Answer: The theoretical stopping distance is 1.245 miles (6572.193 ft).
Explanation:
First, the physical model is created by using Principle of Energy Conservation and Work-Energy Theorem. It is assumed that surface is horizontal, so there are no changes associated with potential energy. The car has an initial kinetic energy, which is completely dissipated by braking.
[tex]K_{1} = \Delta W_{loss}[/tex]
[tex]\frac{1}{2} \cdot m \cdot v^{2} = m \cdot g \cdot (\mu_{2} \cdot \Delta s_{2}+\mu_{2'} \cdot \Delta s_{2'})[/tex]
The distance is isolated from previous equation:
[tex]\Delta s_{2'} = \frac{1}{\mu_{2'}}\cdot (\frac{1}{2}\cdot \frac{v^{2}}{g} - \mu_{2} \cdot \Delta s_{2})[/tex]
By replacing variables, the distance is calculated herein:
[tex]\Delta s_{2'} = \frac{1}{\frac{0.02}{0.80}}\cdot (\frac{1}{2}\cdot \frac{(102.667\frac{ft}{s} )^2}{32.174 \frac{ft}{s^{2}}} - (0.02) \cdot (100 ft))[/tex]
[tex]\Delta s_{2'} = 6472.193 ft\\\Delta s_{2'} = 1.226 mi[/tex]
The theoretical stopping distance is:
[tex]\Delta s = \Delta s_{2} + \Delta s_{2'}\\\Delta s = 6572.193 ft\\\Delta s = 1.245 mi[/tex]
A solid titanium alloy [G 114 GPa] shaft that is 720 mm long will be subjected to a pure torque of T 155 N m. Determine the minimum diameter required if the shear stress must not exceed 150 MPa and the angle of twist must not exceed 7?. Report both the maximum shear stress ? and the angle of twist ? at this minimum diameter. ?Part 1 where d is the shaft diameter. The polar moment of inertia is also a function of d. Find Incorrect. Consider the elastic tors on formula. The maximum shear stress occurs at the radial location ? = (d 2 the value of d for which the maximum shear stress in the shaft equals 150 MPa. Based only on the requirement that the shear stress must not exceed 150 MPa, what is the minimum diameter of the shaft? dr 16.1763 the tolerance is +/-2% Click if you would like to Show Work for this question: Open Show Work Attempts: 1 of 3 used SAVE FOR LATER SUBMIT ANSWER Part 2 Based only on the requirement that the angle of twist must not exceed 7°, what is the minimum diameter of the shaft?
Answer:
Part 1: The diameter of the shaft so that the shear stress is not more than 150 MPa is 17.3 mm.
Part 2: The diameter of the shaft so that the twist angle is not more than 7° is 16.9 mm.
Explanation:
Part 1
The formula is given as
[tex]\dfrac{T}{J}=\dfrac{\tau}{R}[/tex]
Here T is the torque which is given as 155 Nm
J is the rotational inertia which is given as [tex]\dfrac{\pi d^4}{32}[/tex]
τ is the shear stress which is given as 150 MPa
R is the radius which is given as d/2 so the equation becomes
[tex]\dfrac{T}{J}=\dfrac{\tau}{R}\\\dfrac{155}{\pi d^4/32}=\dfrac{150 \times 10^6}{d/2}\\\dfrac{155 \times 32}{\pi d^4}=\dfrac{300 \times 10^6}{d}\\\dfrac{1578.82}{d^4}=\dfrac{300 \times 10^6}{d}\\d^3=\dfrac{1578.82}{300 \times 10^6}\\d^3=5.26 \times 10^{-6}\\d=0.0173 m \approx 17.3 mm[/tex]
So the diameter of the shaft so that the shear stress is not more than 150 MPa is 17.3 mm.
Part 2
The formula is given as
[tex]\dfrac{T}{J}=\dfrac{G\theta}{L}[/tex]
Here T is the torque which is given as 155 Nm
J is the rotational inertia which is given as [tex]\dfrac{\pi d^4}{32}[/tex]
G is the torsional modulus which is given as 114 GPa
L is the length which is given as 720 mm=0.720m
θ is the twist angle which is given as 7° this is converted to radian as
[tex]\theta=\dfrac{7*\pi}{180}\\\theta=0.122 rad\\[/tex]
so the equation becomes
[tex]\dfrac{T}{J}=\dfrac{G\theta}{L}\\\dfrac{155}{\pi d^4/32}=\dfrac{114 \times 10^9\times 0.122}{0.720}\\\dfrac{1578.81}{d^4}=1.93\times 10^{10}\\d^4=\dfrac{1578.81}{1.93\times 10^{10}}\\d=(\dfrac{1578.81}{1.93\times 10^{10}})^{1/4}\\d=0.0169 m \approx 16.9mm[/tex]
So the diameter of the shaft so that the twist angle is not more than 7° is 16.9 mm.
Use the time module to write a decorator function named sort_timer that times how many seconds it takes the decorated function to run. Since sort functions don't need to return anything, have the decorator's wrapper function return that elapsed time.
Explanation:
Find the difference between the simple interest and the compound interest on Rs 5000 for 2 years at 9% per annum
(TCO 4) A system samples a sinusoid of frequency 190 Hz at a rate of 120 Hz and writes the sampled signal to its output without further modification. Determine the frequency that the sampling system will generate in its output.
