Suppose an op amp has a midband voltage gain of 500,000. If the upper cutoff frequency is 15 Hz, what does the frequency response look like? (Malvino, 20150123) Malvino, A. (20150123). Electronic Principles, 8th Edition [VitalSource Bookshelf version]. Retrieved from vbk://9781259200144 Always check citation for accuracy before use.
Answers
Answer:
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.
Explanation:
Related Questions
The tangent function is defined as tan(theta) = sin(theta)/cos(theta). This expression can be evaluated to solve for the tangent as long as the magnitude of cos(theta) is no too near to 0. Assume that theta is given in degrees, and write the MATLAB statements to evaluate tan(theta) as long as the magnitude of cos(theta) is greater than or equal to 10e-2. If the magnitude of cos(theta) is less than 10e-2, write out an error message instead.
Answers
Answer:
The code is as attached here.
Explanation:
The code is as given below
theta = input(' Enter the value of theta?');
y = sin(theta*pi()/180);
z = cos(theta*pi()/180);
if z < 0.01
fprintf('The value of theta is very low')
else
t=round(y/z,2);
disp(['The value of tangent theta is ',num2str(t)]);
end
In MATLAB, evaluate tangent of theta if the magnitude of cosine is
[tex]> = 10[/tex]⁻²; otherwise, display an error message.
Here are the MATLAB statements to evaluate [tex]\( \tan(\theta) \)[/tex] as long as the magnitude of [tex]\( \cos(\theta) \)[/tex] is greater than or equal to [tex]\( 10^{-2} \)[/tex], and display an error message if the magnitude of [tex]\( \cos(\theta) \)[/tex] is less than [tex]\( 10^{-2} \)[/tex]:
```matlab
% Define theta in degrees
theta_deg = input('Enter the value of theta in degrees: ');
% Convert theta to radians
theta_rad = deg2rad(theta_deg);
% Calculate cosine of theta
cos_theta = cos(theta_rad);
% Check if the magnitude of cos(theta) is greater than or equal to 10^-2
if abs(cos_theta) >= 1e-2
% Evaluate tangent of theta
tan_theta = sin(theta_rad) / cos_theta;
disp(['tan(theta) = ', num2str(tan_theta)]);
else
% Display error message
disp('Error: The magnitude of cos(theta) is too small. Cannot evaluate tan(theta).');
end
```
This script prompts the user to enter the value of [tex]\( \theta \)[/tex] in degrees. It then converts [tex]\( \theta \)[/tex] to radians and calculates [tex]\( \cos(\theta) \)[/tex]. If the magnitude of [tex]\( \cos(\theta) \)[/tex] is greater than or equal to [tex]\( 10^{-2} \)[/tex], it evaluates [tex]\( \tan(\theta) \)[/tex] using the given formula and displays the result. Otherwise, it displays an error message indicating that the magnitude of [tex]\( \cos(\theta) \)[/tex] is too small to evaluate [tex]\( \tan(\theta) \).[/tex]
Consider a subsonic engine inlet (i.e., a diffuser) with an inlet area of 1.5 m² and exit-to-inlet area ratio of 1.44. Air enters the diffuser flowing at 5 kg/s with a static pressure of 0.350 bar and a velocity of 110 m/s. The gases exit the diffuser at a static pressure of 0.375 bar and a velocity of 80 m/s. Assume that the ambient (external) pressure on the walls of the diffuser is 0.340 bar.
(a) For these conditions, what is the force (in units of lbf and N) transmitted to the structure holding the diffuser? Provide both the direction and magnitude of the force and use a picture to help present your answer.
(b) Based solely on your results, comment on whether the force acting on the diffuser would help speed up or slow down a vehicle that used this diffuser as part of a jet propulsion system.
Answers
Answer:
Ai=2300 in² , Ao=Ai*1.44=3312 in²
m=10 lbm/s
Pi=5 psia , Po=5.4 psia , Pa=5.5 psia
Vi=120 m/s , Vo=78 m/s
a) Force =m(Vo-Vi) = -190.5 N = -42.82 lbf (towards the inlet)
b) since force is negative it will slow down the system.
Explanation:
100 kg of refrigerant-134a at 200 kPa iscontained in a piston-cylinder device whose volume is 12.322 m3. The piston is now moved until the volume is one-half its original size. This is done such that thepressure of the refrigerant-134a does not change. Determine (a) the final temperatureand (b) the change in the specific internal energy.
