An example of LEED Green Associate will be D. The director of the museum has a strong general knowledge of the major periods and styles of art represented in the museum.
The LEED Green Associate simply affirms a professional's comprehension of the green building practices and principles.
It should be noted that the LEED Green Associate shows the general knowledge of the green building practices.
In conclusion, the correct option is the director of the museum has a strong general knowledge of the major periods and styles of art represented in the museum.
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how many types of lavatory there is?
Answer:
there are generally two types of toilet bowl types- round and elongated.
Are spent lead acid batteries exempt from hazardous waste regulations if they are recycled?
Answer: Spent lead acid batteries (SLABs) are considered hazardous waste under the Resource Conservation and Recovery Act (RCRA) in the United States. However, they are exempt from the regulations that apply to other hazardous waste if they are recycled properly.
According to the Environmental Protection Agency (EPA), when spent lead acid batteries are recycled, they are considered a "universal waste" under RCRA. Universal wastes are certain types of hazardous waste that are considered less hazardous than other hazardous wastes and are subject to less stringent regulations. The regulations for universal waste are intended to encourage the collection and proper recycling or disposal of these materials.
To be considered for this exemption, the spent lead-acid batteries must be managed in accordance with the regulations for universal waste, which include proper labeling, storage, and transportation, as well as the requirement that the batteries be sent to a facility that is authorized to manage universal waste.
It is important to note that individual states may have different regulations regarding management and disposal of spent lead-acid batteries, so it is best to check with your state's environmental agency for specific information.
if the tire’s radius is 23 cm , what is d , the magnitude of the spot’s displacement after 2.0 seconds?
In a rolling tire, the magnitude of the spot's displacement is defined as the distance traveled by a point on the circumference of the tire with respect to a stationary observer. The magnitude of the spot's displacement is 2.88 m after 2.0 seconds.
The tire's radius is given as 23 cm and the magnitude of the spot's displacement is to be calculated after 2.0 seconds. Let's first calculate the total distance covered by the spot of the tire's circumference using the formula of distance = speed × time.
The formula for the distance covered by a point on a rotating wheel is given as:
d = 2πrN, Where r = the radius of the tire and N = the number of revolutions made by the tire during the given time period. Turning 23 cm to meters, we get r = 0.23 m.
The speed of the tire is the tangential speed of the wheel, which can be expressed as follows: v = rω, Where v = the linear speed, r = the radius of the wheel, and ω = the angular velocity.
We can determine the number of rotations (N) made by the tire using the following formula:
N = ωt / 2π.
Now, to determine the distance covered by the point on the wheel, we substitute the values in the given formula:
d = 2πrN= 2πr(ωt/2π)= rωt= r(2π/t)t= 2s.
Substituting the values, we get:
d = 0.23 m × ω × (2s) = 0.23 m × (2π/t) × 2 s= 2.88 m.
Thus, the magnitude of the spot's displacement is 2.88 m after 2.0 seconds.
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With respect to expanding opportunities for ICT environmental contributions, what term is associated with the issues such as LEED Certified Buildings, Smart Motors, and Industrial Robots?
Nowadays, you can link practically all of your home's appliances to a single device and operate it from anywhere. A better use of time is made possible through productivity apps.
What opportunities for ICT environmental contributions?The most pressing climatic concerns facing the globe can be addressed with the use of ICTs, which can also help with the transition to a circular economy that is much needed.
Access to biometric gadgets, food management software, and exercise routines easily leads to improved health. Communication with friends and family is simpler and less expensive.
Therefore, Additionally, they can be employed to keep track of, lesson, and adjust to climate change.
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A circular secondary clarifier (a.k.a. a final clarifier) is to be designed for an activated sludge treatment plant serving a municipality. The state regulatory agency criteria for final clarifiers used for activated sludge are: Peak SOR = 1,200 gal/day-ft2 Average SOR = 500 gal/day-ft2 Peak solids loading 50 lb/day-ft Peak weir loading-30,000 gal/day.ft - The average flow to the aeration basin prior to junction with the recycle line is 3.5 MGD - The recycled sludge flow is 30 percent of the average influent flow rate and is constant throughout the day. 0.3 The mixed liquor suspended solids (MLSS) 2,500 mg/L. - The ratio of the peak hourly influent flow to the average hourly flow is 2.0 Qpsayo..g = 2.0 Determine a. The clarifier diameterb. The peak weir loading if peripheral effluent weirs are used.
a. The clarifier diameter is approximately 174 feet.
b. The peak weir loading is approximately 0.009 lb/day-ft.