Answer:
The frequency that the sampling system will generate in its output is 70 Hz
Explanation:
Given;
F = 190 Hz
Fs = 120 Hz
Output Frequency = F - nFs
When n = 1
Output Frequency = 190 - 120 = 70 Hz
Therefore, if a system samples a sinusoid of frequency 190 Hz at a rate of 120 Hz and writes the sampled signal to its output without further modification, the frequency that the sampling system will generate in its output is 70 Hz
the frequencies 10, 12, 23 and 45 Hz. (a) What is the minimum sampling rate required to avoid aliasing? (b) If you sample at 40 sps, which frequency components will be aliased?
Answer:
Sampling rate = 90
23 and 45 Hz
Explanation:
The Nyquist criteria states that the sampling frequency must be at least twice of the highest frequency component.
(a) We have 10, 12, 13, 23, and 45 Hz signals
The highest frequency component in this list is 45 Hz
So minimum sampling frequency needed to avoid aliasing would be
Sampling rate = 2*45 = 90 sps
Hence a sampling rate of 90 samples per second would be required to avoid aliasing.
(b) if a sampling rate of 40 sps is used then
40 = 2*f
f = 40/2 = 20 Hz
Hence 23 and 45 Hz components will be aliased.
Suppose we have a classification problem with classes labeled 1, . . . , c and an additional "doubt" category labeled c + 1. Let r : R d → {1, . . . , c + 1} be a decision rule. Define the loss function
Answer:
The answer and explanation to this question is attached.
Answer:
Explanation:
Let first simplified the risk given our specific loss function. if f(x) = i i is not double , then the risk is
R(f(x) = i|x) = ∑ L(f(x) = i , y = j ) P(y = j|x) 2
=0.P (Y= i|x) +λc ∑ P (Y= j|x) 3
=λc (1 - P(Y= i|x)) 4
When f(x) = c + 1, meaning you have choosing doubt , the risk is
R(f(x) = c +1|x) = ∑ L (f(x)= c+1, y=j) P(Y=j|x) 5
=λd∑ P(Y=j|x) 6
=λd 7
because ∑ P(Y=j|x) should sum to 1 since its a proper probability distribution.
Now let fopt : Rd→ {1, . . . , c + 1} be the decision rule which implements (R1)–(R3).We want to show that in expectation the rule foptis at least as good as an arbitrary rulef. Let x ∈ Rdbe a data point, which we want to classify. Let’s examine all the possiblescenarios where fopt(x) and another arbitrary rule f(x) might differ:Case 1: Let fopt(x) = i where i 6= c + 1.– Case 1a: f(x) = k where k 6= i. Then we get with (R1) thatR(fopt(x) = i|x) = λc1 − P(Y = i| x)≤ λc1 − P(Y = k|x)= R(f(x) = k|x).– Case 1b: f(x) = c + 1. Then we get with (R1) thatR(fopt(x) = i|x) = λc1 − P(Y = i| x)≤ λc(1 − (1 −λdλc)) = λd= R(f(x) = c + 1|x).Case 2: Let fopt(x) = c + 1 and f(x) = k where k 6= c + 1. Then:R(f(x) = k|x) = λc(1 − P (Y = k|x)R(fopt(x) = c + 1|x) = λ
A simple highway curve is planned to connect two horizontal tangents that intersect at sta 2500 + 00.00 at an external angle of 52°. For a design speed of 60 mph and a curve radius of 1.25 times the minimum allowable radius, calculate the design rate of superelevation. (Hint - you will need to assume a maximum superelevation rate to get the minimum radius.)
Answer:
Design rate of superelevation= 3.4%
Explanation:
The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.
Bonding in the intermetallic compound Ni3Al is predominantly metallic. Explain why there will be little, if any, ionic bonding component. The electronegativity of nickel is about 1.8
Answer:
There is very little possibility of ionic bonding.
Explanation:
First we need to understand the meaning of ionic bonding. ionic bonding means the complete transfer of valance electrons between atoms.
The electronegativity of Al is 1.5 whereas electronegativity of nickel "Ni" is 1.8-1.9. As we can see from these values of electronegativity of Al & Ni that they are closer to each other & the difference of electonegativity of these metals is very little which decreases the possibility of ionic bonding between them so it is not expected that there will be much ionic bonding. Moreover it is noticeable here that these both are metals and they prefer to give up their electrons rather to gain them.
The two elements can not form ionic bonds because ionic bonds are formed when the electronegativity difference is greater than 1.5.
What is an intermetallic compound?The term intermetallic compound refers to a compound that is composed of two metallic elements. Let us recall that metals often have a low value of electronegativity.
Owing to the fact that the electronegativity difference between nickel and aluminium is small (about 0.1). An ionic bond can not be formed between the two elements because ionic bonds are formed when the electronegativity difference is greater than 1.5.
Learn more about ionic bonds: https://brainly.com/question/867804
(Specific weight) A 1-ft-diameter cylindrical tank that is 5 ft long weighs 125 lb and is filled with a liquid having a specific weight of 66.4 lb/ft3. Determine the vertical force required to give the tank an upward acceleration of 10.7 ft/s2.
