What treat got its name from the machine that manufactures them.
Answers
One treat that got its name from the machine that manufactures them is called "PEZ." The name of this small candy that comes in a variety of fruity flavors is derived from the German word "pfefferminz," meaning peppermint.
Pez candy was first introduced in Vienna, Austria in 1927 by Eduard Haas III as an alternative to smoking.The original Pez candy dispensers were sold in metal tins and had no character heads on them. The company released its first dispenser in 1952, which featured a space gun design. However, it wasn't until the introduction of the character heads in the early 1960s that Pez became a popular collector's item.
There are now hundreds of different Pez dispensers, featuring characters from popular TV shows, movies, and even political figures. Despite the many changes and innovations in the Pez company, the candy itself remains largely unchanged, and continues to be a popular treat for people of all ages.Pez is still produced using the same process it has used for decades, in which sugar, corn syrup, and flavorings are combined and then compressed into the small, brick-shaped pieces. These pieces are then placed into individual plastic cartridges that are loaded into the iconic Pez dispensers.
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Related Questions
A single-phase transformer has 480 turns on the primary and 90 turns on the secondary. The mean length of the flux path in the core is 1.8 m and the joints are equivalent to an airgap of 0.1 mm. The value of the magnetic field strength for 1.1 T in the core is 400 A/m, the corresponding core loss is 1.7 W/kg at 50 Hz and the density of the core is 7800 kg/m3. If the maximum value of the flux density is to be 1.1 T when a p.D. Of 2200 V at 50 Hz is applied to the primary, calculate: a. The cross-sectional area of the core; b. The secondary voltage on no load; c. The primary current and power factor on no load.
Answers
Answer:
a) cross sectional area of the core = 0.0187 m²
b) The secondary voltage on no-load = 413 V
c) The primary currency and power factor on no load is 1.21 A and 0.168 lagging respectively.
Explanation:
See attached solution.
A cold store is designed with walls built from brick walls 102. 5mm thick. The interior is insulated with 60mm of mineral wool and 90 mm of foam insulation which is then faced with 20mm of plywood. Finally the plywood is covered with a 1mm thick stainless steel cladding. The effective surface area of the exterior is 320m2 and 70% of the total heat loss is calculated to be lost through this exterior surface. The exterior wall has a surface heat transfer coefficient of 22 W/m2K and the inner wall surface heat transfer coefficient is 14 W/m2K For brick λ = 0. 12 W/mk For mineral wool insulation λ = 0. 03 W/mK
For foam insulation λ = 0. 042 W/mK
For plywood λ = 0. 2 W/mk For stainless Steel λ = 25 W/mK The cold store is to be maintained at -25°C when the outside ambient temperature can reach 38°C. A) Derive and explain the Fourier equation for heat transfer through a flat surface. B) i) modify the Fourier equation to meet the requirements of the multi-layer surface, ii) hence determine the required cooling load of the refrigeration unit
Answers
The Fourier equation for heat transfer through a flat surface is given by:Q = (k * A * ΔT) / d.Using the given values and calculations for the different layers,
you can substitute the appropriate values into the equations to find the required cooling load of the refrigeration unit. where:Q is the rate of heat transfer,k is the thermal conductivity of the material,A is the surface area, ΔT is the temperature difference across the surface, and d is the thickness of the material. To modify the Fourier equation for the multi-layer surface of the cold store, we need to consider the heat transfer through each layer. We can calculate the overall heat transfer coefficient (U) for the multi-layer wall using the following formula:1/U = (1/h1) + (Δx1/k1) + (Δx2/k2) + ... + (1/hn)
where: h1, h2, ... hn are the surface heat transfer coefficients of each layer,Δx1, Δx2, ... are the thicknesses of each layer, and k1, k2, ... are the thermal conductivities of each layer. To determine the required cooling load of the refrigeration unit, we can use the formula:Q = U * A * ΔT where: Q is the rate of heat transfer,U is the overall heat transfer coefficient of the wall, A is the surface area, and ΔT is the temperature difference between the inside and outside of the cold store.Using the given values and calculations for the different layers, you can substitute the appropriate values into the equations to find the required cooling load of the refrigeration unit.
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A horizontal force P is applied to a 130 kN box resting on a 33 incline. The line of action of P passes through the center of gravity of the box. The box is 5m wide x 5m tall, and the coefficient of static friction between the box and the surface is u=0.15. Determine the smallest magnitude of the force P that will cause the box to slip or tip first. Specify what will happen first, slipping or tipping.