To determine the clarifier diameter, we can use the formula:
D = sqrt[(Qp / (2.26 * SOR * (1 + R))] where Qp is the peak influent flow rate, SOR is the peak surface overflow rate, and R is the recycle ratio. Plugging in the given values, we get:
D = sqrt[(2.0 * 3.5 MGD / (2.26 * 1200 gal/day-ft2 * (1 + 0.3)))]
D ≈ 174 ft
To determine the peak weir loading, we can use the formula:
W = Qp / (1 + R) where W is the peak weir loading. Plugging in the given values, we get:
W = 2.0 * 3.5 MGD / (1 + 0.3)
W ≈ 105,263 gal/day-ft
To convert the weir loading to solids loading, we can use the formula:
SL = W * MLSS * 8.34 lb/gal where SL is the solids loading and MLSS is the mixed liquor suspended solids concentration. Plugging in the given values, we get:
SL = 105,263 gal/day-ft * 2,500 mg/L * 8.34 lb/gal / (1,000,000 mg/L)
SL ≈ 0.009 lb/day-ft
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In the planning process of the product development life cycle what is it important to inventory
A. Goals, mission, costs, and strengths
B. Goals, mission, capabilities, and constraints
C. Goals, mission, capabilities, cost, and strengths
D. Goals, mission, capabilities, cost and constraints
Your Answer would be A I believe.
two delta-connected loads are connected in parallel and powered by a balanced yconnected system. the smaller of the two loads draws 10 kva at a lagging pf of 0.75, and the larger draws 25 kva at a leading pf of 0.8. the line voltage is 400 v. calculate (a) the power factor at which the source is operating; (b) the total power drawn by the two loads; (c) the phase current of each load.
The power factor at which the source is operating is approximately 0.79. The total power drawn by the two loads is 27.5 kW. The phase current of Load 1 is approximately 25 A, and the phase current of Load 2 is approximately 62.5 A.
(a) Calculating the power factor at which the source is operating:
To find the power factor at which the source is operating, we need to calculate the apparent power (S) drawn by the loads.
Apparent power is given by the formula:
S = V * I, where V is the line voltage and I is the current.
The apparent power for each load can be calculated using the formula:
S = P / pf, where P is the real power.
For Load 1:
P1 = 10 kVA * 0.75 = 7.5 kW (since power factor is given in lagging, we use the same power value)
S1 = P1 / pf1 = 7.5 kW / 0.75 = 10 kVA
For Load 2:
P2 = 25 kVA * 0.8 = 20 kW (since power factor is given in leading, we use the same power value)
S2 = P2 / pf2 = 20 kW / 0.8 = 25 kVA
The total apparent power drawn by the loads is:
Stotal = S1 + S2 = 10 kVA + 25 kVA = 35 kVA
The power factor (pf) of the source can be calculated as:
pf = Ptotal / Stotal, where Ptotal is the total real power drawn.
Since Ptotal = P1 + P2 = 7.5 kW + 20 kW = 27.5 kW,
pf = 27.5 kW / 35 kVA ≈ 0.7857 (or approximately 0.79)
Thus, the answer is approximately 0.79.
(b) Calculating the total power drawn by the two loads:
The total real power drawn by the loads (Ptotal) is the sum of the individual real powers:
Ptotal = P1 + P2 = 7.5 kW + 20 kW = 27.5 kW
Thus, the answer is 27.5 kW.
(c) Calculating the phase current of each load:
To calculate the phase current (I) of each load, we can use the formula:
I = S / V, where S is the apparent power and V is the line voltage.
For Load 1:
I1 = S1 / V = 10 kVA / 400 V ≈ 25 A
For Load 2:
I2 = S2 / V = 25 kVA / 400 V ≈ 62.5 A
The phase current of Load 1 is approximately 25 A, and the phase current of Load 2 is approximately 62.5 A.