Answers
Without specific property data for refrigerant-134a at the given conditions, the final temperature and change in specific internal energy cannot be calculated directly. Typically, such tasks require referencing property tables or using equations of state for refrigerant-134a.
Explanation:The question concerns the process of compressing a refrigerant-134a in a piston-cylinder device while keeping the pressure constant. Given that 100 kg of refrigerant-134a at 200 kPa is initially contained within a volume of 12.322 m3 and then is compressed to half its volume without changing the pressure, we are tasked with determining the final temperature and the change in specific internal energy of the refrigerant.
Since the pressure remains constant during the compression, this scenario can be classified as an isobaric process. However, without specific data such as the initial temperature or specific internal energy values for refrigerant-134a at the given states, we cannot directly calculate the final temperature and change in specific internal energy. Typically, these calculations would require consulting refrigerant-134a property tables or equations of state appropriate for refrigerant-134a to find values at the specified initial and final volumes while maintaining constant pressure.
To find the final temperature, one would typically use the relationship between pressure, volume, and temperature given by the ideal gas law or the specific equations pertaining to refrigerant-134a. The change in specific internal energy could then be determined using specific heat capacities if assuming an ideal gas behavior, or more accurately, through the refrigerant-134a property tables looking at the specific internal energy values at the initial and final states.
George and Ellen Rottweiler encourage their adult daughter Guinevere to break her engagement and continue living in their home, saying, "You're so bright and attractive; you can find a better guy than this." A family systems theorist would term this as a(n) ________.
Answers
Answer:
Negative feedback
Explanation:
In Biology, negative feedback refers to the counteraction of an effect by its own influence on the process producing it. For instance, the presence of a high level of a particular hormone in the blood may inhibit further secretion of that hormone.
In other words, in negative feedback, the result of a certain action may inhibit further performance of that action
While it is important to gain information on the prospect, it is relatively unimportant to gain any information on the prospect’s organization prior to initiating sales dialogue.ANS:FPTS:1DIF:Difficulty: EasyREF:p. 124-125OBJ:LO: 5-5
Answers
Answer:
FALSE.
Explanation:
This statement can be considered false, because it is important to know essential aspects of the prospect before starting the sales dialogue.
In addition to obtaining basic information about the client, it is relevant for the company to know the prospect's organization.
We can define the prospect as the one who, for the company, has the ideal customer profile, but who has not yet shown interest in consuming its products and services.
Therefore, knowing information about it, about its basic and complex characteristics will help the organization to develop sales and marketing strategies aimed at attracting the prospect.
A road has a crest curve, where the PVI station is a 71 35. The road transitions from a 2.1% grade to a -3.4% grade. The highest point of the curve is at station 74 10. What are the PVC and PVT stations
Answers
Answer:
Stat PVC = Stat(82+98.5)
Stat PVT = Stat(59+71.5)
Explanation
PVI = 71 + 35
Let G1 = Grade 1; G2 = Grade 2
G1 = +2.1% ; G2 = -3.4%
Highest point of curve at station = 74 + 10
General equation of a curve:
[tex]y = ax^{2} +bx+c\\dy/dx=2ax+b\\[/tex]
At highest point of the curve [tex]dy/dx=o[/tex]
[tex]2ax+b=0\\x=-b/2a\\x=G1L/(G2-G1)\\x=L/2 +(stat 74+10)-(stat 71+35)\\x=L/2 + 275[/tex]
[tex]-G1L/(G2-G1) = (L/2 + 275)/100\\L = -2327 ft\\Station PVC = Stat(71+35)+(-2327/2)\\\\Stat PVC = 7135-1163.5\\Stat PVC = Stat(82+98.5)\\[/tex]
Station PVT
[tex]Station PVT = Stat PVI + (L/2)\\Station PVT = Stat(71+35)+(-2327/2)\\Station PVT = 7135-1163.5\\Stat PVT = Stat(59+71.5)[/tex]
. A two-dimensional fluid motion is represented by a point vortex of strength Γ set at distance c from an infinite straight solid boundary. Write expressions for the velocity potential and stream function in Cartesian coordinates. Derive an expression for velocity on the boundary. Draw the streamlines and plot the velocity distribution on the boundary when Γ = π and c = 1.
Answers
Answer:
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.
Explanation:
A one-dimensional plane wall of thickness 2L=100mm experiences uniform thermal energy generation of q dot=1000 W/m^3 and is convectively cooled at x=+-50mm by an ambient fluid characterized by T infinity=20degreesC. If the steady-state temperature distribution within the wall is T(x)=a(L^2-x^2)+b where a=10 degrees C/m^2 and b=30 degrees C, what is the thermal conductivity of the wall? What is the value of the convection heat transfer coefficient, h?
