Answer:
When are resistors in series? Resistors are in series whenever the flow of charge, called the current, must flow through devices sequentially. For example, if current flows through a person holding a screwdriver and into the Earth, then
R
1
in Figure 1(a) could be the resistance of the screwdriver’s shaft,
R
2
the resistance of its handle,
R
3
the person’s body resistance, and
R
4
the resistance of her shoes.
Figure 2 shows resistors in series connected to a voltage source. It seems reasonable that the total resistance is the sum of the individual resistances, considering that the current has to pass through each resistor in sequence. (This fact would be an advantage to a person wishing to avoid an electrical shock, who could reduce the current by wearing high-resistance rubber-soled shoes. It could be a disadvantage if one of the resistances were a faulty high-resistance cord to an appliance that would reduce the operating current.)
Two electrical circuits are compared. The first one has three resistors, R sub one, R sub two, and R sub three, connected in series with a voltage source V to form a closed circuit. The first circuit is equivalent to the second circuit, which has a single resistor R sub s connected to a voltage source V. Both circuits carry a current I, which starts from the positive end of the voltage source and moves in a clockwise direction around the circuit.
Figure 2. Three resistors connected in series to a battery (left) and the equivalent single or series resistance (right).
To verify that resistances in series do indeed add, let us consider the loss of electrical power, called a voltage drop, in each resistor in Figure 2.
According to Ohm’s law, the voltage drop,
V
, across a resistor when a current flows through it is calculated using the equation
V
=
I
R
, where
I
equals the current in amps (A) and
R
is the resistance in ohms
(
Ω
)
. Another way to think of this is that
V
is the voltage necessary to make a current
I
flow through a resistance
R
.
So the voltage drop across
R
1
is
V
1
=
I
R
1
, that across
R
2
is
V
2
=
I
R
2
, and that across
R
3
is
V
3
=
I
R
3
. The sum of these voltages equals the voltage output of the source; that is,
V
=
V
1
+
V
2
+
V
3
.
This equation is based on the conservation of energy and conservation of charge. Electrical potential energy can be described by the equation
P
E
=
q
V
, where
q
is the electric charge and
V
is the voltage. Thus the energy supplied by the source is
q
V
, while that dissipated by the resistors is
q
V
1
+
q
V
2
+
q
V
3
.
Explanation:
How many kg moles of Sodium Sulphate will contain 10 kg of
Sodium?
70.40mol cuz
1g sodium sulfate = 0.00704mol
take 10kg × 1000 = 10,000g
10,000g × 0.00704
final answer 70.40mol
(as per my thinking)
Answer:
70.40mol cuz
1g sodium sulfate = 0.00704mol
take 10kg × 1000 = 10,000g
10,000g × 0.00704
final answer 70.40mol
What type of engineer makes sure equipment is safe and operational
Answer:
mechanical engineer is the best answer
According to O*NET, what is the most common level of education among Licensing Examiners and Inspectors?
You've asked an Incomplete question, lacking options. I answered based on the existing O*NET report.
Answer:
high school diploma
Explanation:
According to the Occupational Information Network (O*NET), most people who are Licensing Examiners and Inspectors typically have a high school diploma.
In other words, they do not seek to acquire a post-secondary school education.
Answer:
B
Explanation:
According to edge its answer B
associate's degree or on-the-job experience
got it right as a lucky guess as the O*net site is updated but edge doesn't bother to update their questions or links.
Click this link to view O*NET’s Education section for Licensing Examiners and Inspectors. According to O*NET, what is the most common level of education among Licensing Examiners and Inspectors?
bachelor’s degree
associate's degree or on-the-job experience .......This is the correct answer.
some college, no degree
associate degree
Laminated steel parts are designed:
A. to reduce noise and vibration.
B.for higher energy absorption.
C.to allow good part formability,
D. to increase tensile strength.
Answer:
to increase tensile strength.
Explanation:
Steel as part of the materials used in the construction industries, or other manufacturing industries could be defective if it was not appropriately manufactured. This defects could be due to foreign materials in it, scratches, blisters etc which would lead to the tensile strength being reduced.