Answers
Answer:
SECTION LEARNING OBJECTIVES
By the end of this section, you will be able to do the following:
Distinguish between static friction and kinetic friction
Solve problems involving inclined planes
Section Key Terms
kinetic friction static friction
Static Friction and Kinetic Friction
Recall from the previous chapter that friction is a force that opposes motion, and is around us all the time. Friction allows us to move, which you have discovered if you have ever tried to walk on ice.
There are different types of friction—kinetic and static. Kinetic friction acts on an object in motion, while static friction acts on an object or system at rest. The maximum static friction is usually greater than the kinetic friction between the objects.
Imagine, for example, trying to slide a heavy crate across a concrete floor. You may push harder and harder on the crate and not move it at all. This means that the static friction responds to what you do—it increases to be equal to and in the opposite direction of your push. But if you finally push hard enough, the crate seems to slip suddenly and starts to move. Once in motion, it is easier to keep it in motion than it was to get it started because the kinetic friction force is less than the static friction force. If you were to add mass to the crate, (for example, by placing a box on top of it) you would need to push even harder to get it started and also to keep it moving. If, on the other hand, you oiled the concrete you would find it easier to get the crate started and keep it going.
Figure 5.33 shows how friction occurs at the interface between two objects. Magnifying these surfaces shows that they are rough on the microscopic level. So when you push to get an object moving (in this case, a crate), you must raise the object until it can skip along with just the tips of the surface hitting, break off the points, or do both. The harder the surfaces are pushed together (such as if another box is placed on the crate), the more force is needed to move them.
Using leftover paint colors is acceptable in a paint shop and will help cut down on waste.
True or false
Answers
Answer:
True
Explanation:
they put it through a process to be able to reuse it
as the angle of the ramp is increased the force parallel increases /decreases / remains the same
Answers
As the angle of the ramp is increased, the force parallel increases. Hence, option (a) can be considered as the correct answer.
When the angle of a ramp is increased, the force parallel to the ramp, also known as the parallel component of the gravitational force, does increase. This is because the component of gravity acting parallel to the ramp increases with the angle. However, it's important to note that the total gravitational force acting on an object remains constant regardless of the angle of the ramp.As the angle of the ramp increases, the force required to push or pull an object up the ramp against gravity increases. This is due to the increase in the vertical component of the gravitational force, which opposes the motion up the ramp. The parallel force required to overcome this increased vertical force also increases.
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In 2002 Acme Chemical purchased a large pump for
$112,000. Acme keys their cost estimating for these
pumps to the industrial pump index, with a baseline of
100 established in 1992. The index in 2002 was 212.
Acme is now (2010) considering construction of a nev
addition and must estimate the cost of the same type
and size of pump. If the industrial pump index is
currently 286, what is the estimated cost of the new
pump?
Answers
Answer:
$151094
Explanation:
Solution
Recall that:
Acme Chemical in 2002 purchased a large pump worth of = 112,000
The estimation for the pump to the industrial pump index is =100
The index in 2002= 212
The current index is = 286
k = is the reference year for which cost or price is known.
n = the year for which cost or price is to be estimated (n>k).
Cn = the estimated cost or price of item in year n.
Ck = the cost or price of item in reference year k.
Cn = Ck * (In / Ik )
Now,
We find the estimated cost of the new pump which is stated as follows:
Cn = (112,000 * 286) /212
=32032000/212
=$151094
Therefore, the estimated cost of the new pump is $151094
Please show NEC article 310.10(H)
Answers
Conductors in Parallel is covered in 310.10(H). Often when working with higher ampacities, we find it to be cumbersome and expensive to keep increasing the size of the wire and conduit. The solution is often to use multiple runs that are connected to a common location on each end. Notice in the 2011 edition of the code that much of this section is highlighted gray, indicating new or revised text. The change here was that the previous code stated that you were permitted to parallel conductors 1/0 AWG and larger; however, it didn’t specifically prohibit you from paralleling smaller conductors, which was the intent and the way it was enforced in all my years of enforcement. However, that’s not what the actual language said, and ambiguous language cancause enforcement issues; therefore, in the 2011 code it was made clear that you are only allowed to parallel conductors 1/0 AWG and larger.
which one of these reduce fraction?
Answers
Determine the magnitude of force at the pin A and in the cable BC needed to support the 410-lb load. Neglect the weight of the boom AB. (Figure 1) Determine the magnitude of force at the pin A. Express your answer to three significant figures and include the appropriate units. Determine the force in the cable BC. Express your answer to three significant figures and include the appropriate units.
Answers
The magnitude of force at pin A is 410 lbs and the force in cable BC is 0 lbs.