Answers
Answer:
note:
solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment
Answer:
A) Thermal conductivity of wall = 50 W/m.c
B) value of the convection heat transfer coefficient, h = 5 W/m².C
Explanation:
I've attached all explanations
A freshwater jet boat takes in water through side vents and ejects it through a nozzle of diameter D = 75 mm; the jet speed is Vj. The drag on the boat is given by Fdrag =kV2, where V is the boat speed. Find an expression for the steady speed, V, in terms of water density rho, flow rate through the system of Q, constant k, and jet speed Vj. A jet speed Vj = 15 m=s produces a boat speed of V = 10 m=s. (a) Under these conditions, what is the new flow rate Q? (b) Find the value of the constant k. (c) What speed V will be produced if the jet speed is increased to Vj = 25 m=s? (d) What will be the new flow rate?
Answers
To find an expression for the jet boat's steady speed, equate the water jet force (ρQVj) with the drag force (kV^2). Solve for V to get the equation for speed in terms of ρ, Q, k, and Vj. Then use known values for Vj and V to find Q and k and repeat the process for a different Vj to get new V and Q values.
Explanation:To find an expression for the steady speed V of a freshwater jet boat in terms of the water density ρ, flow rate Q, constant k, and jet speed Vj, we can apply Newton's second law, assuming that the force provided by the water jet equals the drag force on the boat when at steady speed. The water jet force can be expressed as the rate of change of momentum of the water, which is ρQVj (since Q is the mass flow rate ρQ is the momentum flow rate), and the drag force is given as kV^2. Equating these two forces gives us:
ρQVj = kV^2
Solving for V will give us the steady speed expression we are looking for:
V = √(ρQVj / k)
To calculate the new flow rate Q given Vj = 15 m/s and V = 10 m/s, we plug these values into the expression obtained from above:
Q = kV^2 / (ρVj)
With given values, we would have:
Q = k * 10^2 / (ρ * 15)
The constant k can be determined using the known conditions and solving for k.
k = ρQVj / V^2
For (c) and (d), the same equations can be applied with the jet speed Vj changed to 25 m/s to find the new boat speed and flow rate.
Water is to be boiled at sea level in a 30-cm-diameter stainless steel pan placed on top of a 3-kW electric burner. If 60 percent of the heat generated by the burner is transferred to the water during boiling, determine the rate of evaporation of water.
Answers
Answer:
mevaporation=˙Qhfg=1. 8 kJ /s2269. 6 kJ /kg=0 . 793×10−3kg/ s=2. 855 kg /h
Explanation:
The properties of water at 1 atm and thus at the saturation temperature of 100C are hfg =2256.4 kJ/kg (Table A-4). The net rate of heat transfer to the water is ˙Q=0 . 60×3 kW=1 . 8 kWNoting that it takes 2256.4 kJ of energy to vaporize 1 kg of saturated liquid water, therate of evaporation of water is determined to be mevaporation=˙Qhfg=1. 8 kJ /s2269. 6 kJ /kg=0 . 793×10−3kg/ s=2. 855 kg /h
The rate of evaporation will be "2.871 Kg/hour".
Evaporation of water:According to the question,
Rate of heat supplied:
= 60% of 3 kW
= 1.8 kW
= 1.8 KJ/s
Vaporization of water, [tex]\Delta H = 2257 \ KJ/Kg[/tex]
Time taken will be:
= [tex]\frac{2257}{1.8}[/tex]
= [tex]1254 \ s[/tex]
= [tex]\frac{1254}{3600} \ hour[/tex]
= [tex]0.3482 \ hour[/tex]
hence,
The rate of evaporation,
= [tex]\frac{Mass \ of \ water}{Time}[/tex]
= [tex]\frac{1}{0.3482}[/tex]
= [tex]2.871 \ Kg/hour[/tex]
Thus the above answer is right.
Find out more information about evaporation here:
https://brainly.com/question/4406110
In a typical transmission line, the current I is very small and the voltage V is very large. A unit length of the line has resistance R.
For a power line that supplies power to 10 000 households, we can conclude that:
a) ????V < ????^2 R
b) ????^2 R = 0
c) ????V = ????^2 R
d) ????V > I^2 R
e) ???? = V/R
Answers
Question:
In a typical transmission line, the current I is very small and the voltage V is very large. A unit length of the line has resistance R.