In-order to overcome this, there is need to carry out a technique called lamination. This technique is the process of manufacturing the steel in multiple layers so that the composite materials improves the strength of the steel and its stability. This helps to prevent its failure when used in construction.
A power washer is being used to clean the siding of a house. Water enters at 20 C, 1 atm, with a velocity of 0.2 m/s .A jet of water exits at 20 C, 1 atm, with a velocity of 20 m/s an elevation of 5 m. At steady state, the magnitude of the heat transfer rate from the power unit to the surroundings is 10% of the power input. Determine the power input to the motor, kW.
A power washer is being used to clean the siding of a house. Water at the rate of 0.1 kg/s enters at 20°c and 1 atm, with the velocity 0.2m/s. The jet of water exits at 23°c, 1 atm with a velocity 20m/s at an elevation of 5m. At steady state, the magnitude of the heat transfer rate from power unit to the surroundings is 10% of the power input. Determine the power input to the motor in kW.
Answer:
Net power of 1.2 KW is being extracted
Explanation:
We are given;
Mass flow rate; m' = 0.1kg/s
Inlet temperature; T1 = 20°C = 293K
Inlet pressure; P1 = 1 atm = 10^(5) pa
Inlet velocity; v1 = 0.2 m/s
Exit Pressure; P2 = 1 atm = 10^(5) pa
Exit Temperature; T2 = 1 atm = 296K
Exit velocity; V2 = 20m/s
Change in elevation; h = Z2 - Z1 = 5m
We are told that the magnitude of the heat transfer rate from the power unit to the surroundings is 10% of the power input.
Thus;
Q = -0.1W
From Bernoulli equation;
Q - W = ∆Potential energy + ∆Kinetic energy + ∆Pressure energy
Where;
∆Potential energy = mg(z2 - z1)
∆Kinetic energy = ½m(v2² - v1²)
∆Pressure energy = mc_p(T2 - T1)
Thus;
-0.1W - W = [m'g(z2 - z1)] + [½m'(v2² - v1²)] + [m'c_p(T2 - T1)]
Where C_p is specific heat capacity of water = 4200 J/Kg.k
Plugging in the relevant values, we have;
-1.1W = (0.1 × 9.81 × 5) + (½ × 0.1(20² - 0.2²)) + (0.1 × 4200 × (296 - 293))
-1.1W = 4.905 + 19.998 + 1260
-1.1W = 1284.903
W = -1284.903/1.1
W ≈ -1168 J/s ≈ -1.2 KW
The negative sign means that work is extracted from the system.
A heat pump is to be used to heat a house during the winter, as shown in Fig. 6–52. The house is to be maintained at 21 ℃ at all times. The house is estimated to be losing heat at a rate of 135,000 kJ/h when the outside temperature drops to -5 ℃. Determine the minimum power required to drive this heat pump.
Answer:
[tex]3.32\ \text{kW}[/tex]
Explanation:
[tex]T_c[/tex] = Outside temperature = [tex]-5^{\circ}\text{C}[/tex]
[tex]T_h[/tex] = Temperature of room = [tex]21^{\circ}\text{C}[/tex]
[tex]Q_h[/tex] = Heat loss = 135000 kJ/h = [tex]\dfrac{135000}{3600}=37.5\ \text{kW}[/tex]
Coefficient of performance of heat pump
[tex]\text{COP}=\dfrac{1}{1-\dfrac{T_c}{T_h}}\\\Rightarrow \text{COP}=\dfrac{1}{1-\dfrac{273.15-5}{273.15+21}}\\\Rightarrow \text{COP}=11.3[/tex]
Input power
[tex]W_i=\dfrac{Q_h}{\text{COP}}\\\Rightarrow W_i=\dfrac{37.5}{11.3}\\\Rightarrow W_i=3.32\ \text{kW}[/tex]
The minimum power required to drive this heat pump is [tex]3.32\ \text{kW}[/tex].
During her soccer game, Brittany hears her coach tell her team to look for better shot selection. The offensive strategy her coach is attempting to get Brittany and her teammates to use more is shooting when the:
Answer:
B: Team has the ball near center line
Explanation:
i guessed so...