To determine the magnitude of force at pin A and in cable BC, we need to use the principle of equilibrium. Since the system is in equilibrium, the sum of all forces acting on it must be zero.
First, let's find the force at pin A. Since there are only two forces acting on point A, the force in the cable AB and the force in the cable AC must be equal and opposite to the force of the load. Thus, the force at pin A is 410 lbs.
Now, to find the force in cable BC, we need to consider the forces acting on point B. There are three forces acting on point B, the force in the cable AB, the force in the cable BC, and the force of tension in the cable CD. Since the system is in equilibrium, the sum of all forces acting on point B must be zero. Thus,
force in AB - force in BC - force of tension in CD = 0
We know that the force in AB is 410 lbs, and the force at pin A is also 410 lbs. Therefore, the force of tension in CD must also be 410 lbs. Thus,
410 lbs - force in BC - 410 lbs = 0
Solving for the force in BC, we get:
force in BC = 410 lbs - 410 lbs = 0 lbs
Therefore, the force in cable BC is zero. This makes sense because cable BC is slack and not under tension.
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The magnitude of force at pin A is 410 lbs and the force in cable BC is 0 lbs.
To determine the magnitude of force at pin A and in cable BC, we need to use the principle of equilibrium. Since the system is in equilibrium, the sum of all forces acting on it must be zero.
First, let's find the force at pin A. Since there are only two forces acting on point A, the force in the cable AB and the force in the cable AC must be equal and opposite to the force of the load. Thus, the force at pin A is 410 lbs.
Now, to find the force in cable BC, we need to consider the forces acting on point B. There are three forces acting on point B, the force in the cable AB, the force in the cable BC, and the force of tension in the cable CD. Since the system is in equilibrium, the sum of all forces acting on point B must be zero. Thus,
force in AB - force in BC - force of tension in CD = 0
We know that the force in AB is 410 lbs, and the force at pin A is also 410 lbs. Therefore, the force of tension in CD must also be 410 lbs. Thus,
410 lbs - force in BC - 410 lbs = 0
Solving for the force in BC, we get:
force in BC = 410 lbs - 410 lbs = 0 lbs
Therefore, the force in cable BC is zero. This makes sense because cable BC is slack and not under tension.
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Ferroconcrete is reinforced concrete that combines concrete and ________. A. Lead c. Copper b. Iron d. Aluminum.
Answers
Answer:
B. Iron
Explanation:
took the test.
Which handles all application operations between a user and an organization's back end business systems?
Answers
The application layer, also known as the presentation layer, is responsible for handling all application operations between a user and an organization's back end business systems.
What do you mean by application layer?
The application layer is a layer in the Open Systems Interconnection (OSI) model, which is a standard model for the design of computer networks. The OSI model consists of seven layers, each of which is responsible for a different aspect of the communication process between computer systems.
The application layer is the highest layer in the OSI model and is closest to the end user. It is responsible for providing the services that directly support the user applications, such as file transfers, email, and virtual terminal access.
The application layer, also known as the presentation layer, is responsible for handling all application operations between a user and an organization's back end business systems. It acts as an intermediary between the user interface and the back end systems, ensuring that the user's requests are properly processed and that the resulting data is formatted and presented in a user-friendly manner.
This layer is responsible for implementing the logic required to perform actions such as data validation, authentication, and authorization, and it is also responsible for maintaining the user's session state and presenting the user with appropriate error messages in the event of any issues.
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Hot water enters the tube-side of a two shell pass, four tube pass shell-and-tube heat exchanger at 87°C at a flow rate of 2.5 kg/s. Unused engine oil enters the shell side of the same heat exchanger at 27°C at a mass flow rate of 2.8 kg/s. Each shell section has 18 tubes within it. Each tube has an inner diameter of 24 mm, a wall thickness of 1.1 mm, is 4 m in length including the hairpin turn, and is made of a copper alloy (v=336 W m^-1 K^-1 ). At the given inlet conditions, it is known that for this two-shell heat exchanger the overall hot-side heat transfer coefficient is 2100 W m^-2 K^-1 . Assuming no significant amount of external heat is lost or gained by the heat exchanger.
Determine:
a. the cold-side overall heat transfer coefficient
b. the outlet temperature of the engine oil
c. the outlet temperature of the water
d. the rate of heat transfer between the streams.
Answers
To determine the requested values, we can use the heat exchanger equations and principles. Let's solve the problem step by step:
a. The overall heat transfer coefficient for the cold side can be calculated using the formula:
1/U_c = 1/h_o + R_fo + R_w + R_fi + 1/h_i
where h_o and h_i are the oil-side and water-side convective heat transfer coefficients, respectively. R_fo and R_fi are the fouling resistances on the oil and water sides, and R_w is the wall resistance.