For a power line that supplies power to 10 000 households, we can conclude that
a) IV < I²R
b) I²R = 0
c) IV = I²R
d) IV > I²R
e) I = V/R
Answer:
d) IV > I²R
Explanation:
In a typical transmission line, the current I is very small and the voltage V is very high as to minimize the I²R losses in the transmission line.
The power delivered to households is given by
P = IV
The losses in the transmission line are given by
Ploss = I²R
Therefore, the relation IV > I²R holds true, the power delivered to the consumers is always greater than the power lost in the transmission line.
Moreover, losses cannot be more than the power delivered. Losses cannot be zero since the transmission line has some resistance. The power delivered to the consumers is always greater than the power lost in the transmission.
Water at 60°F passes through 0.75-in-internal diameter copper tubes at a rate of 1.2 lbm/s. Determine the pumping power per ft of pipe length required to maintain this flow at the specified rate.
The density and dynamic viscosity of water at 70°F are rho = 62.30 lbm/ft^3 and μ = 6.556 x 10^-4 lbm/ft*s. The roughness of copper tubing is 5 x 10^-6 ft.
The pumping power per ft of pipe length required to maintain this flow at the specified rate = _________ W (per ft length)
Answers
Answer:
The pumping power per ft of pipe length required to maintain this flow at the specified rate 0.370 Watts
Explanation:
See calculation attached.
- First obtain the properties of water at 60⁰F. Density of water, dynamic viscosity, roughness value of copper tubing.
- Calculate the cross-sectional flow area.
- Calculate the average velocity of water in the copper tubes.
- Calculate the frictional factor for the copper tubing for turbulent flow using Colebrook equation.
- Calculate the pressure drop in the copper tubes.
- Then finally calculate the power required for pumping.
Write a program with two inputs, current price and last month's price (both integers). Then, output a summary listing the price, the change since last month, and the estimated monthly mortgage computed as (currentPrice * 0.045) / 12. in coral
Answers
Complete Question
Sites like Zillow get input about house prices from a database and provide nice summaries for readers. Write a program with two inputs, current price and last month's price (both integers). Then, output a summary listing the price, the change since last month, and the estimated monthly mortgage computed as (currentPrice * 0.045) / 12. Ex: If the input is 200000 210000, the output is: This house is $200000. The change is $-10000 since last month. The estimated monthly mortgage is $750.
Use Coral Programming Language
Answer:
// Program is written in Coral Programming Language
// Comments are used for explanatory purpose
// Program starts here
// Variable declaration
int currentprice
int prevprice
int change
float mortgage
Put "Enter current price to output" to output
currentprice = Get next input
Put "Enter last month price to output" to output
prevprice = Get next input
// Calculate Change since last month
change = currentprice - prevprice
// Calculate Monthly Mortgage
mortgage = currentprice * 0.045 / 12
// Print Results
Put "This house is $" to output
Put currentprice to output
Put "\n" to output
Put "This change is $" to output
Put change to output
Put "\n" to output
Put "This house is $" to output
Put currentprice to output
Put "since last month\n" to output
Put "This estimated monthly mortgage is $" to output
Put mortgage to output
// End of Program
A cylindrical specimen of some metal alloy 11.2 mm (0.4409 in.) in diameter is stressed elastically in tension. A force of 15600 N (3507 lbf) produces a reduction in specimen diameter of 5 × 10-3 mm (1.969 × 10-4 in.). Compute Poisson's ratio for this material if its elastic modulus is 100 GPa (14.5 × 106 psi).
Answers
Answer:
attached below
Explanation:
A 3-kg block rests on top of a 2-kg block supported by, but not attached to, a spring of constant 40 N/m. The upper block is suddenly removed. Determine (a) the maximum speed reached by the 2-kg block, (b) the maximum height reached by the 2-kg block.
Answers
This Physics question is directed towards High School students and pertains to the maximum speed and height a 2-kg block achieves when released from a compressed spring, following the removal of a 3-kg block initially resting on it.
Explanation:The subject of this question is Physics, and it is appropriate for a High School grade level. The question is regarding the oscillatory motion and energy conversion of a block attached to a spring system. Specifically, it involves understanding the concepts of potential energy stored in a spring, conservation of mechanical energy, and kinematics of simple harmonic motion.
To solve part (a) of the problem for the 2-kg block, you'll need to use the conservation of energy principle. Initially, when the 3-kg block is also on the spring, the potential energy stored in the compressed spring is equal to the kinetic energy the 2-kg block will have when it reaches its maximum speed. After the upper block is removed, the spring force will only accelerate the 2-kg block.