The section should span between 10.9 and 13.4 cm (4.30 and 5.30 inches) as measured from the end supports and should be able to carry a minimum load of 13 N applied evenly to the top of the span. The maximum load for each member is 4.5 Newtons in compression and tension. Each member is 4.2 cm long (ball center to ball center). All joints must be fully constrained. Determine the max value of P before the truss fails. The section that can hold the most weight will receive bonus points.
Answer:
hello below is missing piece of the complete question
minimum size = 0.3 cm
answer : 0.247 N/mm2
Explanation:
Given data :
section span : 10.9 and 13.4 cm
minimum load applied evenly to the top of span : 13 N
maximum load for each member ; 4.5 N
lets take each member to be 4.2 cm
Determine the max value of P before truss fails
Taking average value of section span ≈ 12 cm
Given minimum load distributed evenly on top of section span = 13 N
we will calculate the value of by applying this formula
= [tex]\frac{Wl^2}{12} = (0.013 * 0.0144 )/ 12[/tex] = 1.56 * 10^-5
next we will consider section ; 4.2 cm * 0.3 cm
hence Z (section modulus ) = BD^2 / 6
= ( 0.042 * 0.003^2 ) / 6 = 6.3*10^-8
Finally the max value of P( stress ) before the truss fails
= M/Z = ( 1.56 * 10^-5 ) / ( 6.3*10^-8 )
= 0.247 N/mm2
What did Congress do in 1787 to settle land disputes among the settlers?
Answer:
On July 13, 1787, Congress enacts the Northwest Ordinance, structuring settlement of the Northwest Territory and creating a policy for the addition of new states to the nation. ... In 1781, Virginia began by ceding its extensive land claims to Congress, a move that made other states more comfortable in doing the same
Answer:
They divided up the Northwest Territory into acre squares and sold them.
Explanation:
If a poems has a regular rhythm throughout the poem, it has: PLEASE HELP MEH I WILL GIVE YOU BRAINLEIST!!
A. tone
B. imagery
C. irony
D. meter
Answer:
D, meter.
Explanation:
Rhythm is associated with meter, which identifies units of stressed and unstressed syllables.
If a poem has a regular rhythm throughout the poem, it has a meter. Option D is correct.
What is meter in poem?Meter, which distinguishes between stressed and unstressed syllables, is related to rhythm. The fundamental rhythmic framework of a stanza or a line of poetry is known as meter.
The number of feet in the poem serves as a measure of the poem's meter, which is the rhythm of the language.
Many traditional poem forms call for a certain verse meter or a group of meters that alternate in a specified pattern. Prosody refers to both the study of meters and other types of versification, as well as their practical application.
Therefore, option D is correct.
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Technician A says that the B-pillars aid in resisting roof crush.
Technician B says that the IIHS rates vehicles to resist roof crush to five times the weight of the vehicle.
Who is right?
A only
O B only
Both A and B
Neither Anor B
Answer:
The answer is "Both A and B" are right
Explanation:
During the previous twenty years car producers have made significant advances in planning vehicle structures that give more noteworthy tenant insurance in planar accidents (Lund and Nolan 2003). Be that as it may, there has been little agreement with respect to the significance of rooftop strength in rollover crashes, just as the best strategy for surveying that strength. In 2006 one-fourth of lethally harmed traveler vehicle tenants were associated with crashes where vehicle rollover was considered the most hurtful occasion (Protection Establishment for Expressway Wellbeing, 2007). Numerous lethally harmed tenants in rollovers are unbelted, and some are totally or mostly launched out from the vehicle (Deutermann 2002).
There is difference concerning how underlying changes could influence launch hazard or the danger of injury for inhabitants who stay in the vehicle, paying little mind to belt use.
Answer:A Only
Explanation:
I Took the test
Air at 1 atm and 25◦C blows across a large concrete surface 20 m wide maintained
at 60◦C. The flow velocity is 6 m/s. Calculate the convection heat loss from the
surface.
This is heat transfer convection, mechanical engineering
please solve this question guys I'm gonna really really be appreciate it for you guys
Answer:
Air at 1 atm and 25◦C blows across a large concrete surface 20 m wide maintained
at 60◦C. The flow velocity is 6 m/s. Calculate the convection heat loss from the
surface.
