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]
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]
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.
The term _______________refers to the science of using fluids to perform work.
Answer:
Hi, there your answer is hydraulics
Explanation:
Find at the terminals of the circuit
Explain the advantages of using register indirect addressing mode over direct addressing mode with an 8051 assembly code. Also provide an example where it is inefficient to code using register indirect addressing mode.
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
A steel plate of width 120mm and thickness of 20mm is bent into a circular arc radius of 10. You are required to calculate the maximum stress induced and the bending moment which will give the maximum stress. You are given that E=2*10^5
Answer:
Hence the magnitude of the pure moment m will be [tex]2\times 10^5.[/tex]
Explanation:
Width of steel fleet = 120 mm The thickness of steel fleet = 10 mm Let the circle of radius = 10 mNow,
We know that,
[tex]\frac{M}{I} = \frac{E}{R}[/tex]
Thus, [tex]M =\frac{EI}{R}[/tex]
Here
R = 10000 mm
[tex]I=\frac{1}{12}\times 120\times 10^{3}\\= 10^{4} mm^{4}[/tex]
[tex]E=2\times 10^{5}n/mm^{2}\\\\E=2\times 10^{5}n/mm^{2}\\\\M={(2\times 10^{5}\times 10^{4})/{10000}}\\\\M=2\times 10^{5}[/tex]
Hence, the magnitude of the pure moment m will be [tex]2\times 10^5.[/tex]
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.
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?
Greek engineers had the unenviable task of moving largecolumns from the quarries to the city.One engineer,Chersiphron, tried several different techniques to do this. Onemethodwas to cut pivot holes into the ends of the stone andthen use oxen to pull the column. The 5-ft diameter columnweighs 14,000 lbs, andthe team of oxen generates a constantpull force of 2,000 lbs on the center of the cylinder,G. Knowingthat the column starts from rest and rolls without slipping,determine:(a) the velocity of its center,G,after it has moved7ft,and (b) the minimum static coefficient of friction that will keepit from slipping.
Answer:
yea
Explanation:
okkudjeheud email and delete the electronic version of this communication
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
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
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.
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.
After adding refrigerant to a nonpressurized container used with a system-dependent recovery
device, the pressure of the container will:
Select one:
A. Increase to the saturation point.
B. Remain at 0 psig.
C. Increase in proportion to the amount of refrigerant added.
D. Decrease in proportion to the refrigerant added.
There are different aspect to system-dependent recovery equipment. The pressure of the container will Increase to the saturation point.
System-dependent is known to be a passive recovery system. It is known as the recovery of refrigerant from a system by using the refrigeration system's internal pressure and/or compressor so as to simulate the recovery process.
System-dependent equipment is not used used with appliances that has more than 15 pounds of refrigerant.
When refrigerant pressure increases, saturation temperature is known to also increases and as pressure decreases, the saturation temperature will also decrease. Saturation is known to have a pressure-temperature relationships for all refrigerants.
Learn more about saturation from
https://brainly.com/question/4529762
A 2*8 inches wood member has a length =11 ft it's density is =30 Lb/ft^3 find the actual weight of this member in Lb
Answer:
359.046 Ib
Explanation:
density = m / v ---- ( 1 )
m = mass
v = volume = l * b * h
Given data :
density = 30 Ib / ft^3
l = 2 inches = 0.1667 ft
b = 8 inches = 0.6667 ft
h = 11 ft
v ( volume ) = 0.1667 * 0.6667 * 11 = 1.22 ft^3
back to equation 1
m = 30 * 1.22 = 36.6 Ib
Actual weight ( F ) = m * g
= 36.6 * 9.81 = 359.046 Ib.
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 )
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
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
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.
lời mở đầu cho môn lí thuyết tài chính tiền tệ
Answer:
No puedo entenderte solo en español
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:
A spring having a stiffness k is compressed a distance δ. The stored energy in the spring is used to drive a machine which requires power P. Determine how long the spring can supply energy at the required rate.
