Answer:
Explanation:
[tex]\dot x (t) - ax(t)=bu(t)\\\\e^{-at}\dot x = e^{-at}ax = \frac{d}{dt} (e^{-at})=e^{-at}bu\\\\\int\limits^t_0 \frac{d}{d\tau}(e^{-a\tau}x(\tau))d\tau=e^{-at}x(t)-x(0)=\int\limits^t_0 e^{-a\tau}bu(\tau))d\tau\\\\x(t)=e^{at}x(0)+ \int\limits^t_0 e^{-a(t-\tau)}bu(\tau))d\tau[/tex]
Similarly,
[tex]e^{-At}\dot x(t) -e^{-At}Ax(t) = \frac{d}{dt} (e^{-At}x(t))=e^{-At}Bu(t)\\\\\int\limits^t_0 \frac{d}{d\tau}(e^{-A\tau}x(\tau))d\tau=e^{-At}x(t)-e^{-A.0}x(0)=\int\limits^t_0 e^{-A\tau}Bu(\tau))d\tau\\\\since\:\:\:e^{-A.0}=I \:\:\: and\:\:\:[e^{-At}]^{-1}=e^{At}\\\\x(t)=e^{At}x(0)+ e^{At}\int\limits^t_0 e^{-A\tau}Bu(\tau))d\tau\\\\x(t)=e^{At}x(0)+ \int\limits^t_0 e^{A(t-\tau)}Bu(\tau))d\tau[/tex]
For initial condition x(0) = 0
[tex]x(t)= \int\limits^t_0 e^{A(t-\tau)}Bu(\tau))d\tau[/tex]
A Scotch-yoke mechanism is used to convert rotary motion into reciprocating motion. As the disk rotates at the constant angular rate , a pin A slides in a vertical slot causing the slotted member to displace horizontally according to x = r sin(t) relative to the fi xed disk center O. Determine the expressions for the velocity and acceleration of a point P
Answer:
The question continues ; Determine the expressions for the velocity and acceleration of a point P as a function of time t, and determine the maximum velocity of point P during one cycle. Use the values r = 75mm and w = pie-rads/s
Explanation:
The diagram and the detailed step by step explanation is as shown in the attachment
In the Scotch-yoke mechanism, if the displacement is given by x = r sin(t), the velocity v = r cos(t) and acceleration a = -r sin(t). The negative sign in acceleration indicates that it is in the opposite direction to displacement.
Explanation:The Scotch-yoke mechanism which converts rotary motion into reciprocating motion can be analyzed using principles of kinematics. If we have the displacement given by x = r sin(t), the velocity and acceleration can be derived from this displacement equation.
The velocity (v) is the time derivative of the displacement function, i.e., v = dx/dt = r cos(t).
The acceleration (a) is the time derivative of the velocity function, so a = dv/dt = -r sin(t).
The negative sign signifies that acceleration is in the opposite direction to displacement.
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A manometer containing a fluid with a density of 60 lbm/ft3 is attached to a tank filled with air. If the gage pressure of the air in the tank is 9.4 psig and the atmospheric pressure is 12.5 psia, the fluid-level difference between the two columns, h, in feet is
Answer:
The fluid level difference in the manometer arm = 22.56 ft.
Explanation:
Assumption: The fluid in the manometer is incompressible, that is, its density is constant.
The fluid level difference between the two arms of the manometer gives the gage pressure of the air in the tank.
And P(gage) = ρgh
ρ = density of the manometer fluid = 60 lbm/ft³
g = acceleration due to gravity = 32.2 ft/s²
ρg = 60 × 32.2 = 1932 lbm/ft²s²
ρg = 1932 lbm/ft²s² × 1lbf.s²/32.2lbm.ft = 60 lbf/ft³
h = fluid level difference between the two arms of the manometer = ?
P(gage) = 9.4 psig = 9.4 × 144 = 1353.6 lbf/ft²
1353.6 = ρg × h = 60 lbf/ft³ × h
h = 1353.6/60 = 22.56 ft
A diagrammatic representation of this setup is presented in the attached image.
Hope this helps!
An engineering student claims that a country road can be safely negotiated at 65 mi/h in rainy weather. Because of the winding nature of the road, one stretch of level pavement has a sight distance of only 510 ft. Assuming practical stopping distance, comment on the student
Answer:
Negotiated speed should be lower. Perception/reaction time is too less than design values.