Since no values are provided for h_o and R_fo, we can assume reasonable estimates. Let's assume h_o = 1500 W/m^2 K and R_fo = 0.0005 m^2 K/W (common values for oil-side heat exchangers). R_w can be calculated using the formula:
R_w = ln(r_o/r_i) / (2πLk)
where r_o and r_i are the outer and inner radii of the tube, L is the tube length, and k is the thermal conductivity of the tube material.
Substituting the given values into the formulas, we can calculate 1/U_c.
b. To find the outlet temperature of the engine oil, we can use the energy balance equation:
m_o * Cp_o * (T_o,out - T_o,in) = Q
where m_o is the mass flow rate of the oil, Cp_o is the specific heat capacity of the oil, T_o out is the outlet temperature of the oil, T_o in is the inlet temperature of the oil, and Q is the rate of heat transfer between the streams.
We can rearrange the equation to solve for T_o, out.
c. Similarly, for the outlet temperature of the water, we can use the energy balance equation:
m_w * Cp_w * (T_w,in - T_w,out) = Q
where m_w is the mass flow rate of the water, Cp_w is the specific heat capacity of the water, T_w in is the inlet temperature of the water, T_w, out is the outlet temperature of the water, and Q is the rate of heat transfer between the streams.
Rearranging the equation will allow us to solve for T_w, out.
d. The rate of heat transfer between the streams can be calculated using the formula:
Q = m_w * Cp_w * (T_w,in - T_w,out) = m_o * Cp_o * (T_o,out - T_o,in)
Substituting the known values, we can determine the rate of heat transfer.
By solving these equations using the given values and the assumptions made, we can obtain the desired results.
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A television set connected to a 120-v outlet consumes 21 w of power. (a) how much current flows through the television? (b) how much time does it take for 10 million electrons to pass through the tv?
Answers
The current flowing through the television is 0.175 A and it will take 9.14 x 10^-12 s for 10 million electrons to pass through the TV.
(a) The current flowing through the television can be calculated using the formula: I = P/V,
where I is the current in amps,
P is the power in watts,
and V is the voltage in volts.
Plugging in the given values, we get:
I = 21 W / 120 V = 0.175 A
Therefore, the current flowing through the television is 0.175 A.
(b) The amount of time it takes for 10 million electrons to pass through the TV can be calculated using the formula: t = Q / I,
where t is the time in seconds,
Q is the charge in coulombs,
and I is the current in amps.
The charge of 10 million electrons is:
Q = (10 million electrons) x (1.6 x 10^-19 C / electron) = 1.6 x 10^-12 C
Plugging in the values for Q and I, we get:
t = (1.6 x 10^-12 C) / (0.175 A) = 9.14 x 10^-12 s
Hence, the current flowing through the television is 0.175 A and it will take 9.14 x 10^-12 s for 10 million electrons to pass through the TV.
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A force of 100 kip is acting at angle of 60 with horizontal axis. What is horizontal component of the force? 100* Cos60 100* Sin60 100* Sin30 100* Cos3
Answers
The horizontal component of a force of 100 kip acting at an angle of 60 degrees with the horizontal axis is 50 kip.
To determine the horizontal component of a force, we use trigonometric functions. In this case, we can use the cosine function to find the horizontal component. The cosine of an angle is defined as the ratio of the length of the adjacent side to the length of the hypotenuse in a right triangle.
In the given scenario, the force of 100 kip can be represented as the hypotenuse of a right triangle, with the horizontal component being the adjacent side. The angle between the force and the horizontal axis is 60 degrees. By using the cosine function, we can calculate the horizontal component as the product of the force magnitude (100 kip) and the cosine of the angle (cos 60 degrees):
Horizontal component \(= 100 kip \times cos(60 \textdegree) = 100 kip \times 0.5 = 50 kip.\)
Therefore, the horizontal component of the force is 50 kip.
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In heat treating to harden steel ball bearings (c 500 J/kg K, 7800 kg/m3 , k 50 W/m K), it is desirable to increase the surface temperature for a short time without significantly warming the interior of the ball. This type of heating can be accomplished by sudden immersion of the ball in a molten salt bath with T 1300 K and h 5000 W/m2 K. Assume that any location within the ball whose temperature exceeds 1000 K will be hardened. Estimate the time required to harden the outer millimeter of a ball of diameter 20 mm, if its initial temperature is 300 K.