This is heat transfer convection, mechanical engineering
please solve this question guys I'm gonna really really be appreciate it for you guys
calculate the quantities of materials required for the first class brickwork in 1:6 cement mortar for 10 cu.m. assume the suitable data.
The quantity of bricks required increases with the surface area of the wall, but the thickness of a masonry wall does not affect the total quantity of bricks used in the wall
True or False
Answer:
false
Explanation:
A product (that has not been yet invented) that involves lighting up when something happens?
Answer:
Maybe when there is a fire there can be fire drones that can take it out. and it can also resuce people who are stuck there.
Explanation:
A converging nozzle has an exit area of 0.001 m2. Air enters the nozzle with negligible velocity at a pressure of 1 MPa and a temperature of 360 K. For isentropic flow of an ideal gas with k = 1.4 and the gas constant R = Ru/MW = 287 J/kg-K, determine the mass flow rate in kg/s and the exit Mach number for back pressures of (a) 500 kPa and (b) 784 kPa.
Answer:
a) for back pressures of (a) 500 kPa
- mass flow rate is 2.127 kg/s
- exit Mach number is 1.046
b) for back pressures of (a) 784 kPa
- mass flow rate is 1.793 kg/s
- exit Mach number is 0.6
Explanation:
Given that;
A₂ = 0.001 m²
P₁ = 1 MPa
T₁ = 360 K
k = 1.4
P₂ = 500 Kpa
(1000/500)^(1.4-1 / 1.4) = 360 /T₂
2^(0.4/1.4) = 360/T₂
1.219 = 360 / T₂
T₂ = 360 / 1.219
T₂ = 295.32 K
CpT₁ + V₁²/2000 = CpT₂ + V₂²/2000
we substitute
CpT₁ + V₁²/2000 = CpT₂ + V₂²/2000
1.005 × 360 = 1.005 × 295.32 + v₂²/2000
v₂ = 360.56 m/s²
p₂v₂ = mRT₂
500 × (0.001 × 360.56) = m × 0.287 × 295.32
m = 2.127 kg/s
so Mach Number = V₂ / Vc
Vc = √( kRT) = √( 1.4 × 287 × 295.32) = 344.47 m/s
So Mach Number = V₂ / Vc = 360.56 / 344.47 = 1.046
Therefore for back pressures of (a) 500 kPa
- mass flow rate is 2.127 kg/s
- exit Mach number is 1.046
b)
AT P₂ = 784 kPa
(1000/784)^(1.4-1 / 1.4) = 360/T₂
T₂ = 335.82 K
now
V₂²/2000 = 1.005( 360 - 335.82)
V₂ = 220.45 m/s
P₂V₂ = mRT₂
784 × (0.001 × 220.45) = m( 0.287) ( 335.82)
172.83 = 96.38 m
m = 172.83 / 96.38
m = 1.793 kg/s
just like in a)
Vc = √( kRT) = √( 1.4 × 287 × 335.82) = 367.32 m/s
Mach Number = V₂ / Vc = 220.45 / 367.32 = 0.6
Therefore for back pressures of (a) 784 kPa
- mass flow rate is 1.793 kg/s
- exit Mach number is 0.6
Following are the
Given:
[tex]A_2 = 0.001\ m^2\\\\P_1 = 1\ MPa\\\\T_1 = 360\ K\\\\k = 1.4\\\\P_2 = 500\ Kpa\\\\[/tex]
To find:
Flow rate of mass, and Mach number
Solution:
For point a)
Using formula:
[tex]\to P^{\frac{r-1}{r}} \alpha T\\\\\to (\frac{1000}{500})^(\frac{1.4-1}{1.4}) = \frac{360}{T_2}\\\\\to 2^(\frac{0.4}{1.4}) = \frac{360}{T_2}\\\\\to 1.219 = \frac{360}{T_2}\\\\\to T_2 = \frac{360}{1.219}\\\\\to T_2 = 295.32\ K\\\\[/tex]
[tex]\to C_pT_1 + \frac{V_1^2}{2000} = C_pT_2 + \frac{V_2^2}{2000}\\\\\to 1.005 \times 360 = 1.005 \times 295.32 + \frac{v_2^2}{2000}\\\\\to v_2 = 360.56 \ \frac{m}{s^2} \\\\\to p_2v_2 = mRT_2\\\\\to 500 \times (0.001 \times 360.56) = m \times 0.287 \times 295.32\\\\\to m = 2.127\ \frac{kg}{s}\\\\[/tex]
Mach Number [tex]= \frac{V_2}{V_c}\\\\[/tex]
[tex]\to V_c = \sqrt{( kRT)} = \sqrt{( 1.4 \times 287 \times 295.32)} = 344.47 \ \frac{m}{s}\\\\[/tex]
[tex]\to \frac{V_2}{V_c} = \frac{360.56}{344.47} = 1.046[/tex]
For point b)