To find the maximum speed reached by the 2-kg block, you would calculate the kinetic energy equivalent to the potential energy stored in the spring and set it equal to \\((1/2)mv^2\\). For part (b), you could use the conservation of energy principle again to find the maximum height the 2-kg block reaches by equating the initial kinetic energy to the potential energy at the maximum height (\\(mgh\\)).
A closed tank contains ethyl alcohol to a depth of 66 ft. Air at a pressure of 23 psi fills the gap at the top of the tank. Determine the pressure at a closed valve attached to the tank 10 ft above its bottom
Answers
Answer:
639.4psi
Explanation:
Pressure at the closed valve = Air pressure+ (density*gravity*height)
=23psi+(49.27lb/ft^3*32.17ft/s^2*56ft^3)
=23psi+88760.89psft
=23psi+(88760.89/144)psi
=23+616.4
=639.4psi
A piston-cylinder device contains 0.15 kg of air initially at 2 MPa and 350 °C. The air is first expanded isothermally to 500 kPa, then compressed polytropically with a polytropic exponent of 1.2 to the initial pressure, and finally compressed in an isobaric process to the initial state. Determine the boundary work for each process and the net work for the cycle?
Answers
Answer:
Isothermal expansion W₁ =-37198.9 J
Polytropic Compression W₂ =-34872.82 J
Isobaric Compression W₃ = -6974.566 J
The net work for the cycle = -79046.29 J
Explanation:
Mass of air = 0.15 kg = 150 g
Molar mass = 28.9647 g/mol
Number of moles = 150 g /28.9647 g/mol = 5.179 moles of air
PV = nRT therefrore V = nRT/(P) = 5.179*8.314*(350+273.15)/(2×10⁶) = 0.0134167 m³
For isothermal expansion we have
P₁V₁ = P₂V₂ or V₂ = P₁V₁/P₂ = 2×10⁶*0.0134167 / (5×10⁵) = 0.0536668 m³
Therefore work done
W₁ = -nRTln(V₂/V₁) = -26833ln(4) = -37198.9 J
Stage 2
Compression polytropically we have
[tex]\frac{P_2}{P_3} = (\frac{V_3}{V_2} )^n[/tex] where P₃ = 2 MPa
Therefore V₃ = [tex](\frac{1}{4} )^{\frac{1}{1.2} }*V_2[/tex] = 1.6904×10⁻² m³
Work = W₂ = [tex]\frac{P_2V_2-P_3V_3}{n-1}[/tex] = -34872.82 J
[tex]\frac{P_2}{P_3} = (\frac{T_2}{T_3} )^\frac{n}{n-1}[/tex] or T₃ = [tex]T_2*(\frac{P_3}{P_2})^\frac{n-1}{n}[/tex] = 785.12 K
Isobaric compression we have thus
Work done W₃ = P(V₁ -V₃) = -6974.566 J
Total work = W₁ + W₂ + W₃ = -37198.9 J + -34872.82 J + -6974.566 J = -79046.29 J
Two sections of a pressure vessel are to be held together by 5/8 in-11 UNC grade 5 bolts. You are told that the length of the bolts is 1.5 in, the length of the threaded portion of the bolts is 0.75 in, and that their elastic modulus is E=30 Mpsi. The total load on the joint is 36 kip and the stiffness of the members is given as km=8.95 Mlbf/in. What is the minimum number of bolts that should be used to guard against excess proof strength with a factor of safety of np=1.2? Be sure to make an estimate for the preload.
Answers
Answer:
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.
Explanation:
Consider the general form of the Reynolds transport theorem (RTT) given by dBsys dt = d dt ∫CV rhob dV + ∫CS rhobV› r·n › dA where V › r is the velocity of the fluid relative to the control surface. Let Bsys be the mass m of a closed system of fluid particles. We know that for a system, dm/dt = 0 since no mass can enter or leave the system by definition. Use the given equation to derive the equation of conservation of mass for a control volume.
Answers
Answer:
Explanation:
note:
solution is attached due to error in mathematical equation. please find the attachment
Calculate the magnitude of the force FB in the back muscles that is needed to support the upper body plus the box and compare this with his weight. The mass of the upper body is 55.0 kg and the mass of the box is 30.0 kg.
