1 = -68 MPa
Tan 2a = C/{(A-B)/2}
Tan 2a = 30/(45+60)/2
a = 14.87 degree
Principal stress
p1 = (45+60)/2 + (45-60)/2 cos 2a + 30 sin2a = 53 MPa
b) Shear stress in plane
Sqrt ({(45-(-60))/2}^2 + 30) = 60.5 MPa
Average = (45-(-60))/2 = -7.5 MPa
The following true stresses produce the corresponding true plastic strains for a brass alloy: True Stress (psi) True Strain 49300 0.11 61300 0.21 What true stress is necessary to produce a true plastic strain of 0.25
Answer:
64640.92 psi
Explanation:
True stress ( psi ) True strain
49300 0.11
61300 0.21
Determine the true stress necessary to produce a true plastic strain of 0.25
бT1 = 49300
бT2 = 61300
бT3 = ?
∈T1 = 0.11
∈T2 = 0.21
∈T3 = 0.25
note : бTi = k ∈Ti^h
∴ 49300 = k ( 0.11 )^h ----- ( 1 )
61300 = k ( 0.21)^h ------ ( 2 )
solving equations 1 and 2 simultaneously
49300/61300 = ( 0.11 / 0.21 )^h
0.804 = (0.52 )^h
next step : apply logarithm
log ( 0.804 ) = log(0.52)^h
h = log 0.804 / log (0.52)
= 0.33
back to equation 1
49300 = k ( 0.11 )^0.33
k = 49300 / (0.11)^0.33
= 102138
therefore бT3 = K (0.25)^h
= 102138 ( 0.25 )^ 0.33
= 64640.92
A column carries 5400 pounds of load and is supported on a spread footing. The footing rests on coarse sand. Design the smallest square footing (to the next 3-inch increment) that will safely carry the column load. The footing will be 1 ft 9 in. deep and will be constructed of cast-in-place concrete.
Answer:
Following are the responses to the given question:
Explanation:
The term _______________refers to the science of using fluids to perform work.
Answer:
Hi, there your answer is hydraulics
Explanation:
How did the development of John Hadley's octant into a sextant enhance its measuring capabilities?
O A.
navigators could use the instrument in stormy conditions
OB.
navigators could measure angles to the nearest minute
O C. navigators could measure angles up to 60°
OD.
navigators could measure angles up to 120°
Answer:
b
Explanation:
Technician A says lever action pushes a rod into the brake booster and master cylinder
when the driver pushes on the brake pedal. Technician B says the produces hydraulic
pressure in the master cylinder. Who is correct?
A manufacturing facility with a wastewater flow of 0.011 m3/sec and a BOD5 of 590 mg/L discharges into the Cattaraugus Creek. The creek has a 10-year, 7-day low flow of 1.7 m3/sec. Upstream of the facility, the BOD5 of the creek is 0.6 mg/L. The BOD rate constants k are 0.115 d-1. Determine: [A] UBOD of wastewater. [B] UBOD of creek. [C] What is the initial ultimate BOD after mixing
This question is incomplete, the complete question is;
A manufacturing facility with a wastewater flow of 0.011 m³/sec and a BOD₅ of 590 mg/L discharges into the Cattaraugus Creek. The creek has a 10-year, 7-day low flow of 1.7 m³/sec. Upstream of the facility, the BOD₅ of the creek is 0.6 mg/L. The BOD rate constants k are 0.115 d⁻¹ for the wastewater and 3.7 d⁻¹ for the creek. The temperature of both the creek and tannery of wastewater is 20°C. Determine: [A] UBOD of wastewater. [B] UBOD of creek. [C] What is the initial ultimate BOD after mixing
Answer:
a) Ultimate BOD of wastewater is 1349.188 mg/L
b) Ultimate BOD of creek is 0.6 mg/L
c) the initial ultimate BOD after mixing is 9.27 mg/L
Explanation:
Given the data in the question;
Q[tex]_{wastewater[/tex] = 0.011 m³/s
BOD[tex]_{wastewater[/tex] = 590 mg/L
Q[tex]_{creek[/tex] = 1.7 m³/sec
BOD[tex]_{creek[/tex] = 0.6 mg/L
time t = 5
rate constants k for wastewater = 0.115 d⁻¹
rate constants k for creek = 3.7 d⁻¹
a) UBOD of wastewater.
The Ultimate BOD of wastewater is;
BOD[tex]_{wastewater[/tex] = L₀[tex]_{wastewater[/tex]( 1 - [tex]e^{-kt[/tex] )
where BOD[tex]_{wastewater[/tex] is the BOD of wastewater after 5 days, L₀[tex]_{wastewater[/tex] is the ultimate BOD of wastewater, k is the rate constant of wastewater and t is the time( days ).
we make L₀[tex]_{wastewater[/tex] the subject of formula
BOD[tex]_{wastewater[/tex] = L₀[tex]_{wastewater[/tex]( 1 - [tex]e^{-kt[/tex] )
L₀[tex]_{wastewater[/tex] = BOD[tex]_{wastewater[/tex] / ( 1 - [tex]e^{-kt[/tex] )
so we substitute
L₀[tex]_{wastewater[/tex] = 590 / ( 1 - [tex]e^{(-0.115*5)[/tex] )
L₀[tex]_{wastewater[/tex] = 590 / ( 1 - [tex]e^{(-0.575)[/tex] )
L₀[tex]_{wastewater[/tex] = 590 / ( 1 - 0.5627 )
L₀[tex]_{wastewater[/tex] = 590 / 0.4373
L₀[tex]_{wastewater[/tex] = 1349.188 mg/L
Therefore, Ultimate BOD of wastewater is 1349.188 mg/L
b) UBOD of creek
The Ultimate BOD of creek is;
BOD[tex]_{creek[/tex] = L₀[tex]_{creek[/tex]( 1 - [tex]e^{-kt[/tex] )
we make L₀[tex]_{creek[/tex] the subject of formula
L₀[tex]_{creek[/tex] = BOD[tex]_{creek[/tex] / (1 - [tex]e^{-kt[/tex] )
we substitute
L₀[tex]_{creek[/tex] = 0.6 / ( 1 - [tex]e^{(-3.7*5)[/tex] )
L₀[tex]_{creek[/tex] = 0.6 / ( 1 - [tex]e^{(-18.5)[/tex] )
L₀[tex]_{creek[/tex] = 0.6 / ( 1 - (9.2374 × 10⁻⁹) )
L₀[tex]_{creek[/tex] = 0.6 / 0.99999
L₀[tex]_{creek[/tex] = 0.6 mg/L
Therefore, Ultimate BOD of creek is 0.6 mg/L
c) the initial ultimate BOD after mixing;
Lₐ = [( Q[tex]_{wastewater[/tex] × L₀[tex]_{wastewater[/tex] ) + ( Q[tex]_{creek[/tex] × L₀[tex]_{creek[/tex] )] / [ Q[tex]_{wastewater[/tex] + Q[tex]_{creek[/tex] ]
we substitute
Lₐ = [( 0.011 × 1349.188 ) + ( 1.7 × 0.6 )] / [ 0.011 + 1.7 ]
Lₐ = [ 14.841068 + 1.02 ] / 1.711
Lₐ = 15.861068 / 1.711
La = 9.27 mg/L
Therefore, the initial ultimate BOD after mixing is 9.27 mg/L