Answers
To estimate the time required to harden the outer millimeter of a steel ball bearing, we can use the concept of heat transfer. The rate of heat transfer can be calculated using the equation, Where Q is the heat transferred, h is the convective heat transfer coefficient, A is the surface area, T_surface is the surface temperature, and T_initial is the initial temperature.
Given that the diameter of the ball is 20 mm, the radius (r) would be 10 mm or 0.01 m. Therefore, A = 4 * π * (0.01)^2 = 0.001256 m^2 Using the given values, we have h = 5000 W/m^2 K,
T_surface = 1300 K, and
T_initial = 300 K.
To harden the steel, it is required to reach a temperature above 1000 K. Let's assume it takes approximately 700 J/kg to raise the temperature of the outer millimeter of the ball by 1 K. The mass of the outer millimeter of the ball can be calculated as:
\(m = ρ * V\)
Where c is the specific heat capacity of the steel, assumed to be 500 J/kg K.\(Time = 6.27 / (0.11 * 500)\)
= 0.113 s Therefore, it would take approximately 0.113 seconds to harden the outer millimeter of the steel ball bearing.
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True or false
Consumer is the end user
Client is usually a company
Consumer is usually a company
Client is the end user
Consumer is the end user
Client is usually a company
Consumer is usually a company
Client is the end user
Answers
Answer:
Explanation:
a) True
b) False
c) False
d) False
e) True
f) False
g) False
h)False
Plz mark as brainliest
Flip-flops are normally used for all of the following applications, except ________. logic gates data storage frequency division counting
Answers
Flip-flops are normally used for all of the following applications, except logic gates.
What are Flip flops?Flip flops are known to be tools that are used for counting. They come in different ranges.
Note that Flip flops are one that can be seen on counters, storage registers, and others and as such, Flip-flops are normally used for all of the following applications, except logic gates.
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Technician A says that a conditionally-exempt small quantity generator is required to use hazardous waste shipment manifests.
Technician B says that maintaining hazardous waste records may be required for at least three years.
Who is right?
A)
Aonly
B)
B only
C) Both A and B
D) Neither A nor B
Answers
The statements made by both technicians are correct. Thus, the correct option for this question is C, i.e. both A and B.
What is a Technician?A technician may be characterized as someone whose job involves skilled practical work with scientific equipment, for example in a laboratory. The role of these people is to repair, install, replace, and service different systems and equipment.
The statement of the technician states that a conditionally-exempt small quantity generator is required to use hazardous waste shipment manifests.
While the statement technician B states that maintaining hazardous waste records may be required for at least three years. Both options are correct with respect to maintaining and degrading hazardous wastes.
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estion 11 points A 2-litre four-stroke Indirect injection diesel engine is designed to run at 4500 rpm with a power output of 45 kW, the volumetric efficiency is found to be 80 per cent. The bufe is 0.071 kg/m3 and the fuel has a calorific value of 42 Ming. The ambient conditions for the test were 20 and 1 bar. Calculate the ideal Otto cycle efficiency the compression ratio is 12 and specifichent ratio is 1.4 O a 1629% Ob 26.9% Oc0962 Od: 62.9% Moving to the next question prevents changes to this answer con 11025 MacBook Pro INS 5 %. OP E R G D65 F
Answers
The ideal Otto cycle efficiency for the given conditions of a 2-liter four-stroke indirect injection diesel engine running at 4500 rpm, with a power output of 45 kW, a volumetric efficiency is of 80%.
To calculate the ideal Otto cycle efficiency, we need to use the given information and apply the relevant formulas. Here are the steps to determine the efficiency:
Step 1: Calculate the compression ratio (r):
The compression ratio is given as 12.
Step 2: Calculate the specific heat ratio (γ):
The specific heat ratio is given as 1.4.
Step 3: Calculate the air-fuel ratio (AFR):
The air-fuel ratio can be calculated using the equation:
AFR = (1/bufe) * (AFR_stoichiometric)
The buffer (bu) is given as 0.071 kg/m³.
The stoichiometric air-fuel ratio (AFR_stoichiometric) can be calculated using the equation:
AFR_stoichiometric = (AFR_fuel) * (mass_fuel/mass_air)
The calorific value of the fuel is given as 42 Ming (million British thermal units per gallon). However, the given unit is not standard, so we need to convert it to a standard unit like MJ/kg.
Assuming the given unit is 42 MJ/kg, we can calculate the stoichiometric air-fuel ratio.