Answers
Answer:
807.5N
Explanation:
The combined mass (m) on the back muscle is 55kg + 30kg = 85kg
Acceleration due to gravity (g) = 9.8m/s²
Therefore the force FB = ma = 85*9.8
FB= 807.5N
Write a program to calculate the property tax. Property tax is calculated on 92% of the assessed value of the property. For example, if the assessed value is $200,00.00, the property tax is on 184,000.00. Assume that the property tax rate is $1.05 for each $100 of the assessed value. Your program should prompt the user to enter the assessed value of the property. Store the output in a file in the following format. (Here is a sample output.
Answers
Answer:
#include<stdio.h>
void main()
{
// using file pointer to print output to txt file
FILE *fptr;
float assessedValue, taxableAmount, taxRate = 1.05, propertyTax;
/* open for writing */
fptr = fopen("output.txt", "w");
if (fptr == NULL)
{
printf("File does not exists \n");
return;
}
// prompting user to enter assessed value and storing it in assessedValue variable
printf("Enter the Assessed Value of property : ");
scanf("%f", &assessedValue);
//writing assessed value to output.txt file using fprintf file i/o function
fprintf(fptr, "AssessedValue : $ %.2f\n", assessedValue);
//calculating taxableAmount based on given condition in the question
taxableAmount = (assessedValue * 0.92);
//writing taxable Amount to output.txt file using fprintf file i/o function
fprintf(fptr, "TaxableAmount: $ %.2f\n", taxableAmount);
//writing tax Rate to output.txt file using fprintf file i/o function
fprintf(fptr, "Tax Rate for each $100.00: $ %.2f\n", taxRate);
//calculating propertyTax based on given condition in the question
propertyTax = ((taxableAmount/100)*taxRate);
//writing property Tax Amount to output.txt file using fprintf file i/o function
fprintf(fptr, "propertyTax: $ %.2f\n", propertyTax);
//closing file using fclose function
fclose(fptr);
}
Explanation :
I used Turbo C compiler to compile and run the C program. The below program compiles and at the run time, automatically, prints output to a file called output.txt.
When you compile the program, remember to check the BIN folder in Turbo c folder of C drive where your turbo c has been installed.
Output:
Assessed value: $200000
Taxable amount: $184000
Tax Rate for each $100.00: $1.05
A three-phase, 600 MVA, 13.8KV AC generator has a synchronous reactance of 2.0 per unit. The generator is connected to a system for which the specified bases are 100MVA and 345KV. a) Find the per-unit value of the generator synchronous reactance on the specified bases. b) Find the ohmic value of the synchronous reactance. Problem 4 A single-phase source is connected to an electrical load. The load
Answers
Answer:
(a) 0.00053
(b) 0.1 mΩ
Explanation:
New per-unit reactance is given as:New Per-unit reactance = [tex]2* \frac{100}{600} * (\frac{13.8}{345}) ^{2} = 0.00053[/tex]
Ohmic reactance : [tex]\frac{13.8^2}{600} = 0.31 ohm[/tex]Ohmic per unit : 0.31 * 0.00053 = 0.1 mΩGiven Information:
Zpu_old = 2 pu
Sbase_new = 100 MVA
Sbase_old = 600 MVA
kV_old = 13.8 kV
kV_new = 345 kV
Required Information:
Zpu_new = ?
ZΩ = ?
Answer:
Zpu_new = 0.000533 pu
ZΩ = 0.634 Ω
Explanation:
a) Find the per-unit value of the generator synchronous reactance on the specified bases.
When the base kVA and base kV are changed then we use following relation to update the per unit values.
Zpu_new = Zpu_old*(Sbase_new/Sbase_old)*(kV_old/kV_new)²
Zpu_new = 2*(100x10⁶/600x10⁶)*(13.8x10³/345x10³)²
Zpu_new = 0.000533 pu
b) Find the ohmic value of the synchronous reactance.
ZΩ = Zbase*Zpu_new
Where Zbase is calculated as
Zbase = (kVbase)²/Sbase
Zbase = (345x10³)²/100x10⁶
Zbase = 1190.25 Ω
ZΩ = Zbase*Zpu_new = 1190.25*0.000533 = 0.634 Ω
The pistons of a V-6 automobile engine develop 226.5 hp. If the engine driveshaft rotational speed is 4700 RPM and the torque is 248 ft·lbf, determine the shaft power, in hp, and the percentage of the developed power that is transferred to the driveshaft.
Answers
Answer:
221.929 hp
98.19%
Explanation:
See attached picture.
hree large plates are separated bythin layers of ethylene glycol and water. The top plate moves to the right at 2m/s. At what speed and in what direction must the bottom plate be moved to hold the center plate stationary?
Answers
Answer: For the center plate to remain stationed in one position without rotating, the bottom plate has to move to the left at a speed of 2m/s, so as to cancel the force acting on it from the top.