[tex]\to P_2 = 784\ kPa\\\\\to (\frac{1000}{784})^{(\frac{0.4}{1.4})} = \frac{360}{T_2}\\\\\to T_2 = 335.82\ K\\\\[/tex]
now
[tex]\to \frac{V_2^2}{2000} = 1.005( 360 - 335.82)\\\\\to V_2 = 220.45 \frac{m}{s}\\\\\to V_c = \sqrt{( kRT)} = \sqrt{( 1.4 \times 287 \times 335.82)} = 367.32\ \frac{ m}{s}\\\\\to c= \frac{220.45}{ 367.32} = 0.6\\[/tex]
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Electronic components are often mounted with good heat conduction paths to a finned aluminum base plate, which is exposed to a stream of cooling air from a fan. The sum of the mass times specific heat products for a base plate and components is 5000 J/K, and the effective heat transfer coefficient times surface area product is 10 W/K. The initial temperature of the plate and the cooling air temperature are
Answer:
Hello your question is incomplete below is the complete question
Electronic components are often mounted with good heat conduction paths to a finned aluminum base plate, which is exposed to a stream of cooling air from a fan. The sum of the mass times specific heat products for a base plate and components is 5000 J/K, and the effective heat transfer coefficient times surface area product is 10 W/K. The initial temperature of the plate and the cooling air temperature are 295 K when 300 W of power are switched on. 1) Find the plate temperature after 10 minutes.
answer ; 311.36 k
Explanation:
Given data :
sum of mass * specific heat products for a base plate and components ( Mcp )
= 5000 J/K
effective heat transfer coefficient * surface area ( hA ) = 10 W/K
Initial temperature of plate and cooling air temperature( Tc ) = 295 k
power ( Q = W ) = 300 W
a) Determine plate temperature after 10 minutes
10 mins = 600 secs ( t )
heat supplied = change in temp + heat loss
Q * t = mCp ( ΔT ) + hA ( ΔT ) t
300*600 = 5000 * ( T -295 ) + 10 ( T -295 ) * 600
therefore ; T - 295 = 16.363
T = 311.36 K
Connecting rods undergo a process to alleviate manufacturing stresses from forging, a process known as ______.
Connecting rods undergo a process to alleviate manufacturing stresses from forging, a process known as cold forging.
What is forging?Forging is a metalworking process that involves shaping metal with localized compressive forces. A hammer or a die is used to deliver the blows.
Forging is frequently classified according to temperature: cold forging, warm forging, or hot forging.
The goal of forging is to make metal parts. Metal forging produces some of the most durable manufactured parts available when compared to other manufacturing methods.
Minor cracks and empty spaces in the metal are filled as the metal is heated and pressed.
Cold forging is the process of deforming a metal material at room temperature using extremely high pressure. The slug is placed in a die and compressed by a press until it fits into the desired shape.
Thus, the answer is cold forging.
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3. 1 4 1 5 9
this is pi
Answer:
FOLLOWED BY... 2 6 5 3 5 8 9
UwU
determine the values of the viscous damping coefficient c for which the system has a damping ratio of 0.5 and 1.5
Answer:
7.47 lb. s/ft
22.42 lb. s/ft
Explanation:
Without mincing words let us dive straight into the solution to the above problem.
STEP ONE: calculate or determine the frequency.