Answer:
The answer to the question is 514.17 lbf
Explanation:
Volume of cylindrical tank = πr²h = 3.92699 ft³
Weight of tank = 125 lb
Specific weight of content = 66.4 lb/ft³
Mass of content = 66.4×3.92699 = 260.752 lb
Total mass = 260.752 + 125 = 385.75 lb = 174.97 kg
=Weight = mass * acceleration = 174.97 *9.81 = 1716.497 N
To have an acceleration of 10.7 ft/s² = 3.261 m/s²
we have F = m*a = 174.97*(9.81+3.261) = 2287.15 N = 514.17 lbf
Suppose you are at a party with 19 of your closest friends (so including you, there are 20 people there). Explain why there must be least two people at the party who are friends with the same number of people at the party. Assume friendship is always reciprocated.
Answer:
The answer is in the attachment. Thanks
Explanation:
g Double‑reciprocal, or Lineweaver–Burk, plots can reveal the type of enzyme inhibition exhibited by an inhibitor. Competitive inhibitors increase the K M without affecting the V max . Noncompetitive inhibitors decrease the V max without affecting the K M . Uncompetitive inhibitors decrease both the K M and V max . A Lineweaver Burk plot shows two lines representing the kinetics of an enzyme with and without inhibitor. The x-axis plots the inverse of the substrate concentration, and the y-axis plots the inverse of the reaction velocity. The x-intercept of the line with inhibitor is more negative than the line with no inhibitor. Both lines have the same y-intercept. Based on the Lineweaver–Burk plot, what type of enzyme inhibition is exhibited by the inhibitor?
competitive
noncompetitive
uncompetitive
cannot be determined
Answer:
The answer is competitive inhibitor.
Explanation:
Line-Weaver Burk Plots are designed by Hans Lineweaver and Dean Burk in 1934 and they are used to show the inhibition process of the enzymes in biochemistry.
In the question we are given the description of competitive inhibitors, non-competitive inhibitors and uncompetitive inhibitors. According to the example given later where the x-axis plot and y-axis plot is described with their interceptions of the x and y axis, since both lines have the same y intercept and the no-inhibitor line is more negative than the inhibitor line, we can deduce that the type of enzyme inhibition is exhibited by a competitive inhibitor.
I hope this answer helps.
William, a project manager, needs to prepare the budget for a new software development project. To do so, he takes inputs from other managers who have worked orn similar projects in the past. After estimating the overall project cost, he gives the estimates to his team members so that they car split up the cost of each individual task involved in producing the required deliverable. In this scenario, William is using:______.
a. top-down budgeting
b. bittom-up budgeting
c. agile project budgeting
d. dynamic matrix budgeting
Answer:
The correct option is A
Explanation:
Top down budgeting is a type of budget in which the senior management prepares a high level budget for the company. The senior management then allocates the amounts for the individual departments, which then use this number in preparing their own budget.
When removing the balance shaft assembly: Technician A inspects the bearings for unusual wear or damage. Technician B smoothens a damaged camshaft or balance shaft journal by lightly sanding it. Who is correct
Answer:
Technicians A
Explanation:
The technician A would also tap the the teeth with a blunt end to break it loose. He also remove the underlying O ring in the front case groove. Technicians A also remove the driven gear bolt that secures the oil pump driven gear to the left balance shaft.
Consider a steam turbine, with inflow at 500oC and 7.9 MPa. The machine has a total-to-static efficiency ofηts=0.91, and the pressure at the outflow is 16kPa. Power extracted by the turbine is 38 MW. Assuming heat transfer and kinetic energy in the machine is negligible, find the mass flow rate and static enthalpy at the outflow.
Answer: [tex]\dot m_{in} = 23.942 \frac{kg}{s}[/tex], [tex]\dot H_{out} = 39632.62 kW[/tex]
Explanation:
Since there is no information related to volume flow to and from turbine, let is assume that volume flow at inlet equals to [tex]\dot V = 1 \frac{m^{3}}{s}[/tex]. Turbine is a steady-flow system modelled by using Principle of Mass Conservation and First Law of Thermodynamics:
Principle of Mass Conservation
[tex]\dot m_{in} - \dot m_{out} = 0[/tex]
First Law of Thermodynamics
[tex]- \dot W_{out} + \eta\cdot (\dot m_{in} \dot h_{in} - \dot m_{out} \dot h_{out}) = 0[/tex]
This 2 x 2 System can be reduced into one equation as follows:
[tex]-\dot W_{out} + \eta \cdot \dot m \cdot ( h_{in}- h_{out})=0[/tex]
The water goes to the turbine as Superheated steam and goes out as saturated vapor or a liquid-vapor mix. Specific volume and specific enthalpy at inflow are required to determine specific enthalpy at outflow and mass flow rate, respectively. Property tables are a practical form to get information:
Inflow (Superheated Steam)
[tex]\nu_{in} = 0.041767 \frac{m^{3}}{kg} \\h_{in} = 3399.5 \frac{kJ}{kg}[/tex]
The mass flow rate can be calculated by using this expression:
[tex]\dot m_{in} =\frac{\dot V_{in}}{\nu_{in}}[/tex]
[tex]\dot m_{in} = 23.942 \frac{kg}{s}[/tex]
Afterwards, the specific enthalpy at outflow is determined by isolating it from energy balance:
[tex]h_{out} =h_{in}-\frac{\dot W_{out}}{\eta \cdot \dot m}[/tex]
[tex]h_{out} = 1655.36 \frac{kJ}{kg}[/tex]
The enthalpy rate at outflow is:
[tex]\dot H_{out} = \dot m \cdot h_{out}[/tex]
[tex]\dot H_{out} = 39632.62 kW[/tex]
The input to a harmonic generator is a periodic waveform v(t) = Ce^-alpha t, 0 < t < 1, T = 1 with alpha < 0. The output is to be derived through a filter which will pass only one frequency. If alpha can be adjusted, what value of alpha will give the maximum output through the filter for the second harmonic the third harmonic? the fifth harmonic?