The center plate will not move when the bottom plate is moving left in a speed of 2m/s to counter the speed of the top plate, because a body will continue to be at rest if all the forces acting towards the body are equal. The center plate will be at rest because we have directed equal force from the top and bottom of the plate.
A fire hose nozzle has a diameter of 1.125 in. According to some fire codes, the nozzle must be capable of delivering at least 250 gal/min. If the nozzle is attached to a 3-in.-diameter hose, what pressure must be maintained just upstream of the nozzle to deliver this flowrate?
Answers
Answer:
[tex]P_{1} = 403,708\,kPa\,(58.553\,psi)[/tex]
Explanation:
Let assume that changes in gravitational potential energy can be neglected. The fire hose nozzle is modelled by the Bernoulli's Principle:
[tex]\frac{P_{1}}{\rho\cdot g} = \frac{P_{2}}{\rho \cdot g} + \frac{v^{2}}{2\cdot g}[/tex]
The initial pressure is:
[tex]P_{1} = P_{2}+ \frac{1}{2}\cdot \rho v^{2}[/tex]
The speed at outlet is:
[tex]v=\frac{\dot Q}{\frac{\pi}{4}\cdot D^{2}}[/tex]
[tex]v=\frac{(250\,\frac{gal}{min} )\cdot (\frac{3.785\times 10^{-3}\,m^{3}}{1\,gal} )\cdot(\frac{1\,min}{60\,s} )}{\frac{\pi}{4}\cdot [(1.125\,in)\cdot(\frac{0.0254\,m}{1\,in} )]^{2} }[/tex]
[tex]v\approx 24.592\,\frac{m}{s}\,(80.682\,\frac{ft}{s} )[/tex]
The initial pressure is:
[tex]P_{1} = 101.325\times 10^{3}\,Pa+\frac{1}{2}\cdot (1000\,\frac{kg}{m^{3}} )\cdot (24.592\,\frac{m}{s} )^{2}[/tex]
[tex]P_{1} = 403,708\,kPa\,(58.553\,psi)[/tex]
Answer:
P1 = 42.93 psi
Explanation:
For incompressible fluid, we know that;
A1V1 = A2V2
Making V1 the subject, we obtain;
V1 = A2V2/A1
Now A2V2 is the volumetric flow rate (V') .
Thus; V1 = V'/A1
A1 = πD²/4
Thus, V1 = 4V'/πD²
V' = 250 gal/min
But the diameter is in inches, let's convert to inches³/seconds.
Thus, V' = 250 x 3.85 = 962.5 in³/s
Substituting the relevant values to obtain,
V1 = (4 x 962.5)/(π x 3²) = 136.166 in/s.
Now let's convert to ft/s;
V1 = 136.166 x 0.0833 = 11.34 ft/s
Also for V2;
V2 = (4 x 962.5)/(π x 1.125²) = 968.29 in/s.
Now let's convert to ft/s;
V2 = 968.29 x 0.0833 = 80.66 ft/s
Setting bernoulli equation between the hose and the exit, we obtain;
(p1/γ) + (V1²/2g) = V2²/2g
Where V1 and V2 are intial and final velocities and γ is specific weight of water which is 62.43 lb/ft³ and g i acceleration due to gravity which is 32.2 ft/s²
Making p1 the subject, we obtain;
p1 = (γ/2g)(V2² - V1²)
p1 = (62.43/(2x32.2))(80.66² - 11.34²)
p1 = 6182.35 lb/ft²
So Converting to psi, we have;
p1 = 6182.35/144 = 42.93 psi
A parallel plate capacitor has a separation of 2x10 m and free space between the plates. A 10 V battery is connected across the plates and then removed without disturbing the charge on the plates. The plates are now allowed to come together toa separation of 10% m without disturbing the charge on the plates. Fringing fields can be ignored. A) What is the voltage across the plates? B) How has the energy stored in the capacitor changed?
Answers
Answer:
a) 5 V.
b) Energy also become half.
Explanation:
See attached picture.
Water flows in a tube that has a diameter of D= 0.1 m. Determine the Reynolds number if the average velocity is 10 diameters per second. (b) Repeat the calculations if the tube is a nanoscale tube with a dimeter of D= 100 nm.