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ký hiệu của vật liệu cơ khí
Answer:
by hìuuf5ëcz
Explanation:
6tgiïuggd
The flat belts are moving at the speeds shown. Plot both the velocity profile within the oil film and the shearstress distribution. The pressures at A and B are atmospheric. Take mo
Shear stress values and the velocity profile are v=0.4m/s,
t=45 N/m^ 2
Explanation:
A) Velocity profile=0.6m/s, and 0.2m/s
relative velocity=0.6m/s-0.2m/s
v=0.4m/s
∵ thickness=4 mm
t=4x10^ -3m
B) Shear stress t=R v/t
=0.45x0.4/4x10^-3
t=45 N/m^ 2
Shear velocity, also known as friction velocity, is a way to express a shear stress in velocity units. Comparing actual speeds, like the flow in a stream, to ones that reflect shear between layers of flow is a useful technique in fluid mechanics.
Due to its close connection to earthquakes and the downslope movement of earth materials, the resulting shear has a significant impact on nature. Either solids or liquids can experience shear stress.
Using equation, one can determine the shear velocity.
U*= √r/p in the general case.
INCOMPLETE QUESTION:
The flat belts are moving at the speeds shown. Plot both the velocity profile within the oil film and the shear stress distribution. The pressures at A and B are atmospheric. Take u^0=0.45N.s/m2 and P^0=920KG/M3
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.
. consider the following statements: a.) liquid moves faster in a turbulent flow than in a laminar flow; b.) turbulent flow produces more noise than laminar flow; c.) laminar flow does not change in time (it is a quasi-steady state), while turbulent flow is unsteady; d.) streamlines of a laminar flow stick together, but diverge in a turbulent case.
Real-world fluid flows can exhibit a combination of laminar and turbulent characteristics, depending on factors such as flow velocity, fluid properties, and the presence of obstacles or disturbances.
a) Liquid moves faster in a turbulent flow than in a laminar flow.
This statement is generally true. In a turbulent flow, the fluid moves in a chaotic manner, characterized by eddies and swirling motion. This turbulent motion enhances the mixing and transport of fluid particles, which can result in higher overall flow velocities compared to laminar flow, where fluid moves smoothly in parallel layers.
b) Turbulent flow produces more noise than laminar flow.
This statement is generally true. Turbulent flow is associated with fluctuating velocities and pressure fluctuations, which can generate noise. The irregular motion and collisions of fluid particles in turbulent flow create vibrations and sound waves, resulting in higher noise levels compared to the smooth, ordered flow of laminar flow.
c) Laminar flow does not change in time (it is a quasi-steady state), while turbulent flow is unsteady.
This statement is generally true. Laminar flow is characterized by smooth and ordered flow with parallel streamlines. It remains relatively constant over time, assuming steady conditions. On the other hand, turbulent flow is characterized by irregular and unsteady motion with constantly changing velocities and pressure fluctuations. Turbulent flow exhibits fluctuations in time and space due to the presence of eddies and vortices.
d) Streamlines of a laminar flow stick together, but diverge in a turbulent case.
This statement is generally true. In laminar flow function, adjacent fluid particles move smoothly and in parallel along the streamlines without significant mixing or cross-stream diffusion. As a result, the streamlines tend to remain close together and maintain their initial separation. In turbulent flow, however, the chaotic motion and mixing of fluid particles cause streamlines to diverge and become highly convoluted, leading to intermingling and spreading of fluid particles. Real-world fluid flows can exhibit a combination of laminar and turbulent characteristics, depending on factors such as flow velocity, fluid properties, and the presence of obstacles or disturbances.
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can someone tell me what is the most happy song and the most sad: alguien me puede decir cual es la cancion mas feliz y la triste.
It is subjective to determine the "most happy" and "most sad" song as it varies from person to person. Different songs can evoke different emotions in different people.
However, some songs that are commonly considered to be happy include "Happy" by Pharrell Williams, "Don't Stop Believin'" by Journey, and "I Will Always Love You" by Whitney Houston. Some songs that are commonly considered to be sad include "Everybody Hurts" by R.E.M., "Tears in Heaven" by Eric Clapton, and "My Heart Will Go On" by Celine Dion.
What makes a song happy?The emotions that a song evokes in a listener are subjective, meaning that they can vary from person to person. For example, a song that one person finds to be happy and uplifting, another person may find to be sad or melancholy. This is because emotions are personal experiences that are shaped by a person's individual life experiences, beliefs, and attitudes.
Therefore, Regarding the songs mentioned, "Happy" by Pharrell Williams, "Don't Stop Believin'" by Journey, and "I Will Always Love You" by Whitney Houston are considered happy songs because they have upbeat tempos, positive lyrics, and a cheerful sound that can make people feel good.