Step 4: Calculate the air density (ρ):
The air density can be calculated using the ideal gas law:
ρ = (P * M_air) / (R * T)
The given ambient conditions are:
Temperature (T) = 20 °C = 293 K
Pressure (P) = 1 bar = 100 kPa
Step 5: Calculate the air mass flow rate (m_dot_air):
The air mass flow rate can be calculated using the equation:
m_dot_air = (V_dot_air) * ρ
The volumetric efficiency (η_vol) is given as 80%. The volumetric flow rate of air (V_dot_air) can be calculated using the equation:
V_dot_air = (η_vol) * (V_dot_air_actual)
The actual volumetric flow rate of air (V_dot_air_actual) can be calculated using the equation:
V_dot_air_actual = (rpm) * (V_cyl) / 2
The given engine parameters are:
Engine displacement (V_cyl) = 2 liters = 0.002 m³
Engine speed (rpm) = 4500
Step 6: Calculate the fuel mass flow rate (m_dot_fuel):
The fuel mass flow rate can be calculated using the equation:
m_dot_fuel = (m_dot_air) / (AFR)
Step 7: Calculate the heat input (Q_in):
The heat input can be calculated using the equation:
Q_in = (m_dot_fuel) * (CV_fuel)
The calorific value of the fuel (CV_fuel) is given as 42 MJ/kg.
Step 8: Calculate the heat output (Q_out):
The heat output can be calculated using the equation:
Q_out = (m_dot_air) * (Cp_air) * (T3 - T2)
The specific heat capacity of air at constant pressure (Cp_air) can be calculated using the equation:
Cp_air = γ * (R_air) / (γ - 1)
The gas constant for air (R_air) is known as 287 J/(kg·K).
T2 is the temperature at the end of the compression stroke and can be calculated using the equation:
T2 = (r) ^ (γ - 1)
T3 is the temperature at the end of the combustion process and can be calculated using the equation:
T3 = (r) ^ γ
Step 9: Calculate the ideal Otto cycle efficiency (η_cycle):
The ideal Otto cycle efficiency can be calculated using the equation:
η_cycle = 1 - (Q_out / Q_in)
Now let's substitute the given values into these formulas and calculate the result:
Step 1: Compression ratio (r) = 12
Step 2: Specific heat ratio (γ) = 1.4
Step 3: Air-fuel ratio (AFR) = (1/bufe) * (AFR_stoichiometric)
Step 4: Air density (ρ) = (P * M_air) / (R * T)
Step 5: Air mass flow rate (m_dot_air) = (V_dot_air) * ρ
V_dot_air = (η_vol) * (V_dot_air_actual)
V_dot_air_actual = (rpm) * (V_cyl) / 2
Step 6: Fuel mass flow rate (m_dot_fuel) = (m_dot_air) / (AFR)
Step 7: Heat input (Q_in) = (m_dot_fuel) * (CV_fuel)
Step 8: Heat output (Q_out) = (m_dot_air) * (Cp_air) * (T3 - T2)
Cp_air = γ * (R_air) / (γ - 1)
T2 = (r) ^ (γ - 1)
T3 = (r) ^ γ
Step 9: Ideal Otto cycle efficiency (η_cycle) = 1 - (Q_out / Q_in)
First, we determine the compression ratio and specific heat ratio. Then, we calculate the air-fuel ratio using the buffer and the stoichiometric air-fuel ratio based on the given calorific value of the fuel.
Next, we calculate the air density using the ideal gas law and determine the air mass flow rate based on the volumetric efficiency, engine displacement, and speed.
We then determine the fuel mass flow rate based on the air mass flow rate and air-fuel ratio. The heat input is calculated using the fuel mass flow rate and calorific value of the fuel.
To calculate the heat output, we use the air mass flow rate, specific heat capacity of air, and temperatures at the end of the compression and combustion processes.
Finally, we calculate the ideal Otto cycle efficiency by dividing the heat output by the heat input and subtracting it from 1.
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A typical discounted price of a AAA battery is $0.75. It is designed to provide 1.5 volts and 1.0 amps for about an hour. Now we multiply volts and amps to obtain power of 1.5 watts from the battery. Thus, it costs $0.75 for 1.5 Watt-hours of energy. How much would it cost to deliver one kilo Watt-hour? How does this compare with the cost of energy from your local electric utility at $0.10 per kilo Watt-hour?
Answers
Answer:
gh fjh,vx j ahj ds djv dk
Explanation:
doing for points
The Cost of 1 kilowatt hour of energy at the rate of $0.75 per 1.5 watt hour is $500 which is 5000 times greater than the cost of energy at $0.10 per kilowatt hour.