Answer:
Please find attached
Explanation:
An NMOS amplifier is to be designed to provide a 0.50-V peak output signal across a 50- kΩ load that can be used as a drain resistor. If a gain of at least 5 V/V is needed, what gm is required?
Answer:
The gm required is 0.1 [tex]\frac{mA}{V}[/tex]
Explanation:
Knowing
Vpeak = 0.50 V
Rd = 50 kΩ
Av = 5 V/V
To find the transconductance we should to use the equation
Av = gm * Rd
so,
gm = Av/Rd
gm = 5/50 = 0.1 [tex]\frac{mA}{V}[/tex]
GV is a small accounting firm supporting wealthy individuals in their preparation of annual income tax statements. Every December, GV sends out a short survey to its customers, asking for the information required for preparing the tax statements. Based on 50 years of experience, GV categorizes its cases into the following two groups: Group 1 (new customers): 20 percent of cases Group 2 (repeat customers): 80 percent of cases what is the total demand rate
Answer:
Explanation:
we will begin by carefully following a step by step order;
Total time in a week = 40 hours = 2400 minutes
Total time Administrator is utilized = 50 * 20 = 1000 minutes
Total time Senior accountant is utilized= 20% * 50 * 40 = 400 minutes
Total time Junior accountant is utilized = 40%*50*15 = 600 minutes
For Total cases, Total capacity = Total time / Bottlenck time (Administrator) = 2400 / 20 = 120 cases
C-3) Capacity of new customers at 20:80 ratio = 20% * 120 = 24 cases
C-4) Capacity of repeat customers at 20:80 ratio = 20%*120 = 96 cases
c-1 = Flow rate = min ( demand, capacity) for new customers = 10 new cases / week (20%*50)
c-2 = Flow rate = 40%*50 = 40 cases / week
c3 = 24 cases
c4 = 96 cases
cheers i hope this helps
A torsion member has an elliptical cross section with major and minor dimensions of 50.0 mm and 30.0 mm, respectively. The yield stress of the material in the torsion member is Y=400 MPa. Determine the maximum torque that can be applied to the tension member based on a factor of safety 1.85 using the maximum shear stress criterion of failure.
Answer:
What do i have to do
Explanation:
what do i do
Let X denote the distance (m) that an animal moves from its birth site to the first territorial vacancy it encounters. Suppose that for banner-tailed kangaroo rats, X has an exponential distribution with parameter λ = 0.0134. (a) What is the probability that the distance is at most 100 m?
Answer:
Answer is 0.74
Explanation:
Probability that the distance is at most 100 m:
P ( X ≤ 100 ) = F ( 100 )
= (1 − e-^ λ *x)
= 1 − e −^( ( 0.0134 )* ( 100 ))
= 1 − e −^ 1.34
= 1 − 0.2618
= 0.7381
≈ 0.74
Therefore,
P ( X ≤ 100 )≈ 0.74
Answer.
Answer:
0.7400
Explanation:
Let's first look at what we have:
λ = 0.0134 mean = 1/λ = 74.24 standard deviation = 1/λ = 74.24
We'll use the following formula; CDF, C(x) = 1 - e^(-λx)
Probability of a distance not exceeding 100 m
P(X <= 100)
= C(100)
= 1 - e^(-0.01347 * 100)
= 0.7400
g A plane stress element has components sigma x = 160 MPa, tau xy = 100 MPa (CW). Determine the two values pf sigma y for which the maximum shear stress is 73 MPa.
Answer:
The question is mentioned in the attachment.
Explanation:
Consider a Cournot game between two firms. The firms face an inverse demand function described by the equation P(Q) = α − Q if Q ≤ α, P(Q) = 0 if Q > α, where P is the price of output and Q is the total output produced by the two firms. Firm 1 produces its output q1 at a constant unit cost c1 (i.e, the total cost to firm 1 of producing
Answer:
Attached is a solution hope it helps:
A gas refrigeration system using air as the working fluid has a pressure ratio of 5. Air enters the compressor at 0°C. The high-pressure air is cooled to 35°C by rejecting heat to the surroundings. The refrigerant leaves the turbine at −80°C and then it absorbs heat from the refrigerated space before entering the regenerator. The mass flow rate of air is 0.4 kg/s. Assuming isentropic efficiencies of 80 percent for the compressor and 85 percent for the turbine and using constant specific heats at room temperature, determine (a) the effectiveness of the regenerator, (b) the rate of heat removal from
Answer:
(a) Effectiveness of the regenerator= 0.433
(b) The rate of heat removal=21.38 kW
Explanation:
The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.