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11. Two technicians are discussing tire rotation. Technician A says that you always follow the tire-rotation procedure outlined in the owner's manual or online service information. Technician B says that the modified "X" rotation pattern is seldom used. Which technician is correct? A. Both Technicians A and B B. Technician B only C. Neither Technician A nor B D. Technician A only
Answer:
Technician A
Explanation:
Technician A is correct. Technician B is wrong, as different drive vehicles (FWD, RWD, AWD, 4WD) each require a certain rotation for optimal tire wear.
The Specific weight of an unknown liquid is 12400N/m³. What mass of the liquid is contained in a volume of 500 cm³ ? use; (a) The Standard value of gravity. (6) The minimum value of gravity on the earth (c) The maximum value of gravity on the earth
The mass of the liquid is 61.029 N. The mass of the liquid is 60.636 N in the second scenario.
What is density?The density of a solid, liquid, or gas describes how closely packed the particles are. Density is defined as the amount of mass per unit volume.
The mass of a liquid can be calculated using its volume and density. The formula is:
mass = density x volume
We are given the density of the unknown liquid as 12400 N/m³ and the volume as 500 cm³.
We need to convert the volume from cm³ to m³ before we can use the formula:
500 cm³ = 0.0005 m³
(a) Using the standard value of gravity (9.81 m/s²):
mass = density x volume x gravity
= 12400 N/m³ x 0.0005 m³ x 9.81 m/s²
= 61.029 N
Therefore, the mass of the liquid is 61.029 N.
(b) Using the minimum value of gravity on Earth (9.78 m/s²):
mass = density x volume x gravity
= 12400 N/m³ x 0.0005 m³ x 9.78 m/s²
= 60.636 N
Therefore, the mass of the liquid is 60.636 N.
(c) Using the maximum value of gravity on Earth (9.83 m/s²):
mass = density x volume x gravity
= 12400 N/m³ x 0.0005 m³ x 9.83 m/s²
= 61.422 N
Thus, the mass of the liquid is 61.422 N.
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Consider a vertical piston/cylinder system. The piston has a mass of 35 kg and has an unknown radius. There is a pressure gauge to output the pressure inside the cylinder. If the piston is compressed and the pressure gauge reads 300 kPa. What is the area of the piston
Answer:
Area of the piston = 1.143 * 10^-3 m^2
Explanation:
Here, we are tasked with calculating the area of the piston.
Mathematically, the area of the piston =
Piston force/ cylinder pressure
From the question,
Piston force = mg where m is the mass of the piston = 35 kg and g is the acceleration due to gravity = 9.8 m/s^2
Thus, piston force = 35 * 9.8 = 343 N
Now the pressure gauge reads 300 KPa and 1KPa = 300 * 10^3 Pa = 300000 Pa
Thus, the area of the piston = 343/300000 =
0.001143333333 m^2 which is simply 1.143 * 10^-3 m^2
The UHRS platform is optimized for Edge/Internet Explorer only. You can still use your favorite browser, but keep in mind that you may experience technical issues when working on UHRS with a different browser than Edge or Internet Explorer.
UHRS is optimized for...
It is to be noted that all UHRS platforms are optimized for the popular kinds of internet browser applications.
What is a UHRS?The Universal Human Relevance System (UHRS) is a crowdsourcing platform that allows for data labeling for a variety of AI application situations.
Vendor partners link people referred to as "judges" to offer data labeling at scale for us. All UHRS judges are bound by an NDA, ensuring that data is kept protected.
A browser is a software tool that allows you to see and interact with all of the knowledgeon the World Wide Web. Web sites, movies, and photos are all examples of this.
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For the network of Fig. below, draw the output waveform if Vr = 4.3V , R = 1000Omega and Vi=25 sin ot. Assume an Si diode is used. + R D V_{i}; V_{r}; V_{o}
Compressed Air In a piston-cylinder device, 10 gr of air is compressed isentropically. The air is initially at 27 °C and 110 kPa. After being compressed, the air is at 450 °C. Determine
(a) the final pressure in [MPa],
(b) the increase in total internal energy in [kJ], and
(c) the total work required in [kJ].