Battery power = Current × Voltage
Cost of 1.5 Watt-hour = $0.75
Converting Energy to Watt - hour :
1 kilowatt = 1000 watt
1 kilowatt hour = 1000 watt - hour
Hence,
Cost of 1 kilowatt - hour = 1000 watt - hour can be calculated thus :
1.5 Watt-hour = $0.75
1000 Watt-hour = c
Cross multiply :
1.5c = $0.75 × 1000
1.5c = 750
c = 750 / 1.5
c = 500
Therefore, cost of 1 kilowatt - hour of energy will be $500
Comparing the cost of Energy at $500 per Kilo-Watt hour to Cost at $0.10:
$500 / $0.10 = 5000
Therefore, the cost of energy at $500 per kilowatt hour is 5000 times greater than cost at $0.10 per kilowatt hour.
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engineering controls are devices such as self sheathing needles and sharps containers to block or eliminate the sharp risk.
Answers
The given statement "the engineering controls are devices, such as sharps containers and self-sheathing needles, to block or eliminate the sharp risk" is true because engineering control devices are used to avoid the occurrence of hazards.
Engineering controls are devices that protect workers by eliminating hazardous conditions or by placing a barrier between the hazard and the worker. Examples of engineering control devices are exhaust ventilation to capture and exit airborne emissions or machine guards to shield the workers.
It means the engineering controls devices reduce or prevent hazards from coming into contact with workers. So the devices such as sharps containers and self-sheathing needles to block or eliminate the sharp risk are examples of engineering control devices.
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Complete question
engineering controls are devices such as self sheathing needles and sharps containers to block or eliminate the sharp risk.
True
False
"
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Problem the pressure at a given point is 50 mmhg absolute
Answers
Answer:
6.65 kPa.
- 73.3 kPa.
Explanation:
Without much ado let's jump right into the solution to the problem given. It is given that the pressure = 50 mmHg and the solution to this question is to write out the value of the pressure in kpa and kPa gauge.
P(a) = 0.05m × 133 kN/m³ = 6.65 kPa.
The P(gauge) =( [ 0.05 × 13.6 × 9810] ÷ 1000 ) - 80 = - 73.3 kPa.
the four mechanisms of adhesion, mechanical interlocking, diffusion, adsorption and electrostatic attraction respectively [
Answers
The four mechanisms of adhesion are mechanical interlocking, diffusion, adsorption, and electrostatic attraction.
Mechanical interlocking: This mechanism involves physical interlocking of two surfaces, such as the interlocking of fibers in textiles or the interlocking of teeth in a zipper. The irregularities on the surface create a strong bond through intermolecular forces.
Diffusion: Diffusion occurs when molecules from one material penetrate the surface of another material. This mechanism is commonly observed in adhesives that consist of polymers. The adhesive molecules diffuse into the substrate, forming intermolecular bonds and creating adhesion.
Adsorption: Adsorption refers to the attraction of molecules from one material to the surface of another material. Van der Waals forces, dipole-dipole interactions, or chemical bonds can facilitate adsorption. Adhesives often rely on adsorption to adhere to substrates.
Electrostatic attraction: This mechanism involves the attraction between positively and negatively charged particles. It can occur between polar molecules or charged particles on the surfaces. Electrostatic attraction contributes to the adhesion of materials like plastics or rubber.
Adhesion can be achieved through various mechanisms. Mechanical interlocking relies on physical interlocking, diffusion involves the penetration of molecules into the substrate, adsorption occurs through the attraction of molecules to a surface, and electrostatic attraction relies on the attraction between charged particles. Understanding these mechanisms is crucial for developing effective adhesives and improving bonding techniques in various industries
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The four mechanisms of adhesion are mechanical interlocking, diffusion, adsorption, and electrostatic attraction. Each mechanism plays an important role in creating adhesion between two surfaces.
Mechanical interlocking is the process of creating a mechanical bond between two surfaces by physically interlocking them. This can be achieved by creating grooves or hooks in the surface of one of the materials and then pressing the two surfaces together. Diffusion is the process of atoms or molecules from one surface migrating into the other surface, creating a bond. This is typically seen in metal-to-metal bonding, where atoms from one surface diffuse into the other surface and create a solid-state weld. Adsorption is the process of molecules or atoms adhering to the surface of a material.
This can be achieved by creating a chemical bond between the surface of the material and the atoms or molecules. This type of bonding is often used in the creation of coatings and paints. Electrostatic attraction is the process of positively charged ions attracting negatively charged ions. This type of bonding is often seen in the bonding of two different materials, where one material has a positive charge and the other material has a negative charge.