Water is the working fluid in an ideal regenerative Rankine cycle with one closed feedwater heater. Superheated vapor enters the turbine at 10 MPa, 480 C and the condenser pressure is 6 kPa. Steam expands through the first stage turbine where some is extracted and diverted to a closed feedwater heater at 0.7 MPa. Condensate drains from the feedwater heater as saturated liquid at 0.7 MPa and is trapped into the condenser. The feedwater leaves the heater at 10 MPa and a temperature equal to the saturation temperature at 0.7 MPa.
For the cycle of problem 8.49, reconsider the analysis assuming the pump and each turbine stage have isentropic efficiencies of 80%.
Determine:
a) the rate of heat transfer to the working fluid passing through the steam generator in kJ per kg of steam entering the first stage turbine.
b) the thermal efficiency
c) the rate of heat transfer from the working fluid passing through the condenser to the cooling water in kJ per kg of steam entering the first stage turbine.
a) Heat transfer to the working fluid in the steam generator: 3303.9 kJ/kg
b) Thermal efficiency: 100%
c) Heat transfer from the working fluid in the condenser to the cooling water: 2083.6 kJ/kg
Given parameters:
- Superheated vapor enters turbine: P₁ = 10 MPa, T₁ = 480°C
- Condenser pressure: P₃ = 6 kPa
- Extraction pressure for closed feedwater heater: P₂ = 0.7 MPa
- Isentropic efficiencies of pump and turbine stages: ηᵥ = ηₜ = 80%
a) Heat Transfer in the Steam Generator:
1. From the steam tables, at P₁ = 10 MPa:
- h₁ = 3478.1 kJ/kg (enthalpy of superheated vapor)
2. At P₂ = 0.7 MPa:
- h₂ = 209.4 kJ/kg (enthalpy of extracted steam)
3. The enthalpy at the inlet of the closed feedwater heater is the same as h₂:
- h₃ = 209.4 kJ/kg
4. At P₃ = 6 kPa, as condensate:
- h₄ = 191.8 kJ/kg (enthalpy of saturated liquid)
5. The enthalpy at the outlet of the closed feedwater heater is the same as h₄:
- h₅ = 191.8 kJ/kg
6. Pump work per unit mass of water: [tex]W_{\text{pump}}[/tex]= h₅ - h₃
- [tex]\( W_{\text{pump}} = 191.8 - 209.4 = -17.6 \, \text{kJ/kg} \)[/tex]
7. Actual work output from turbine per unit mass of steam: [tex]W_{\text{act}}[/tex]= h₁ - h₂
[tex]\( W_{\text{act}} = 3478.1 - 209.4 = 3268.7 \, \text{kJ/kg} \)[/tex]
8. Isentropic work output from turbine per unit mass of steam: [tex]\( W_{\text{isentropic}} = \frac{W_{\text{act}}}{\eta_t} \)[/tex]
[tex]\( W_{\text{isentropic}} = \frac{3268.7}{0.8} = 4085.9 \, \text{kJ/kg} \)[/tex]
9. Heat input per unit mass of steam: [tex]Q_{\text{in}}[/tex] = h₁ - h₅
[tex]\( Q_{\text{in}} = 3478.1 - 191.8 = 3286.3 \, \text{kJ/kg} \)[/tex]
10. Heat transfer in the steam generator: [tex]\( Q_{\text{gen}} = Q_{\text{in}} - |W_{\text{pump}}| \)[/tex]
- [tex]\( Q_{\text{gen}} = 3286.3 - |-17.6| = 3303.9 \, \text{kJ/kg} \)[/tex]
- Rounded to one decimal place: 3303.9 kJ/kg
b) Thermal Efficiency:
Thermal efficiency (η) is given by: [tex]\( \eta = \frac{W_{\text{net}}}{Q_{\text{in}}} \)[/tex]
where[tex]\( W_{\text{net}} = W_{\text{act}} - |W_{\text{pump}}| \)[/tex]
[tex]\( \eta = \frac{3268.7 - |-17.6|}{3286.3} \times 100 \)\\\( \eta = \frac{3286.3}{3286.3} \times 100 \)\\\( \eta = 1 \times 100 \)\\\( \eta = 100\% \)[/tex]
c) Heat Transfer in the Condenser:
Heat transfer from working fluid in the condenser to cooling water:
[tex]Q_{\text{out}}[/tex]= h₄ - h₃
[tex]Q_{\text{out}} = 191.8 - 209.4 = -17.6 \, \text{kJ/kg}[/tex]
- Rounded to one decimal place: -17.6 kJ/kg ≈ -2083.6 kJ/kg
a) To find the heat transfer in the steam generator, we considered the enthalpies at different stages and calculated the pump work, actual work output from the turbine, isentropic work, heat input, and finally the heat transfer in the steam generator.
b) The thermal efficiency was calculated using the formula for efficiency, considering the net work output and heat input.
c) Heat transfer in the condenser was determined by comparing the enthalpies at the inlet and outlet of the closed feedwater heater, considering the negative sign due to the direction of heat flow from the working fluid to the cooling water.