Note that for air R-287 J/kg.K and c.-716.5 J/kg.K, and ?-
Answer:
(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ
Explanation:
Solution
Recall that:
A 10 gr of air is compressed isentropically
The initial air is at = 27 °C, 110 kPa
After compression air is at = a450 °C
For air, R=287 J/kg.K
cv = 716.5 J/kg.K
y = 1.4
Now,
(a) W efind the pressure on [MPa]
Thus,
T₂/T₁ = (p₂/p₁)^r-1/r
=(450 + 273)/27 + 273) =
=(p₂/110) ^0.4/1.4
p₂ becomes 2390.3 kPa
So, p₂ = 2.39 MPa
(b) For the increase in total internal energy, is given below:
ΔU = mCv (T₂ - T₁)
=(10/100) (716.5) (450 -27)
ΔU =3030 J
ΔU =3.03 kJ
(c) The next step is to find the total work needed in kJ
ΔW = mR ( (T₂ - T₁) / k- 1
(10/100) (287) (450 -27)/1.4 -1
ΔW = 3035 J
Hence, the total work required is = 3.035 kJ
During one eight-hour shift, 750 non-defective parts are desired from a fabrication operation. The standard time for the operation is 15 minutes. Because the machine operators are unskilled, the actual time it takes to perform the operation is 20 minutes, and ,on average, one-fifth of the parts that begin fabrication are scrapped. Assuming that each of the machines used for this operation will not be available for one hour of each shift, determine the number of machines required.
Answer:
No. of Machines Required = 47
Explanation:
First, we calculate the number of parts that can be manufactured by one machine during the shift. For that purpose, we use following formula:
No. of Parts Manufactured by 1 Machine = Total Operating Time/Time taken to perform Operation
where,
Total Operating Time = 8 h - 1 h = (7 h)(60 min/h) = 420 min
Time taken to perform operation = 20 min
Therefore,
No. of Parts Manufactured by 1 Machine = 420 min/20 min
No. of Parts Manufactured by 1 Machine = 21
Now, we will calculate the no. of non-defective parts manufactured by 1 machine. Since, it is given that one-fifth of the parts manufactured by machine are defective. Therefore, the non-defective parts will be: 1 - 1/5 = 4/5 (four-fifth).
No. of Non- Defective Parts Manufactured by 1 Machine = N = (4/5)(21)
N = 16.8 = 16 (Since, the 17th part will not be able to complete in time)
So, the no. of machines required to produce 750 non-defective parts is given by:
No. of Machines Required = No. of non-defective parts required/N
No. of Machines Required = 750/16
No. of Machines Required = 46.9
No. of Machines Required = 47 (Since, 46 machines will not be able complete the job)
A process fluid having a specific heat of 3500 J/kg.K and flowing at 2 kg/s is to be cooled from 80 C to 50 C with chilled water, which is supplied at a temperature of 15 C and a flow rate of 2.5 kg/s. Assuming an overall heat transfer coefficient of 2000 W/m2.K, calculate the required heat transfer area for the following exchanger conditions: (a) parallel flow, (b) counterflow, (c) shell and tube, one shell pass and two tube passes, and (d) cross-flow, single pass, both fluids unmixed. Compare the results of your analysis.
The required heat transfer area for parallel flow is approximately 4.4 square meters. The required heat transfer area for counterflow is approximately 2.9 square meters.
The required heat transfer area for shell and tube, one shell pass and two tube passes, is approximately 2.9 square meters. The required heat transfer area for cross-flow, single pass, both fluids unmixed, is approximately 8.1 square meters. The analysis shows that for this particular case, the cross-flow configuration requires the largest heat transfer area, followed by parallel flow. Counterflow and shell and tube configurations have similar heat transfer areas, which are lower than that of cross-flow. The choice of the heat exchanger configuration depends on various factors such as space availability, cost, and efficiency.
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Bed load is moved by Group of answer choices traction. saltation. suspension. all of the above traction and saltation
Bedload refers to the gravel, pebbles, stones, and boulders that are too large to be picked up and carried in suspension by the flow of water or air.
The bed load of a stream or river is composed of the particles that travel along the bed, rolling, sliding, and bouncing along the bottom of the channel.
In this context, bed load is moved by all of the above: traction, saltation, and suspension.
Traction is the process by which large and heavy particles are rolled along the stream or riverbed by the force of the water or wind.
Saltation occurs when smaller particles, like sand and pebbles, are bounced along the bed by the turbulence of the flow.