Answers
Answer:
a) [tex]Re_{D} = 111896.745[/tex], b) [tex]Re_{D} = 1.119\times 10^{-7}[/tex]
Explanation:
a) The Reynolds number for the water flowing in a circular tube is:
[tex]Re_{D} = \frac{\rho\cdot v\cdot D}{\mu}[/tex]
Let assume that density and dynamic viscosity at 25 °C are [tex]997\,\frac{kg}{m^{3}}[/tex] [tex]0.891\times 10^{-3}\,\frac{kg}{m\cdot s}[/tex], respectively. Then:
[tex]Re_{D}=\frac{(997\,\frac{kg}{m^{3}} )\cdot (1\,\frac{m}{s} )\cdot (0.1\,m)}{0.891\times 10^{-3}\,\frac{kg}{m\cdot s} }[/tex]
[tex]Re_{D} = 111896.745[/tex]
b) The result is:
[tex]Re_{D}=\frac{(997\,\frac{kg}{m^{3}} )\cdot (10^{-6}\,\frac{m}{s} )\cdot (10^{-7}\,m)}{0.891\times 10^{-3}\,\frac{kg}{m\cdot s} }[/tex]
[tex]Re_{D} = 1.119\times 10^{-7}[/tex]
Design a database suitable for a university registrar. This database should include information about students, departments, professors, courses, which students are enrolled in which courses, which professors are teaching which courses, student grades, TA's for a course (TA's are students), which courses a department offers, and any other information you deem appropriate. Note that this question is more free-form than the questions above, and you need to make some decisions about multiplicities of relationships, appropriate types, and even what information needs to be represented.
Answers
Answer:
Hello there, see step by step explanation for answers
Explanation:
A database design for a University Registrar. The following requirements are for designing a Database Schema.
It should include:
1. Information about Students .
2. Information about Departments.
3. Information about Professors.
4. Information about courses.
5 . student Grades.
6. TA's for a course.
7. Department offering different courses.
For Designing Database for Registrar System We need to define various
1. Entity: Student, Course, Instructor, Course offering.
2. Attributes of Entities:
(a) Student entity has Sid, name, program as its attributes
(b)Course has Course_n, title, credits, and syllabus as its attributes
(c) instructor has iid ,name ,dept, title as its attributes
(d) course offering has section_no ,time, room, year ,semester as its attributes
3. Relationship among various entities
(a) enrolls
(b) teaches
© is offered
E-R Diagram for University Registrar
This Diagram shows student entity enrolls various courses which can be having teaches relationship with instructor. Course is being offered by relationship is offered by course offering .
A database design for a University Registrar. The following requirements are for designing a Database Schema.
It should include:
1. Information about Students .
2. Information about Departments.
3. Information about Professors.
4. Information about courses.
5 . student Grades.
6. TA's for a course.
7. Department offering different courses.
For Designing Database for Registrar System We need to define various
1. Entity: Student, Course, Instructor, Course offering.
2. Attributes of Entities:
(a) Student entity has Sid, name, program as its attributes
(b)Course has Course_n, title, credits, and syllabus as its attributes
(c) instructor has iid ,name ,dept, title as its attributes
(d) course offering has section_no ,time, room, year ,semester as its attributes
3. Relationship among various entities
(a) enrolls
(b) teaches
© is offered
E-R Diagram for University Registrar
This Diagram shows student entity enrolls various courses which can be having teaches relationship with instructor. Course is being offered by relationship is offered by course offering .
In a CNC milling machine, the axis corresponding to the feed rate uses a dc servomotor as the drive unit and a rotary encoder as the feedback sensing device. The motor is geared to a leadscrew with a 10:1 reduction (10 turns of the motor for each turn of the leadscrew). If the leadscrew pitch is 6 mm, and the encoder emits 60 pulses per revolution, determine (a) the rotational speed of the motor and (b) pulse rate of the encoder to achieve a feed rate of 300 mm/min.
Answers
Answer:
a) 500 rev/min.
b) 50 Hz.
Explanation:
See the attached pictures.
You are an electrician on the job. The electrical blueprint shows that eight 500-W lamps are to be installed on the same circuit. The circuit voltage is 277V and is protected by a 20-A circuit breaker. A continuous-use circuit can be loaded to only 80% of its rating. Is a 20-A circuit large enough to carry this load
Answers
Answer:
I = 14.44A
Explanation:
calculating 80% of the circuit breaker current
I = (80/100)20A = 16A
eight 500W lamps are to be installed on the circuit. assume they are connected in parallel
total power = 500 x 8 = 4000W
power = voltage x current
current = power/voltage = 4000W/277V = 14.4A
The current obtained is less than 80% of the circuit breaker.
20A circuit breaker is large enough to carry the load.