Units Used: kN = 103 N
Given: k = 5 kN/m; δ = 400 mm; P = 90 W
Answer:
30w
Explanation:
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 answer to the question is what the
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:
A wastewater treatment plant treats 20 MGD of wastewater containing 950 mg/L of suspended solids in a primary clarifier that has a 20% suspended solids removal efficiency. The rate of sludge collection, the flow rate out the bottom of the clarifier, is 0.08 MGD.
a) What is the solids concentration (in mg/L) in the sludge leaving the clarifier?
b) What mass of solids (in kg/y) is removed annually by the primary clarifier?
Answer:
a) 47500 mg/L
b) 5250366.444 kg/year
Explanation:
Given data:
suspended solids removal efficiency = 20%
Flowrate in the primary clarifier ( Q ) = 20 MGD ( change to Liters/day
Q = 20* 10^6 * 3.785412 Liters /day
settled concentration ( St ) = 950mg/L * 0.2 = 190 mg/L
amount of settled solid = Q * St
= ( 20* 10^6 * 3.785412 ) * 190 = 14384.5656 kg/day
∴ Amount going into sludge with a flowrate of 0.08 MGD = 14384.5656 kg/day
a) concentration of solid in sludge ( leaving the clarifier )
= amount of settled solid / flow rate out of the clarifier in liters/day
= 14384.5656 / ( 0.08 * 10^6 * 3.785412 )
= 0.0475 kg/L
= 47500 mg/L
b) Determine mass of solids that is removed annually
= 14384.5656 kg/day * 365 days
= 5250366.444 kg/year
3. Which of these instruments is used to measure wind speed? A. anemometer C. wind sock B. thermometer D. wind vane It is an instrument that can show both the wind speed and direction. A. Anemometer C. Wind sock B. thermometer D. wind vane 4. 5. At what time does air temperature changes? A. from time to time C. in the evening only B. in the afternoon only D. in the morning only
Answer:
wind vane if it can be used to show wind speed and the other is a
Explanation:
please mark 5 star if im right and brainly when ya can
Hot combustion gases, modeled as air behaving as an ideal gas, enter a steady-state turbine at 145 psi, 2700oR, and exit at 29 psi and 1974oR. The power output is 74.26 hp, and heat transfer from the turbine to the surroundings occurs at a rate of 14 BTU/s.
Required:
Determine the power output of the turbine, in hp.
Hot combustion gases, modeled as air behaving as an ideal gas, enter a steady-state turbine at 145 psi, 2700oR, and exit at 29 psi and 1974oR. The power output is 74.26 hp, and heat transfer from the turbine to the surroundings occurs at a rate of 14 BTU/s.
Required:
a)Determine the power output of the turbine, in hp.
b) The Flow rate
Answer:
a) [tex]w=74.26hp[/tex]
b) [tex]m=0.22[/tex]
Explanation:
From the question we are told that:
Initial Pressure [tex]p_1= 145 psi[/tex]
Initial Temperature [tex]T_1 =2700oR=>2240.33^oF[/tex]
Final Pressure [tex]p_2= 29 psi[/tex]
Final Temperature [tex]t_2=1974oR=>1514.33^oF[/tex]
Output Power [tex]w=74.26hp[/tex]
Heat transfer Rate [tex]Q=14BTU/s[/tex]
Generally the equation for Steady flow energy is mathematically given by
[tex]Q-w=m(h_2-h_1)[/tex]
Where
[tex]m=Flow\ rate[/tex]
From Steam table
[tex]h_1=704btu/ib (at\ p_1= 145\ psi,\ T_1 =2700oR=>2240.33^oF )[/tex]
[tex]h_2=401btu/ib (at\ p_2= 29psi\ t_2=1974oR=>1514.33^oF )[/tex]
Therefore
[tex]-14-74.26=m(401-704)[/tex]
[tex]m=\frac{-14-74.26}{(401-704)}[/tex]
[tex]m=0.22[/tex]