Suspension is the process by which even smaller particles, like silt and clay, are held up and transported by the flow of water or air.
When the bedload is moved by a group of answer choices that is traction and saltation.
This happens because in a stream or river, the larger particles are usually too heavy to be lifted by the water,
so they move along the bottom by rolling, sliding, or bouncing.
Meanwhile, the smaller particles are lifted up and transported by the turbulence of the flow, bouncing along the bed as they go.
This combination of processes results in a mix of traction and saltation as the bedload moves along the stream or riverbed.
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An air-standard Diesel cycle engine operates as follows: The temperatures at the beginning and end of the compression stroke are 30 °C and 700 °C, respectively. The net work per cycle is 590.1 kJ/kg, and the heat transfer input per cycle is 925 kJ/kg. Determine the a) compression ratio, b) maximum temperature of the cycle, and c) the cutoff ratio, v3/v2.
This question is incomplete, the complete question is;
An air-standard Diesel cycle engine operates as follows: The temperatures at the beginning and end of the compression stroke are 30 °C and 700 °C, respectively. The net work per cycle is 590.1 kJ/kg, and the heat transfer input per cycle is 925 kJ/kg. Determine the a) compression ratio, b) maximum temperature of the cycle, and c) the cutoff ratio, v3/v2.
Use the cold air standard assumptions.
Answer:
a) The compression ratio is 18.48
b) The maximum temperature of the cycle is 1893.4 K
c) The cutoff ratio, v₃/v₂ is 1.946
Explanation:
Given the data in the question;
Temperature at the start of a compression T₁ = 30°C = (30 + 273) = 303 K
Temperature at the end of a compression T₂ = 700°C = (700 + 273) = 973 K
Net work per cycle \(W_{net\) = 590.1 kJ/kg
Heat transfer input per cycle Qs = 925 kJ/kg
a) compression ratio;
As illustrated in the diagram below, 1 - 2 is adiabatic compression;
so,
Tγ\(^{Y-1\) = constant { For Air, γ = 1.4 }
hence;
⇒ V₁ / V₂ = \((\) T₂ / T₁ \()^{\frac{1}{Y-1}\)
so we substitute
⇒ V₁ / V₂ = \((\) 973 K / 303 K \()^{\frac{1}{1.4-1}\)
= \((\) 3.21122 \()^{\frac{1}{0.4}\)
= 18.4788 ≈ 18.48
Therefore, The compression ratio is 18.48
b) maximum temperature of the cycle
We know that for Air, Cp = 1.005 kJ/kgK
Now,
Heat transfer input per cycle Qs = Cp( T₃ - T₂ )
we substitute
925 = 1.005( T₃ - 700 )
( T₃ - 700 ) = 925 / 1.005
( T₃ - 700 ) = 920.398
T₃ = 920.398 + 700
T₃ = 1620.398 °C
T₃ = ( 1620.398 + 273 ) K
T₃ = 1893.396 K ≈ 1893.4 K
Therefore, The maximum temperature of the cycle is 1893.4 K
c) the cutoff ratio, v₃/v₂;
Since pressure is constant, V ∝ T
So,
cutoff ratio S = v₃ / v₂ = T₃ / T₂
we substitute
cutoff ratio S = 1893.396 K / 973 K
cutoff ratio S = 1.9459 ≈ 1.946
Therefore, the cutoff ratio, v₃/v₂ is 1.946
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:
\(I=3A\)
Explanation:
From the question we are told that:
Number of lamps \(N=4\)
Potential difference \(V=60v\)
Total Resistance of the lamp is \(R= 20ohms\)
Generally the equation for Current I is mathematically given by
\(I=\frac{V}{R}\)
\(I=\frac{60}{20}\)
\(I=3A\)
(a) Find the equivalent capacitance of three capacitors C1 = 2 {eq}\mu {/eq}F, C2 = 3 {eq}\mu {/eq}F, and C3 = 4 {eq}\mu {/eq}F, when they are in series.
(b) Find the charge on each capacitor and the potential difference across each when the combination is connected across a 12V battery.