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who is the strongest avenger i say hulk but who knows at this point
Answers
Answer:
or is the strongest evenger she hulk
Explanation:
?????????
Answer:
Thor!
Explanation:
In Thor: Ragnarok he beat the Hulk in order for Hulk to win thor had to be electrocuted and in Avengers: Endgame Thor is seen holding open the "Floodgates" and withstanding the radiation from a dying star, also the fact that Thor is a god means that he is all powerful and the rightful heir to the throne to Asgard, plus the fact that he has defeated Loki multiple times a feat that not even the Hulk has done.
A spherical ball has 3cm internal diameter and its inner surface temperature is 250˚C and the outer surface temperature is 30˚C. Calculate the outer diameter of the ball if the heat loss from the ball is 1600W and has 2.75W/m.˚C thermal conductivity.
Answers
Answer:
The outer diameter of the ball is 6.2138 cm
Explanation:
The formula to apply is ;
Heat loss ,
\(Q/t=kA*\frac{( T_1-T_2)}{d}\)
where ;
Q/t=total heat loss from the ball = 1600 w
k=coefficient of heat transmission through the ball= 2.75 W/m.˚C
A=area in m² of the ball with the coefficient of heat transmission
T₁=Hot temperature
T₂=Cold temperatures
d=thickness of the ball
Area of spherical ball using internal diameter, 3cm= 0.03 m will be
Radius = half the diameter = 0.03/2 = 0.015
Area = 4 *π*r²
Area = 4*π*0.015² = 0.002827 m²
Apply the formula for heat loss to get the thickness as:
1600 = {2.75 * 0.002827 *(250-30 ) }/d
1600 =1.711/d
1600d = 1.711
d=1.711/1600 = 0.001069 m
d= 0.1069
Using internal radius and the thickness to get outer radius as;
3 + 0.1069 = 3.1069 cm
Outer diameter will be twice the outer radius
2*3.1069 = 6.2138 cm
HURRY
In the urban fire department, if a fire captain has a problem or issue that requires a decision by a higher official and the deputy chief position has not been filler, he or she may take the issue directly to the fire chief. true or false
Answers
Answer:
Yes, it is true.
Tea brewed for iced tea should never be held for more than which of the following time periods?
1 hour
4 hours
8 hours
12 hours
Answers
# Structure Mechanics.
Draw the internal force (axial force, shear force and moment) diagrams of the frame.
Answers
Define solid and hollow masonry units according to ASTM C90.
Answers
Solid masonry units: These are concrete blocks that are made from solid concrete and have no cores or voids. Solid masonry units are designed to provide strength and stability to masonry walls and are often used in load-bearing applications.
Hollow masonry units: These are concrete blocks that have one or more cores or voids that are not filled with mortar. The cores or voids can be arranged in various patterns, such as a single rectangular core or multiple square cores. Hollow masonry units are used in non-load-bearing applications, such as interior walls or decorative features. They are also used in load-bearing applications, where the cores or voids are filled with grout or concrete to provide additional strength and stability.
/*
(Display four patterns using loops) Use nested loops that display the following
patterns in four separate programs:
*/
public class Exercise_05_18_B {
public static void main(String[] args) {
// Display pattern B
System.out.println("Pattern B");
for (int r = 6; r >= 1; r--) {
for (int c = 1; c <= r; c++) {
System.out.print(c + " ");
}
System.out.println();
}
}
}
Answers
The program displays a triangle of numbers arranged in decreasing order from top to bottom using nested loops. The pattern can be modified by changing the loop conditions and statements to print different patterns of numbers, characters.
Can you explain the pattern displayed in the given Java program using nested loops and how it can be modified to display other patterns?
The given code is a Java program that displays a specific pattern using nested loops. The pattern displayed in this program is Pattern B, which is a triangle of numbers arranged in decreasing order from top to bottom.
The outer loop starts with r=6 and decrements by 1 in each iteration until it reaches 1. This controls the number of rows in the pattern.
The inner loop starts with c=1 and increments by 1 in each iteration until it reaches the current value of r. This controls the number of columns in each row and also the values of the numbers printed in each row.
The "println" statement is used to print a newline character after each row is printed, so that the pattern is displayed in a triangular shape.
The program is defined inside a public class named "Exercise_05_18_B", which can be compiled and executed as a separate program.
To display other patterns using loops, you can modify the inner and outer loop conditions and statements to print different patterns of numbers, characters, or symbols. You can create separate programs for each pattern, and use "println" statements to separate each pattern on a new line.
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