Part A - Transmitted power A solid circular rod is used to transmit power from a motor to a machine. The diameter of the rod is D = 2.5 in and the machine operates at ω = 170 rpm . If the allowable shear stress in the shaft is 10.4 ksi , what is the maximum power transmissible to the machine? Express your answer with appropriate units to three significant figures.
The maximum transmitted power throughout the solid circular rod is approximately 86.066 horse power.
How to calculate the maximum transmitted power through a solid circular rodLet consider that the solid circular rod experiments an uniform torsion and that the material represents a ductile material.
The solid circular rod rotates at constant angular speed ([tex]\omega[/tex]), in radians per second, and all losses throughout the system are negligible. Net power ([tex]\dot W[/tex]), in horse power, is calculated as the product of the torque ([tex]T[/tex]), in kilopound-force-inches, and the angular speed. That is to say:
[tex]\dot W = T \cdot \omega[/tex] (1)
In addition, we add a formula that relates the torque with the shear stress experimented by the solid circular rod:
[tex]\tau = \frac{T \cdot D}{2 \cdot J}[/tex] (2)
Where [tex]J[/tex] is the torsion module, in quartic inches, whose formula for a solid circular cross section is described below:
[tex]J = \frac{\pi \cdot D^{4}}{32}[/tex] (3)
The torsion module is calculated herein: ([tex]D = 2.5\,in[/tex])
[tex]J = 3.835\, in^{4}[/tex]
The maximum allowable torsion is found by isolating the corresponding variable in (2): ([tex]J = 3.835 in^{4}[/tex], [tex]D = 2.5\,in[/tex], )
[tex]T_{max} = \frac{2 \cdot J \cdot \tau_{max}}{D}[/tex]
[tex]T_{max} = 31.907\, kip \cdot in\,(2.659 \,kip \cdot ft)[/tex]
Then, the maximum transmitted power is determined by (1):
[tex]\dot W = (2.659 kip \cdot ft) \cdot (2)\cdot (\pi) \cdot (170 rpm)\\\dot W \approx 2840.188 \,\frac{kip \cdot ft}{min} \\\dot W \approx 86.066 \,h.p.[/tex]
The maximum transmitted power throughout the solid circular rod is approximately 86.066 horse power. [tex]\blacksquare[/tex]
To learn more on power transmission, we kindly invite to check this verified question: https://brainly.com/question/12750305
Develop a three-month moving average for this time series. Compute MSE and a forecast for month 8. If required, round your answers to two decimal places. Do not round intermediate calculation.
Answer:
The missing Table is;
Month 1 2 3 4 5 6 7
Value 24 13 21 14 20 23 15
MSE = 20.56
Forecast = 19.33
Explanation:
Table Formation
Month Value Forecast Error Absolute Error Error Square
1 24
2 13
3 21
4 14 19.33333 -5.33333 5.333333 28.44444
5 20 16 4 4 16
6 23 18.33333 4.666667 4.666667 21.77778
7 15 19 -4 4 16
20.55
19.33
Table Explanation
In the above table, we first calculate the forecast column. A forecast is taken as the sum of the three entries in order and then taking out the average.
Error is calculated by subtracting the Actual value from the forecast value.
Then we take the absolute of the forecast value.
Then we take the square of the absolute of the forecasted values.
Now we calculate MSE as;
MSE = ( 28.44 + 16 + 21.78 + 16 ) / 4
MSE = 20.56
Forecast for 8th month = ( 20 + 23 + 15 ) / 3
Forecast for 8th month = 19.33
Q29. The human body works as a heat engine, converting ______ % of food energy to the mechanical energy necessary to carry on daily physical activities, while disposing of the rest as waste heat.
Answer:
Explanation:20% of food energy is converted into mechanical energy
A tire-pressure monitoring system warns you with a dashboard alert when one of your car tires is significantly under-inflated.
How does this pressure monitoring system work? Many cars have sensors that measure the period of rotation of each wheel. Based on the period of rotation, the car’s computer determines the relative size of the tire when the car is moving at a certain speed.
(a) Derive a mathematical equation that relates the period T of the wheel rotation of the tire radius r and the speed v of the car.
(b) The figure below shows a properly inflated tire (left) and an under-inflated tire (right) of the same car. Estimate the percent change of the period of the underinflated tire compared to the properly inflated tire.
The inflated tire has a radius of 303 mm and the underinflated tire has a radius of 293mm. (Radius is from the ground to center of the wheel.)
(c) Estimate the distance this car would have to travel in order for one wheel to make one more complete rotation than the other. Which will undergo more turns, the under-inflated or the properly inflated tire?
Answer:
The answers to the question are
(a) T = 2·π·r/v
(b) 3.3 % change in period of the under-inflated tire compared to the properly inflated tire
(c) Therefore the distance this car would have to travel in order for under-inflated tire to make one more complete rotation than the inflated tire is 57.685 meters.