(a) To find the equivalent capacitance (Ceq) of three capacitors (C1, C2, and C3) in series, use the formula:
1/Ceq = 1/C1 + 1/C2 + 1/C3
Given, C1 = 2μF, C2 = 3μF, and C3 = 4μF. Plug these values into the formula:
1/Ceq = 1/2 + 1/3 + 1/4
1/Ceq = 6/12 + 4/12 + 3/12
1/Ceq = 13/12
Now, take the reciprocal to find Ceq:
Ceq = 12/13 μF
(b) When capacitors are in series, they all carry the same charge (Q). To find the charge, use the formula:
Q = Ceq * V
Given, V = 12V, and Ceq = 12/13 μF:
Q = (12/13) * 12
Q = 144/13 μC
Now, to find the potential difference (V) across each capacitor, use the formula:
V = Q/C
For C1 (2μF):
V1 = (144/13) / 2 = 72/13 V
For C2 (3μF):
V2 = (144/13) / 3 = 48/13 V
For C3 (4μF):
V3 = (144/13) / 4 = 36/13 V
So, the equivalent capacitance is 12/13 μF, and the charge on each capacitor is 144/13 μC. The potential differences across C1, C2, and C3 are 72/13 V, 48/13 V, and 36/13 V, respectively.
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what is a steel alloy
Answer:
Steel alloy is a alloy which inside has alot of elements and are broken down to 2 groups because of there weight; low alloy steels and high alloy steels, which they are both steel alloys. Alot of these have uses in turbine blades of jet engines, and in nuclear reactors.
Explanation:
Why do i pick the most exspensive hobbies
Airsoft and building pcs
Answer: You make more money
Explanation: A pc is like 2k+ to build so imagine all of the profits coming in from your business/work.
What kind of diffraction pattern hould you expect if a SAD aperture include many grain in a polycrytalline pecimen
A TEM technique called selected area diffraction (SAD) is used to collect diffraction patterns from electron beams that have been scattered by sample lattices.
Sad aperture: What is it?In the chosen area diffraction (SAD) technique, an aperture (referred to as the selected area aperture) is inserted into the picture plane of the objective lens to select an area so that we can restrict the region that forms the diffraction pattern.
Why do you use the term SAD in TEM?The crystallographic experimental method known as selected area (electron) diffraction (abbreviated as SAD or SAED) is often carried out using a transmission electron microscope (TEM).
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select three of the prerequisites for becoming a civil engineer
7–25. Determine the maximum shear stress acting at section a-a of the cantilevered strut. 2 kN 4 kN 1250 mm- 250 mm 300 mm 20 mm 70 mm LB 20 mm + 50 mm
The max. shear stress acting based on the given question is given as \(4.85N/mm^2\)
What is Maximum Shear Stress?Maximum Shear Stress is a concept used in solid mechanics for describing the highest stress that a material can resist before it experiences failure. Such force per unit area works on a plane perpendicular to the axis of the object in question.
The most noteworthy shear stress is found at a plane where the stressing force reaches its maximum potential. In certain three-dimensional stress systems, this plane typically maintains an angle of 45 degrees with respect to the principle's stress axis.
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The 10-kg lamp is suspended from the
three equal-length cords. Determine its
smallest vertical distance s from the
ceiling if the force developed in any
cord is not allowed to exceed 50 N.
120°
A
600 mm
120°
Answer:
hello your question is incomplete attached below is the complete diagram
answer : 519 mm
Explanation:
Given the conditions it can be said that the cords are symmetric i.e. AD, BD , CD has same value
hence ; DA = DB = DC = 600 mm
Tension in all cords are the same ( T ) = 50 N
lets assume the angle between the cords and the Z-axis is ∅
∑ Fz = 0
3 * T cos ∅ - 10*9.81 = 0
therefore ; ∅ = 49.16°
looking at attached triangle AD0
tan ∅ = \(\frac{600}{s}\)
Hence ; s = 519 mm
when using the route command at a command prompt, which entry indicates the default route?
Regardless of whether a default static route is set up, the "default-information originate" command is used to advertise a default route into OSPF.
Static routing is the practice of using a manually specified routing entry rather than information from traffic generated by dynamic routing. Although it isn't always the case, network administrators frequently manually configure static routes in addition to adding entries to a routing table. In contrast to dynamic routing, static routes have been fixed and remain the same despite network configuration changes. Static and dynamic routing are not mutually exclusive. A router often combines both dynamic routing and static routing to maximize routing efficiency and provide fall backs in the case that dynamic routing information cannot be transmitted. As a last-resort gateway or in stub networks, static routing may also be used.
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