The frequency can be calculated by making use of the formula given below;
frequency, w = √k/m. Thus, frequency = √ 12 × 15/ [40/32.2] = 12 rad/s.
STEP TWO: Determine or calculate for the viscous damping coefficient, c for damping ratio of 0.5 and 1.5 respectively.
The viscous damping coefficient, c for 0.5 = [ 12 × 0.5 × 2 ×[ 40/32.2] / 2= 7.47 lb. s/ft.
The viscous damping coefficient, c for 1.5= [ 12 × 1.5 × 2 ×[ 40/32.2] / 2 = 22.42 lb. s/ft.
A structural component is fabricated from an alloy that has a plane strain fracture toughness of 45 MPa.m1/2. It has been determined that this component fails at a stress of 300 MPa when the maximum length of a surface crack is 0.95 mm. What is the maximum allowable surface crack length (in mm) without fracture for this same component exposed to a stress of 300 MPa and made from another alloy with a plane-strain fracture toughness of 57.5 MPa.m1/2
Answer:
1.5510 mm
Explanation:
Plane strain fracture toughness = 45 MPa√m
failing stress ( б ) = 300 MPa
maximum length of surface crack ( a )= 0.95 mm
Determine maximum allowable surface crack length ( in mm )
we will make use of this relationship for Design stress equation to determine the value of Y
б = [tex]\frac{k}{y\sqrt{\pi *a} }[/tex] --------- ( 1 )
k = 45 MPa√m
б = 300 MPa
a = 0.95 mm
y = ?
From equation 1 make Y subject of the equation ( also substitute values into equation 1 above )
hence ; y = 2.7457
Now determine maximum allowable surface crack when component is exposed to a stress of 300 MPa and made from another alloy with plane-strain fracture toughness of 57.5 MPa√m
we will apply the equation
б = [tex]\frac{k}{y\sqrt{\pi *a} }[/tex] --------- ( 2 )
K = 57.5 MPa√m
б = 300 MPa
y = 2.7457
a ( maximum allowable surface crack ) = ?
from equation make a subject of the equation
a = [tex]\frac{1}{\pi } (\frac{k}{\alpha y} )^{2}[/tex]
a = [tex]\frac{1}{\pi } (\frac{57.5}{300*2.7457} )^2[/tex] = 1.5510 mm
These are sites that allow you to upload and download media content such as images, audio, and video
Which tool is used for cutting bricks and other masonry materials with precision?
Answer:
A turbo blade has the best features from both other types of blade. The continuous, serrated edge makes for fast cutting while remaining smooth and clean. They are mostly used to cut a variety of materials, such as tile, stone, marble, granite, masonry, and many other building materials.
Explanation:
Answer:
Masonry Saw
Explanation:
Masonry Saws can be used to cut brick, ceramic, tile and or stone.
Consider a condenser in which steam at a specified temperature is condensed by rejecting heat to the cooling water. If the heat transfer rate in the condenser and the temperature rise of the cooling water is known, explain how the rate of condensation of the steam and the mass flow rate of the cooling water can be determined. Also, explain how the total thermal resistance R of this condenser can be evaluated in this case.
Answer:
Q = [ mCp ( ΔT) ] [tex]_{cooling water }[/tex]
(ΔT)[tex]_{cooling water}[/tex] and Q is given
[tex]m_{cooling water}[/tex] = [tex]\frac{Q}{Cp[ T_{out} - T_{in} ] }[/tex]
next the rate of condensation of the steam
Q = [ m[tex]h_{fg}[/tex] ][tex]_{steam}[/tex]
[tex]m_{steam} = \frac{Q}{h_{fg} }[/tex]
Total resistance of the condenser is
R = [tex]\frac{Q}{change in T_{cooling water } }[/tex]
Explanation:
How will the rate of condensation of the steam and the mass flow rate of the cooling water can be determined
Q = [ mCp ( ΔT) ] [tex]_{cooling water }[/tex]
(ΔT)[tex]_{cooling water}[/tex] and Q is given
[tex]m_{cooling water}[/tex] = [tex]\frac{Q}{Cp[ T_{out} - T_{in} ] }[/tex]
next the rate of condensation of the steam
Q = [ m[tex]h_{fg}[/tex] ][tex]_{steam}[/tex]
[tex]m_{steam} = \frac{Q}{h_{fg} }[/tex]
Total resistance of the condenser is
R = [tex]\frac{Q}{change in T_{cooling water } }[/tex]
A road is constructed at the capital cost of $6 million. At the end of Year 10, major improvements are to be made costing $17 million. At the end of Year 25, a replacement and upgrade is to be done at a cost of $29 million. At the end of year 40, the federal government issues a one-time tax credit in the amount of $12 million.
