Answer:
B
Explanation:
Torsion is application of torque to a shaft to turn it about its longitudinal axis. When torque is applied to a shaft the circle remains unchanged in a circular state, its cross section does not warp but remains flat with a straight radial lines but its longitudinal lines changes into an helix intersecting the circular shaft
A car is traveling at 70 mi/h on a level section of road with good, wet pavement. Its antilock braking system (ABS) only starts to work after the brakes have been locked for 100 ft. If the driver holds the brake pedal down completely, immediately locking the wheels, and keeps the pedal down during the entre process, how many feet will it take the car to stop from the point of initial brake application? (The braking efficiency is 80% with the ABS not working and 100% with the ABS working. Use theoretical stopping distance and ignore air resistance. Let frl = 0.02 when the brakes are locked, but compute the frl once the ABS becomes active.)
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]
Set up the differential equation that describes the motion under the assumptions of this section. Solve the differential equation. State whether the motion of the spring-mass system is harmonic, damped oscillation, critically damped, oroverdamped. If the motion is a damped oscillation, rewrite in the form (22).33. The spring-mass system has an attached mass of 10 g. The spring constant is30 g/s 2 . A dashpot mechanism is attached, which has a damping coefficient of 40 g/s. The mass is pulled down and released. At time t = 0, the mass is 3 cmbelow the rest position and moving upward at 5 cm/s.34. A long spring has a mass of 1 slug attached to it. The spring stretches 16/13 ftand comes to rest. The damping coefficient is 2 slug/s. The mass is subjected toan impulsive force at time t = 0, which imparts a velocity of 5 ft/s downward.
Answer:
Explanation:
The principle and application of differential equation is applied and the steps with detailed and appropriate substitution is as carefully shown in the attached files.
m = mass attached
k = spring constant
c = damping coefficient
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.
(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
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
The heat required to raise the temperature of m (kg) of a liquid from T1 to T2 at constant pressure is Z T2CpT dT (1) In high school and in first-year college physics courses, the formula is usually given asQ ΔH m Q mCp ΔT mCpT2 ␣T1 (2)T1(a) What assumption about Cp is required to go from Equation 1 to Equation 2? (b) The heat capacity (Cp) of liquid n-hexane is measured in a bomb calorimeter. A small reaction flask (the bomb) is placed in a well-insulated vessel containing 2.00L of liquid n–C6H14 at T 300 K. A combustion reaction known to release 16.73 kJ of heat takes place in the bomb, and the subsequent temperature rise of the system contents is measured and found to be 3.10 K. In a separate experiment, it is found that 6.14kJ of heat is required to raise the temperature of everything in the system except the hexane by 3.10 K. Use these data to estimate Cp[kJ/(mol K)] for liquid n-hexane at T 300 K, assuming that the condition required for the validity ofEquation 2 is satisfied. Compare your result with a tabulated value.
Answer:
(a)
dQ = mdq
dq = [tex]C_p[/tex]dT
[tex]q = \int\limits^{T_2}_{T_1} {C_p} \, dT[/tex] = [tex]C_p[/tex] (T₂ - T₁)
From the above equations, the underlying assumption is that [tex]C_p[/tex] remains constant with change in temperature.
(b)
Given;
V = 2L
T₁ = 300 K
Q₁ = 16.73 KJ , Q₂ = 6.14 KJ
ΔT = 3.10 K , ΔT₂ = 3.10 K for calorimeter
Let [tex]C_{cal}[/tex] be heat constant of calorimeter
Q₂ = [tex]C_{cal}[/tex] ΔT
Heat absorbed by n-C₆H₁₄ = Q₁ - Q₂
Q₁ - Q₂ = m [tex]C_p[/tex] ΔT
number of moles of n-C₆H₁₄, n = m/M
ρ = 650 kg/m³ at 300 K
M = 86.178 g/mol
m = ρv = 650 (2x10⁻³) = 1.3 kg
n = m/M => 1.3 / 0.086178 = 15.085 moles
Q₁ - Q₂ = m [tex]C_p[/tex]' ΔT
[tex]C_p[/tex] = (16.73 - 6.14) / (15.085 x 3.10)
[tex]C_p[/tex] = 0.22646 KJ mol⁻¹ k⁻¹
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);
}
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:
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.
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]
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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.
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.
Eureka is a telephone and Internet-based concierge services that specializes in obtaining things that are hard to find (e.g. Super Bowl tickets, first edition books from the 1500s, Faberge eggs). It currently employs 60 staff members who collectively provide 24-hour coverage (over three shifts). They answer the phones and respond to requests entered on the Eureka! Web site. Much of their work is spent on the phone and on computers searching on the Internet. The company has just leased a new office building and is about to wire it. They have bids from different companies to install (1) 100-Base T network (2) Wi-Fi network What would you recommend
attached below is the answer
I would recommend installing both a 100-Base T network and a Wi-Fi network in the new office building for Eureka.
Explanation:Based on the information provided, I would recommend installing both a 100-Base T network as well as a Wi-Fi network in the new office building for Eureka.
A 100-Base T network, also known as Ethernet, provides reliable and high-speed wired connections, perfect for the staff members who spend a significant amount of time on their computers searching the Internet. It ensures a stable and fast connection for their work.
On the other hand, a Wi-Fi network allows for flexible and convenient wireless connectivity, which is beneficial for staff members who frequently use their phones or move around the office space. It provides them with the freedom to connect wirelessly wherever they are in the building.
Learn more about Network installation here:https://brainly.com/question/33827192
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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
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
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
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.
Air at 26 kPa, 230 K, and 220 rn/s enters a turbojet engine in flight. The air mass flow rate is 25 kg/s. The compressor pressure ratio is 11, the turbine inlet temperature is 1400 K, and air exits the nozzle 26 kPa. The diffuser and the nozzle processes are isentropic, the compressor and turbine have isentropic efficiencies of 85% and 90%, respectively. There is no pressure drop for flow through the combustor. Kinetic energy is negligible everywhere except at the diffuser inlet and the nozzle exit. On the basis of air-standard analysis, determine: (a)the pressures and temperatures at each state, (b)the rate of heat addition to the air passing through the combustor, (c) the velocity at the nozzle exit.
Answer:
Explanation:
Answer:
Explanation:
Answer:
Explanation:
This is a little lengthy and tricky, but nevertheless i would give a step by step analysis to make this as simple as possible.
(a). here we are asked to determine the Temperature and Pressure.
Given that the properties of Air;
ha = 230.02 KJ/Kg
Ta = 230 K
Pra = 0.5477
From the energy balance equation for a diffuser;
ha + Va²/2 = h₁ + V₁²/2
h₁ = ha + Va²/2 (where V₁²/2 = 0)
h₁ = 230.02 + 220²/2 ˣ 1/10³
h₁ = 254.22 KJ/Kg
⇒ now we obtain the properties of air at h₁ = 254.22 KJ/Kg
from this we have;
Pr₁ = 0.7329 + (0.8405 - 0.7329)[(254.22 - 250.05) / (260.09 - 250.05)]
Pr₁ = 0.77759
therefore T₁ = 254.15K
P₁ = (Pr₁/Pra)Pa
= 0.77759/0.5477 ˣ 26
P₁ = 36.91 kPa
now we calculate Pr₂
Pr₂ = Pr₁ (P₂/P₁) = 0.77759 ˣ 11 = 8.55349
⇒ now we obtain properties of air at
Pr₂ = 8.55349 and h₂ = 505.387 KJ/Kg
calculating the enthalpy of air at state 2
ηc = h₁ - h₂ / h₁ - h₂
0.85 = 254.22 - 505.387 / 254.22 - h₂
h₂ = 549.71 KJ/Kg
to obtain the properties of air at h₂ = 549.71 KJ/Kg
T₂ = 545.15 K
⇒ to calculate the pressure of air at state 2
P₂/P₁ = 11
P₂ = 11 ˣ 36.913
p₂ = 406.043 kPa
but pressure of air at state 3 is the same,
i.e. P₂ = P₃ = 406.043 kPa
P₃ = 406.043 kPa
To obtain the properties of air at
T₃ = 1400 K, h₃ = 1515.42 kJ/Kg and Pr = 450.5
for cases of turbojet engine,
we have that work output from turbine = work input to the compressor
Wt = Wr
(h₃ - h₄) = (h₂ - h₁)
h₄ = h₃ - h₂ + h₁
= 1515.42 - 549.71 + 254.22
h₄ = 1219.93 kJ/Kg
properties of air at h₄ = 1219.93 kJ/Kg
T₄ = 1140 + (1160 - 1140) [(1219.93 - 1207.57) / (1230.92 - 1207.57)]
T₄ = 1150.58 K
Pr₄ = 193.1 + (207.2 - 193.1) [(1219.93 - 1207.57) / (1230.92 - 1207.57)]
Pr₄ = 200.5636
Calculating the ideal enthalpy of the air at state 4;
Лr = h₃ - h₄ / h₃ - h₄*
0.9 = 1515.42 - 1219.93 / 1515.42 - h₄
h₄* = 1187.09 kJ/Kg
now to obtain the properties of air at h₄⁻ = 1187.09 kJ/Kg
P₄* = 179.7 + (193.1 - 179.7) [(1187.09 -1184.28) / (1207.57 - 1184.28)]
P₄* = 181.316
P₄ = (Pr₄/Pr₃)P₃ i.e. 3-4 isentropic process
P₄ = 181.316/450.5 * 406.043
P₄ = 163.42 kPa
For the 4-5 process;
Pr₅ = (P₅/P₄)Pr₄
Pr₅ = 26/163.42 * 200.56 = 31.9095
to obtain the properties of air at Pr₅ = 31.9095
h₅= 724.04 + (734.82 - 724.04) [(31.9095 - 3038) / (32.02 - 30.38)]
h₅ = 734.09 KJ/Kg
T₅ = 710 + (720 - 710) [(31.9095 - 3038) / (32.02 - 30.38)]
T₅ = 719.32 K
(b) Now we are asked to calculate the rate of heat addition to the air passing through the combustor;
QH = m(h₃-h₂)
QH = 25(1515.42 - 549.71)
QH = 24142.75 kW
(c). To calculate the velocity at the nozzle exit;
we apply steady energy equation of a flow to nozzle
h₄ + V₄²/2 = h₅ + V₅²/2
h₄ + 0 = h₅₅ + V₅²/2
1219.9 ˣ 10³ = 734.09 ˣ 10³ + V₅²/2
therefore, V₅ = 985.74 m/s
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
(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
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
For a bronze alloy, the stress at which plastic deformation begins is 275 MPa and the modulus of elasticity is 115 GPa. If the original specimen length is listed above, what is the maximum length to which it may be stretched without causing plastic deformation
Answer:
maximum length to which it may be stretched without causing plastic deformation is 200 + 0.4782 = 2.4782 mm
Explanation:
given data
stress = 275 MPa
modulus of elasticity = 115 GPa
solution
we will apply here modulus of elasticity formula that is
modulus of elasticity = stress ÷ strain ....................1
here we consider original length is 200 mm
so strain is = [tex]\frac{\triangle L}{original\ length }[/tex] ..........2
put here value in equation 1 we get
115 × [tex]10^{9}[/tex] = [tex]\frac{275 \times 10^{6}}{\frac{x}{200} }[/tex]
solve it we get
x = [tex]\frac{11}{23}[/tex]
x = 0.4782
so maximum length to which it may be stretched without causing plastic deformation is 200 + 0.4782 = 2.4782 mm
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
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]
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:
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:
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:
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
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();
}
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.