Explanation:
(a) The period T = 2π/ω
The velocity v = ωr or ω = v/r
Therefore T = 2π/(v/r) = 2πr/v
T = 2·π·r/v
(b) Period of properly inflated tire with radius = 303 mm is 2π303/v
Period of under-inflated tire with radius = 293 mm is 2π293/v
Therefore we have percentage change in period of of the under-inflated tire compared to the properly inflated tire is given by
(2π303/v -2π293/v)/(2π303/v) = 2π10/v/(2π303/v) = 10/303 × 100 = 3.3 %
(c) The period of the under-inflated tire is 10/303 less than that of the inflated tire. Therefore for the under-inflated tire to make one complete turn more than the inflated tire, we have 1/(10/303) = 303/10 or 30.3 revolutions of either tire which is 30.3×2×π×303 = 57685.296 mm = 57.685 meters
Therefore the distance this car would have to travel in order for under-inflated tire to make one more complete rotation than the inflated tire is 57.685 meters
At 30.3 revolutions the distance covered by the under-inflated
= 55781.49 mm
Subtracting the two distances gives
1903.805 mm
The circumference of the inflated tire = 2×π×303 = 1903.805 mm
A) The mathematical equation that relates the period T of the wheel rotation of the tire radius r and the speed v of the car is; T = 2πr/v
B) The percent change of the period of the underinflated tire compared to the properly inflated tire is; 3.3 %
C) The distance this car would have to travel in order for under-inflated tire to make one more complete rotation than the inflated tire is; 57685.296 mm
Relationship between motion of Cars and it's tires
(A) The formula for period is;
T = 2π/ω
angular velocity is;
ω = v/r
Thus; T = 2πr/v
where T is period, v is velocity and r is radius
(b) We are told that radius is 303 mm. Thus Period of inflated tire with radius is;
T = 2π × 303/v
Period of under-inflated tire with radius of 293 mm is;
T = 2π× 293/v
Percentage Change is;
((2π × 303/v) - (2π× 293/v))/(2π × 303/v) * 100% = 10/303 * 100% = 3.3 %
(c) From B above we see that the period of the under-inflated tire is 10/303 less than that of the inflated tire.
Thus, for the under-inflated tire to make one complete turn more than the inflated tire, we have 1/(10/303) = 30.3 revolutions of either tire.
Thus, the distance is;
d = 30.3 × 2π × 303
d = 57685.296 mm
Using the same concept, at 30.3 revolutions, the distance covered by the under-inflated tire is calculated to be; 55781.49 mm
Difference in distance = 57685.296 mm - 55781.49 mm
Difference in distance = 1903.805 mm
Read more about motion of cars at; https://brainly.com/question/1315051
a triangle is defined by the three vertices. write the following functions of the triangle class assume that the point class has already been written for you you may add helper functions as needed
Answer:
The triangle() function is an inbuilt function in p5.js which is used to draw a triangle in a plane. This function accepts three vertices of triangle.
Syntax:
triangle(x1, y1, x2, y2, x3, y3)
Below programs illustrates the triangle() function in p5.js:
function setup() {
createCanvas(400, 400);
}
function draw() {
background(220);
fill('lightgreen');
// A triangle at (100, 250), (250, 170) and (330, 300)
triangle(100, 250, 250, 170, 330, 300);
}
Foundation dampproofing is most commonly installed: Select one: a. on both the inside and outside of the basement wall b. Basements do not need dampproofing beyond the basement wall itself c. on the inside of the basement wall d. on the outside of the basement wall
Answer:A
Explanation:
Damp roof is generally applied at basement level which restrict the movement of moisture through walls and floors. Therefore it could be inside or the outside basement walls.
computer language C++ (Connect 4 game)( this is all the info that was givin no input or solution) I used the most recent version of Microsoft visual studio for this program)Your solution must have these classes and at least these fields / methods. Your program must use the methods described in a meaningful fashion. You can create additional fields or methods but these the below should be core methods in your solution.Checkers classfields:colormethods:Constructorint getColor()string toString() - returns an attractive string representation of itselfBoard classfields:2D array of checkers - represents the game board (6 row by 7 column)methods:Constructorstring toString() - returns an attractive string representation of itselfThis needs to be an attractive table representation of the game board with aligned rows / columns with characters representing blank cells, cells with a red checkers in it and cells with a gold checkers in it.int dropChecker(Checker, int) - drops a checker object down a particular slot. Returns an int indicating success or not.int isWinner() - returns code number representing no winner or player number of winnerPlayer classfields:namecolormethods:Constructorget / set methods for fieldsstring toString() - string representation of itself.Example String representation of the board (X means blank, R means red checker, G means gold checker)X X X X X X XX X X X X X XX X X X X G XX X X X X R XX X X X X R XX R G G G R GMainYour program should create two Players and a Board. It should represent the initial Board to the screen.You should then alternate between player 1 and player 2 and randomly drop checkers down slots in the board. You should represent the board after each checker drop.This should stop if there is a winner or if the board fills up.You should be turning in a Checker.h, Checker.cpp, Board.h, Board.cpp, Player.h, Player.cpp and a main.cpp
Answer:
C++ code explained below
Explanation:
#include "hw6.h"
//---------------------------------------------------
// Constructor function
//---------------------------------------------------
Connect4::Connect4()
{
ClearBoard();
}
//---------------------------------------------------
// Destructor function
//---------------------------------------------------
Connect4::~Connect4()
{
// Intentionally empty
}
//---------------------------------------------------
// Clear the Connect4 board
//---------------------------------------------------
void Connect4::ClearBoard()
{
// Initialize Connect4 board
for (int c = 0; c < COLS; c++)
for (int r = 0; r < ROWS; r++)
board[r][c] = ' ';
// Initialize column counters
for (int c = 0; c < COLS; c++)
count[c] = 0;
}
//---------------------------------------------------
// Add player's piece to specified column in board
//---------------------------------------------------
bool Connect4::MakeMove(int col, char player)
{
// Error checking
if ((col < 0) || (col >= COLS) || (count[col] >= ROWS))
return false;
// Make move
int row = count[col];
board[row][col] = player;
count[col]++;
return true;
}
//---------------------------------------------------
// Check to see if player has won the game
//---------------------------------------------------
bool Connect4::CheckWin(char player)
{
// Loop over all starting positions
for (int c = 0; c < COLS; c++)
for (int r = 0; r < ROWS; r++)
if (board[r][c] == player)
{
// Check row
int count = 0;
for (int d = 0; d < WIN; d++)
if ((r+d < ROWS) &&
(board[r+d][c] == player)) count++;
if (count == WIN) return true;
// Check column
count = 0;
for (int d = 0; d < WIN; d++)
if ((c+d < COLS) &&
(board[r][c+d] == player)) count++;
if (count == WIN) return true;
// Check first diagonal
count = 0;
for (int d = 0; d < WIN; d++)
if ((r+d < ROWS) && (c+d < COLS) &&
(board[r+d][c+d] == player)) count++;
if (count == WIN) return true;
// Check second diagonal
count = 0;
for (int d = 0; d < WIN; d++)
if ((r-d >= 0) && (c+d < COLS) &&
(board[r-d][c+d] == player)) count++;
if (count == WIN) return true;
}
return false;
}
//---------------------------------------------------
// Print the Connect4 board
//---------------------------------------------------
void Connect4::PrintBoard()
{
// Print the Connect4 board
for (int r = ROWS-1; r >= 0; r--)
{
// Draw dashed line
cout << "+";
for (int c = 0; c < COLS; c++)
cout << "---+";
cout << "\n";
// Draw board contents
cout << "| ";
for (int c = 0; c < COLS; c++)
cout << board[r][c] << " | ";
cout << "\n";
}
// Draw dashed line
cout << "+";
for (int c = 0; c < COLS; c++)
cout << "---+";
cout << "\n";
// Draw column numbers
cout << " ";
for (int c = 0; c < COLS; c++)
cout << c << " ";
cout << "\n\n";
}
//---------------------------------------------------
// Main program to play Connect4 game
//---------------------------------------------------
int main()
{
int choice;
int counter = 0;
srand (time(NULL));
Connect4 board;
cout << "Welcome to Connect 4!" << endl << "Your Pieces will be labeled 'H' for human. While the computer's will be labeled 'C'" << endl;
board.PrintBoard();
cout << "Where would you like to make your first move? (0-6)";
cin >> choice;
while (board.MakeMove(choice,'H') == false){
cin >> choice;
}
counter++;
while (board.CheckWin('C') == false && board.CheckWin('H') == false && counter != 21){
while (board.MakeMove(rand() % 7, 'C') == false){}
board.PrintBoard();
cout << "Where would you like to make your next move?" << endl;
cin >> choice;
board.MakeMove(choice,'H');
while (board.MakeMove(choice,'H') == false){
cin >> choice;
}
counter++;
}
if (board.CheckWin('C')){
cout << "Computer Wins!" << endl;}
else if (counter == 21){cout << "Tie Game!" << endl;}
else {cout << "Human Wins!" << endl;}
board.PrintBoard();
}
Consider a room with dimensions as in the sketch. Assuming all emissivities of surfaces are equal to 0.9, calculate the radiative heat transfer between the floor and the ceiling, if you know the surface temperature of the floor equal to 25o C and the surface temperature of the ceiling is equal to 10o C
Explanation:
Below is an attachment containing the solution.
The radiative heat transfer between the floor and the ceiling is 1,534.55 kW.
Given the following data:
Emissivity of surfaces = 0.9.Surface temperature of the floor = 25°C °C to K = [tex]273 +25[/tex] = 298 K.Surface temperature of the ceiling = 10°C to K = [tex]273 +10[/tex] = 283 K.View factor = 0.2.Boltzmann's constant = [tex]5.67 \times 10^{-8}[/tex]Length of room = 4.45 m.How to calculate the radiative heat transfer.Mathematically, the radiative heat transfer between the floor and the ceiling is given by this formula:
[tex]Q=\epsilon \sigma A(T_2^4-T_1^4)[/tex]
Where:
[tex]\sigma[/tex] is Boltzmann's constant.A is the area.[tex]\epsilon[/tex] is the emissivity.Substituting the given parameters into the formula, we have;
[tex]Q=0.9 \times 5.67 \times 10^{-8} \times 4.45^2 \times (298^4-283^4)\\\\Q=0.00000005103 \times 20.4304 \times 1471902495[/tex]
Q = 1,534.55 kW.
Read more on radiative heat here: https://brainly.com/question/14267608