Over a 50-year analysis period (assuming a 10% interest rate) what is the annualized cost to the nearest dollar?
Answer:
The annualized cost is:
$299,272.
Explanation:
a) Data and Calculations:
Year 0 Capital cost of road construction = $6 million
Year 10 Major improvements cost = $17 million
Year 25 Replacement and upgrade cost = $29 million
Year 40 Federal government one-time tax credit = $12 million
Period of project analysis = 50 years
Cost of capital (discount rate) = 10%
Annualized cost at present value costs:
Amount spent Discount Factor Present value
$6 million 1 $6,000,000
$17 million 0.386 6,562,000
$29 million 0.092 2,668,000
($12 million) 0.0222 (266,400)
Total cost $14,963,600
Annualized cost = $14,963,600/50 = $299,272
b) The annualized cost for the road construction project, which is the annualized value of the net present costs of $14,963,600, is divided by 50. Before obtaining the net present costs, the cash outflows, including the tax credit, are discounted to their present values, using the discount rate of 10%. And then, the average of the cost is obtained by dividing the total net present cost into 50 years.
2. What is the main job of a cylinder head?
OA. Contain the rapid increase in combustion chamber temperature
OB. Contain the rapid increase in combustion chamber pressure
OC. Prevent engine oil from getting past the pistons
OD. Hold the Head Gasket in place
Grade/Exit
Answer:
Explanation:
The cylinder head sits on the engine and closes off the combustion chamber. The gap that remains between the cylinder head and the engine is completed by the head gasket. Another task of the cylinder head is to ensure the constant lubrication of the cylinder
advantage of iron meter
Answer:
Less Friction Error
Explanation:
The friction error is very less in the moving
iron instrument because their torque weight
ratio is high.
hope it helps!
Answer:
Universal Use -The MI instrument is independent of the of current and hence for both AC and DC
Explanation:
Mark as Brainlist Answer
the importance of reading a circuit diagram to interpret a wiring diagram?
Answer:
The ability to read electrical schematics is a really useful skill to have. To start developing your schematic reading abilities, it's important to memorize the most common schematic symbols. ... You should also be able to get a rough idea of how the circuit works, just by looking at the schematic.
Explanation:
Two children are playing on a seesaw. The child on the left weighs 50 lbs. And the child on the right weighs 100 lbs. If the child on the left is 3 feet from the pivot, how far must the child on the right be from the pivot fro the seesaw to be balanced.
Given :
Two children are playing on a seesaw. The child on the left weighs 50 lbs. And the child on the right weighs 100 lbs.
To Find :
If the child on the left is 3 feet from the pivot, how far must the child on the right be from the pivot from the seesaw to be balanced.
Solution :
We know, for seesaw to balance :
[tex]m_1gd_1=m_2gd_2[/tex]
Here, [tex]d_1\ and \ d_2[/tex] is distance from origin from pivot point.
Putting all the values in the equation, we get :
[tex]50\times g\times 3=100\times g\times d_2\\\\d_2=1.5\ feet[/tex]
Therefore, distance of right child from pivot to balance the seesaw is 1.5 feet.
Hence, this is the required solution.
Think about all the things that you have learned about wildfire in this module.
In light of climate change, explain how this risk map might look different in 100 years.
Answer:
CLIMATE CHANGE HAS inexorably stacked the deck in favor of bigger and more intense fires across the American West over the past few decades, science has incontrovertibly shown. Increasing heat, changing rain and snow patterns, shifts in plant communities, and other climate-related changes have vastly increased the likelihood that fires will start more often and burn more intensely and widely than they have in the past.
Explanation: