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.
In order to test the feasibility of drying a certain foodstuff, drying data were obtained in a tray dryer with air flow over the top exposed surface having an area of 0.186 m2. The bone dry sample weight was 3.765 kg dry solid. At equilibrium after a long period, the wet sample weight was 3.955 kg (water plus solid). Hence, 3.955-3.765 or 0.190 kg of equilibrium moisture was present. The following table provides the sample weights versus time during the drying test:
Time (hr)
Weight (kg)
Time (hr)
Weight (kg)
Time (hr)
Weight (kg)
0
4.944
2.2
4.554
7.0
4.019
0.4
4.885
3.0
4.404
9.0
3.978
0.8
4.808
4.2
4.241
12.0
3.955
1.4
4.699
5.0
4.150
A) Calculate the free moisture content, X (kg water per kg dry solid) for each data point and plot X versus t;
B) Using the slope, calculate the drying rate, R in kg water per hour per square meter and plot R versus X;
C) Using this drying rate curve, predict the total time to dry the sample from X = 0.20 to X = 0.04. Use numerical integration in the falling rate period. What is the drying rate in the constant rate period and what is X in the constant rate period?
The question involves calculating free moisture content, plotting it against time, determining the drying rate, and predicting the total drying time using numerical integration during the falling rate period. The constant and falling rate periods of drying are crucial in understanding the sample's drying behavior.
Explanation:The student is tasked with calculating the free moisture content ( extit{X}) for various data points during the drying of a foodstuff in a tray dryer, plotting extit{X} versus time (t), determining the drying rate ( extit{R}), and predicting the total drying time using numerical integration for the falling rate period of drying.
Firstly, free moisture content is calculated by finding the difference in sample weight at each time from the bone-dry weight and dividing by the dry solid weight (3.765 kg).Plot this free moisture content versus time to visualize the drying curve.Next, by calculating the slope of the initial linear portion of the plot, the constant drying rate can be obtained. This is the weight of water removed per hour per square meter of exposed surface area.Then determine the drying rate at various intervals.Plot the drying rate versus free moisture content ( extit{X}).Finally, use numerical methods to integrate the drying rate over the moisture content to predict the total drying time from a moisture content of 0.20 to 0.04.The falling rate period is characterized by a decreasing drying rate which requires numerical integration to find the total drying time. The constant rate period is defined by a uniform drying rate and typically occurs before the falling rate period.
The detailed answer provides calculations for free moisture content, drying rate, and prediction of total drying time. It also includes instructions for plotting the moisture ratio vs. time.
Calculating the Free Moisture Content:
To calculate the free moisture content (X) for each data point, subtract the bone dry weight from the wet weight. Then, divide by the bone dry weight.
Calculating Drying Rate and Predicting Total Drying Time:
Determine the drying rate using the slope of the moisture ratio vs. time plot. Then, use numerical integration to predict the time to dry the sample from X = 0.20 to X = 0.04. Identify the drying rate in the constant rate period and X in that period.
Moisture Ratio vs. Time Plot:
Plot the moisture ratio (MR) versus time to observe the constant rate and falling rate periods during the drying.
(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
2. The initially velocity of the box and truck is 60 mph. When the truck brakes such that the deceleration is constant it takes the truck 350 ft to come to rest. During that time the box slides 10 ft and slams into the end of the truck at B. If the coeff of kinetic friction is 0.3, at what speed relative to the truck does the box hit at B? Ans: 5.29 ft/s.
Answer:
Speed with which the box hits the truck at B relative to the truck = 5.29 ft/s
Explanation:
First of, we calculate the deceleration of the truck+box setup using the equations of motion.
x = distance tavelled during the deceleration = 350 ft
u = initial velocity of the truck = 60 mph = 88 ft/s
v = final velocity of the truck = 0 m/s
a = ?
v² = u² + 2ax
0² = 88² + 2(a)(350)
700 a = - 88²
a = - 11.04 ft/s²
But this deceleration acts on the crate as a force trying to put the box in motion.
The motion of the box will be due to the net force on the box
Net force on the box = (Force from deceleration of the truck) - (Frictional force)
Net force = ma
Frictional Force = μmg = 0.3 × m × 32.2 = 9.66 m
Force from the deceleration of the truck = m × 11.04 = 11.04 m
ma = 11.04m - 9.66m
a = 11.04 - 9.66 = 1.4 ft/s²
This net acceleration is now responsible for its motion from rest, through the 10 ft that the box moved, to point B to hit the truck.
x = 10 ft
a = 1.4 ft/s²
u = 0 m/s (box starts from rest, relative to the truck)
v = final velocity of the box relative to the truck, before hitting the truck's wall = ?
v² = u² + 2ax
v² = 0² + 2(1.4)(10)
v² = 28
v = 5.29 ft/s
(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
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
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
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.
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.
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
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.
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
Design a circuit with output f and inputs x1, x0, y1, and y0. Let X = x1x0 and Y = y1y0 represent two 2-digit binary numbers. The output f should be 1 if the numbers represented by X and Y are equal. Otherwise, f should be 0.
Complete question
The complete question is shown on the first uploaded image
Answer:
a The table is shown on the second uploaded image
b The simplest possible sum product is shown on the fifth uploaded image
Explanation:
The explanation is shown on the third fourth and fifth uploaded image
A 2-m3 rigid tank initially contains air at 100 kPa and 22°C. The tank is connected to a supply line through a valve. Air is flowing in the supply line at 600 kPa and 22°C. The valve is opened, and air is allowed to enter the tank until the pressure in the tank reaches the line pressure, at which point the valve is closed. A thermometer placed in the tank indicates that the air temperature at the final state is 77°C. Determine (a) the mass of air that has entered the tank and (b) the amount of heat transfer
Answer:
9.58 Kg of air has entered the tank.
heat entered=3483.76 Kilo.Joule
Explanation:
(A) R=287 Kilo.J/Kg.K
as per initial conditions P=100 Kilo.Pa ,V=2 cubic meter, T=22 C=295.15 K,
using the relation P*V=m*R*T
m=(100*1000*2)/(287*295.15)=2.36 Kg this is the mass that is already present in tank.
after filling tank at 600 Kilo.Pa.
P=600 Kilo Pa T=77 C=350.15 K
P*V=m*R*T
m=(600*1000*2)/(287*350.15)=11.94 Kg
mass that has entered=11.94-2.36=9.58 Kg
(b) using air psychometric property table
specific heat content initial 100 KILO Pa and 22 C=295.576 Kilo.Joule/Kg
specific heat content final 600 Kilo Pa and 77 C=350.194 Kilo.Joule/Kg
heat at initial stage=295.576*2.36=697.56 Kilo.Joule
heat at final stage=350.194*11.94=4181.32 Kilo.Joule
heat entered=4181.32-697.56=3483.76 Kilo.Joule
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.
Imagine two tanks. Tank A is filled to depth h with water. Tank B is filled to depth h with oil. Which tank has the largest pressure? Why? Where in the tank does the largest pressure occur?
Answer:
Tank A, due to higher density of water. At the bottom of the tank.
Explanation:
According to the theory of hydrostatics, pressure change is the product of density, gravity constant and depth. The higher the density, the higher the depth. As oil ([tex]\rho_{oil} = 920 \frac{kg}{m^{3}}[/tex]) has a density lower than in water ([tex]\rho_{water} = 1000 \frac{kg}{m^{3}}[/tex]), the largest pressure occur in tank A. The highest pressure occurs at the bottom of the tank A due to the fluid column.
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
It takes a resistance heater having 2 kW power to heat a room having 5 m X 5 m X 6 m size to heat from 0 to 33 oC at sea level. Calculate the amount of time needed for this heating to occur in min (Please take Patm=101 kPa, give your answer with three decimals, and do NOT enter units!!!).
Answer: 50 minutes
Explanation:
The energy needed to heat air inside the room is the electric energy dissipated by the resistance. It is known after using First Principle of Thermodynamics:
[tex]Q_{in,air} = \dot W_{dis, heater} \cdot \Delta t[/tex]
[tex]\rho_{air} \cdot V_{room} \cdot c_{p,air} \cdot \Delta T = \dot W_{dis,heater} \cdot \Delta t[/tex]
The needed time is:
[tex]\Delta t =\frac{\rho_{air}\cdot V_{room}\cdot c_{p,air} \cdot\Delta T}{\dot W_{dis,heater}}[/tex]
Where [tex]\rho_{air} = 1.20 \frac{kg}{m^{3}}[/tex] and [tex]c_{p,air} = 1.012 \frac{kJ}{kg \cdot ^{\circ} C}[/tex]:
[tex]\Delta t = 3005.640 s (50.094 min)[/tex]
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
The primary mirror of a large telescope can have a diameter of 10 m and a mosaic of 36 hexagonal segments with the orientation of each segment actively controlled. Suppose this unity feedback system for the mirror segments has the loop transfer function
L(s) = Gc (s)G(s) = K / s(s^2 + 2s + 5) .
(a) Find the asymptotes and sketch them in the s-plane.
(b) Find the angle of departure from the complex poles.
(c) Determine the gain when two roots lie on the imaginary axis.
(d) Sketch the root locus.
Answer:
b. Angle of departure=26.56 degrees
c. √5j
Please see attachment for the sketch and step by step guide.
A photovoltaic panel of dimension 2m×4m is installed on the
roof of a home. The panel is irradiated with a solar flux of GS =700W /m2, oriented normal to
the top panel surface. The absorptivity of the panel to the solar
Irradiation is s =0.83, and the efficiency of conversion of the absorbed flux to electrical power
is P /sGsA =0.553−0.001Tp, where Tp is the panel temperature expressed in Kelvins and A is
the solar panel area. Determine the electric power generated for
a. A still summer day, in which Tsur =T[infinity]=35C, h=100W /m2 â‹…K, and
b. A breezy winter day, for which Tsur =T[infinity]=−15oC, h=30W /m2 ⋅K. The panel Emissivity is 0.90.
The electric power generated by a photovoltaic panel is calculated using the panel's dimensions, solar flux, absorptivity, conversion efficiency, and environmental conditions such as temperature and heat transfer coefficient, with different calculations for summer day and breezy winter day scenarios.
Explanation:Calculation of Electric Power Generated by a Photovoltaic Panel
To determine the electric power generated by a photovoltaic panel, we use the given parameters of the panel's dimensions, solar flux, absorptivity, efficiency of conversion, and the various conditions of the environment as provided in the question. On a still summer day with given temperature and heat transfer coefficient, we would need to calculate the temperature of the panel and use it in the efficiency formula to find the electrical power generated. Similarly, calculations would be performed for a breezy winter day. These calculations involve physics principles like heat transfer, solar radiation, and photovoltaic efficiency.
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.
Your program will be a line editor. A line editor is an editor where all operations are performed by entering commands at the command line. Commands include displaying lines, inserting text, editing lines, cutting and pasting text, loading and saving files. For example, a session where the user enters three lines of text and saves them as a new file may appear as:
Answer:
Java program given below
Explanation:
import java.util.*;
import java.io.*;
public class Lineeditor
{
private static Node head;
class Node
{
int data;
Node next;
public Node()
{data = 0; next = null;}
public Node(int x, Node n)
{data = x; next =n;}
}
public void Displaylist(Node q)
{if (q != null)
{
System.out.println(q.data);
Displaylist(q.next);
}
}
public void Buildlist()
{Node q = new Node(0,null);
head = q;
String oneLine;
try{BufferedReader indata = new
BufferedReader(new InputStreamReader(System.in)); // read data from terminals
System.out.println("Please enter a command or a line of text: ");
oneLine = indata.readLine(); // always need the following two lines to read data
head.data = Integer.parseInt(oneLine);
for (int i=1; i<=head.data; i++)
{System.out.println("Please enter another command or a new line of text:");
oneLine = indata.readLine();
int num = Integer.parseInt(oneLine);
Node p = new Node(num,null);
q.next = p;
q = p;}
}catch(Exception e)
{ System.out.println("Error --" + e.toString());}
}
public static void main(String[] args)
{Lineeditor mylist = new Lineeditor();
mylist.Buildlist();
mylist.Displaylist(head);
}
}
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.
214Bi83 --> 214Po84 + eBismuth-214 undergoes first-order radioactive decay to polonium-214 by the release of a beta particle, as represented by the nuclear equation above. Which of the following quantities plotted versus time will produce a straight line?(A) [Bi](B) [Po](C) ln[Bi](D) 1/[Bi]
Answer:
(C) ln [Bi]
Explanation:
Radioactive materials will usually decay based on their specific half lives. In radioactivity, the plot of the natural logarithm of the original radioactive material against time will give a straight-line curve. This is mostly used to estimate the decay constant that is equivalent to the negative of the slope. Thus, the answer is option C.
Answer:
Option C. ln[Bi]
Explanation:
The nuclear equation of first-order radioactive decay of Bi to Po is:
²¹⁴Bi₈₃ → ²¹⁴Po₈₄ + e⁻
The radioactive decay is expressed by the following equation:
[tex] N_{t} = N_{0}e^{-\lambda t} [/tex] (1)
where Nt: is the number of particles at time t, No: is the initial number of particles, and λ: is the decay constant.
To plot the variation of the quantities in function of time, we need to solve equation (1) for t:
[tex] Ln(\frac{N_{t}}{N_{0}}) = -\lambda t [/tex] (2)
Firts, we need to convert the number of particles of Bi (N) to concentrations, as follows:
[tex] [Bi] = \frac {N particles}{N_{A} * V} [/tex] (3)
[tex] [Bi]_{0} = \frac {N_{0} particles}{N_{A} * V} [/tex] (4)
where [tex]N_{A}[/tex]: si the Avogadro constant and V is the volume.
Now, introducing equations (3) and (4) into (2), we have:
[tex] Ln (\frac {\frac {[Bi]*N_{A}}{V}}{\frac {[Bi]_{0}*N_{A}}{V}}) = -\lambda t [/tex]
[tex] Ln (\frac {[Bi]}{[Bi]_{0}}) = -\lambda t [/tex] (5)
Finally, from equation (5) we can get a plot of Bi versus time in where the curve is a straight line:
[tex] Ln ([Bi]) = -\lambda t + Ln([Bi]_{0}) [/tex]
Therefore, the correct answer is option C. ln[Bi].
I hope it helps you!
The presence of free oxygen in the atmosphere is attributed mainly to Volcanic activity Development of plant life, especially algae Formation of the inner core Formation of the outer core
Answer:
Development of plant life, especially algae
Explanation:
The main source of oxygen in the atmosphere is the photosynthesis process that produces sugars and oxygen by utilizing carbon dioxide and water.Tiny oceanic plants called phytoplanktons have the ability to support life in water bodies for plants, animals and fish.The phytoplanktons consume most of the carbon dioxide in air and with the help of energy from the sun, they convert nutrients and carbon dioxide to complex organic compounds which are main source of plant material. Phytoplanktons are responsible for oxygen in water which is approximated to be 50% of world's oxygen.Green-algae is a good example of phytoplankons.
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]
Choose the best data type for each of the following so that any reasonable value is accommodated but no memory storage is wasted. Give an example of a typical value that would be held by the variable, and explain why you chose the type you did.
a. the number of siblings you have
b. your final grade in this class
c. the population of Earth
d. the population of a U.S. county
e. the number of passengers on a bus
f. one player's score in a Scrabble game
g. one team's score in a Major League Baseball game
h. the year an historical event occurred
i. the number of legs on an animal
j. the price of an automobile
Answer:
Explanation:
Part (a):
Statement : The number of siblings you have
Suitable Data type : Byte
Typical Value : From -128 and up to 127
Explanation: Byte data type is the most suitable since it can covers minimum and maximum number of siblings one can have.
Part (b):
Statement : Your final grade in this class
Suitable Data type : Char
Typical Value : 1 byte
Explanation: Grades is in the form of alphabetical letter which is either A, B, C, D, F or E which can be stored in character data type.
Part (c):
Statement : Population of Earth
Suitable Data type : Long
Maximum Value : 9223372036854775807
Explanation: Long Data takes up to 8 bytes and can store up to 9223372036854775807 which can cater for more than 36 billion. The population of earth is only around 7 billion currently making Long data type the most suitable data type to store earth population.
Part (d):
Statement : Population of US Country
Suitable Data type : Integer
Typical Value :2147483647
Explanation: Integer data type takes up to 4 bytes and can store up to 2147483647 making it suitable to store U.S population.
Part (e):
Statement : The number of passengers on bus
Suitable Data type : Byte
Typical Value :From -128 up to 127
Explanation: The typical maximum number of passengers of a bus are only around 72. Byte data type is the most suitable since it can cater the number up to 127.
Part (f):
Statement : Player's score in a Scrabble game
Suitable Data type : Short
Typical Value : 32767
Explanation: The maximum point can be scored in the Scrabble game is only 830 therefore the most suitable data type for this case is the short data type.
Part (g):
Statement : One team's score in a Major League Baseball game
Suitable Data type : Byte
Typical Value : From -128 up to 127
Explanation: The maximum point can be scored in the Base ball game is only 49 therefore the most suitable data type for this case is the Byte data type since it can cater up to 127.
Part (h):
Statement : The year an historical event occurred
Suitable Data type : Short
Maximum Value: 32767
Explanation: The historic event year can be any number from 1 to 2020 therefore the most suitable data type is the short data type.
Part (i):
Statement : The number of legs on an animal
Suitable Data type : Short
Maximum Value: 32767
Explanation: The most number of legs found are 750 legs therefore the most suitable data type is the short data type which can cater up to 32767.
Part (j):
Statement : The Price of an automobile
Suitable Data type : Float
Maximum Value: 340282350
Explanation: The most expensive car is around 15 million therefore the most suitable data type is the float data type which can cater up to 340 million.
As per the question the best data type for following so that a reasonable value is accommodated as per the type of example.
The number of siblings The number of siblings you have, Data type: Byte and has a Value: From -128 and up to 127The final grade in class, Data type: Char, Value: 1 byteStatement Population of the Earth, Data type: Long. Value: 9223372036854775807.The statement of Population of US Country, Data type: Integer, Value 2147483647.The statement: passengers on the bus, Data type: Byte, Value: From -128 up to 127. The statement of scrabble game, type: Short, Value: 32767.Learn more about the data type for each.
brainly.com/question/24871576.
In a tensile test on a steel specimen, true strain = 0.12 at a stress of 250 MPa. When true stress = 350 MPa, true strain = 0.26. Determine the strength coefficient and the strain-hardening exponent in the flow curve equation.
Answer:
The strength coefficient is [tex]625[/tex] and the strain-hardening exponent is [tex]0.435[/tex]
Explanation:
Given the true strain is 0.12 at 250 MPa stress.
Also, at 350 MPa the strain is 0.26.
We need to find [tex](K)[/tex] and the [tex](n)[/tex].
[tex]\sigma =K\epsilon^n[/tex]
We will plug the values in the formula.
[tex]250=K\times (0.12)^n\\350=K\times (0.26)^n[/tex]
We will solve these equation.
[tex]K=\frac{250}{(0.12)^n}[/tex] plug this value in [tex]350=K\times (0.26)^n[/tex]
[tex]350=\frac{250}{(0.12)^n}\times (0.26)^n\\ \\\frac{350}{250}=\frac{(0.26)^n}{(0.12)^n}\\ \\1.4=(2.17)^n[/tex]
Taking a natural log both sides we get.
[tex]ln(1.4)=ln(2.17)^n\\ln(1.4)=n\times ln(2.17)\\n=\frac{ln(1.4)}{ln(2.17)}\\ n=0.435[/tex]
Now, we will find value of [tex]K[/tex]
[tex]K=\frac{250}{(0.12)^n}[/tex]
[tex]K=\frac{250}{(0.12)^{0.435}}\\ \\K=\frac{250}{0.40}\\\\K=625[/tex]
So, the strength coefficient is [tex]625[/tex] and the strain-hardening exponent is [tex]0.435[/tex].
Two soils are fully saturated with liquid (no gas present) and the soils have the same void ratio. One soil is saturated with water and the other is saturated with alcohol. The unit weight of water is approximately 1g/cm3 while that of alcohol is about 0.8 g/cm3. Which soil sample has the larger water content? Why?
Answer:
water sample have more water content
Explanation:
given data
soil 1 is saturated with water
unit weight of water = 1 g/cm³
soil 2 is saturated with alcohol
unit weight of alcohol = 0.8 g/cm³
solution
we get here water content that is express as
water content = [tex]S_s \times \gamma _w[/tex] ....................1
here soil is full saturated so [tex]S_s[/tex] is 100% in both case
so put here value for water
water content = 100 % × 1
water content = 1 g
and
now we get for alcohol that is
water content = 100 % × 0.8
water content = 0.8 g
so here water sample have more water content
The soil sample saturated with water has a larger water content than the one with alcohol because water's unit weight is greater than that of alcohol and both soils have the same void ratio.
Explanation:The soil sample saturated with water will have a larger water content compared to the soil saturated with alcohol. This is because the unit weight of water is greater than that of alcohol. Since both soils are fully saturated and have the same void ratio, the volume of liquid in both cases is identical. Therefore, the soil with the denser liquid (water) will contain more mass of liquid per unit volume, leading to a higher water content.
Calculate the modulus, in GPa, of the bar of a steel alloy that has a measured stress of 321 MPa corresponding to a measured strain of 0.00155 in the linear elastic portion of the curve. Round answer to 3 significant figures and report answer in the format: 123 GPa
Answer:
E = 207 GPa
Explanation:
Young's modulus is given by following equation:
[tex]E = \frac{\sigma}{\epsilon}[/tex]
Where [tex]\sigma[/tex] and [tex]\epsilon[/tex] are stress and strain, respectively.
By replacing terms:
[tex]E = \frac{0.321 GPa}{0.00155}\\E = 207 GPa[/tex]