Explanation:
Given:
- The claimed safe speed V_1 = 65 mi/h
- Sight distance D = 510 ft
- The practical deceleration a = 11.2 ft/s ... according to standards
Find:
Assuming practical stopping distance, comment on the student whether the claim is correct or not
Solution:
- Calculate the practical stopping distance:
d = V_1^2 / 2*a
d = ( 65 * 1.46 )^2 / 2*11.2 = 402.054 ft
- Solve for reaction distance d_r is as follows:
d_r = D - d = 510 - 402.054 = 107.945 ft
- The perception/time reaction is:
t_r = d_r/V_1 = 107.945 / 94.9
t_r = 1.17 sec
Answer: The perception/reaction time t_r = 1.17 s is well below the t = 2.3 s.
Hence, the safe speed should be lower.
A pipe produces successive harmonics at 300 Hz and 350 Hz. Calculate the length of the pipe and state whether it is closed at one end or not. Assume the speed of sound to be 340 m/s.
Answer:
The pipe is open ended and the length of pipe is 3.4 m.
Explanation:
For identification of the type of pipe checking the successive frequencies in both the open pipe and closed pipe as below
Equation for nth frequency for open end pipe is given as
[tex]f_n=\frac{nv}{2L}[/tex]
For (n+1)th value the frequency is
[tex]f_{n+1}=\frac{(n+1)v}{2L}[/tex]
Taking a ratio of both equation and solving for n such that the value of n is a whole number
[tex]\frac{f_{n+1}}{f_n}=\frac{\frac{(n+1)v}{2L}}{\frac{nv}{2L}}\\\frac{350}{300}=\frac{(n+1)}{n}\\350n =300n+300\\50n =300\\n =6\\[/tex]
So n is a whole number this means that the pipe is open ended.
For confirmation the nth frequency for a closed ended pipe is given as
[tex]f_n=\frac{(2n+1)v}{4L}[/tex]
For (n+1)th value the frequency is
[tex]f_{n+1}=\frac{(2n+3)v}{4L}[/tex]
Taking a ratio of both equation and solving for n such that the value of n is a whole number
[tex]\frac{f_{n+1}}{f_n}=\frac{\frac{(2n+3)v}{2L}}{\frac{(2n+1)v}{2L}}\\\frac{350}{300}=\frac{(2n+3)}{(2n+1)}\\700n+350 =600n+900\\100n =550\\n =5.5\\[/tex]
As n is not a whole number so this is further confirmed that the pipe is open ended.
Now from the equation of, with n=6, v=340 m/s and f=300 Hz
[tex]f_n=\frac{nv}{2L}\\300=\frac{6 \times 340}{2L}\\L=\frac{2040}{600}\\L=3.4 m[/tex]
The value of length is 3.4m.
A certain full-wave rectifier has a peak output voltage of 30 V. A 50 mF capacitor-input filter is connected to the rectifier. Calculate the peak-to-peak ripple and the dc output voltage devel-oped across a 600 V load resistance.
In this exercise we have to use electronic knowledge to calculate the voltage value from the resistance. In this way we can conclude that:
Ripple voltage: [tex]8.33mV[/tex] DC output voltage: [tex]19.11 V[/tex]
So from the data given in the text, we have that:
Voltage: [tex]30v[/tex] Load resistance: [tex]600 ohms[/tex] Capacitor filter: [tex]50mF[/tex] Frequency of supply: [tex]120Hz[/tex]
So with the formula given below, we can calculate what is being asked by it:
[tex]Peak \ to \ peak \ ripple = I (load)/(f)(c)[/tex]
They are using the data previously informed and putting it in the given formula, we would be with:
[tex]I (load) = load \ current = 30/600 = 0.05 A\\Peak \ to \ peak \ ripple = 0.05/6\\= 8.33mV[/tex]
The average Dc something produced power for a complete wave exist double :
[tex]Vd_c= 0.637 * 30\\Vd_c =19.11V[/tex]
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Consider a Venturi with a throat-to-inlet area ratio of 0.8, mounted on the side of an airplane fuselage. The airplane is in flight at standard sea level.
If the static pressure at the throat is 2100 lb/ft³, calculate the velocity of the airplane. Note that standard sea-level density and pressure are 1.23 kg/m³ (0.002377 slug/ft³) and 1.01 x 10⁵ N/m2 (2116lb/ft³), respectively.
Defining the radius of area throat to inlet we would have that the proportional relationship would be [tex]\frac{A_2}{A_1}.[/tex]
This equation or relationship is obtained from continuity where
[tex]A_1V_1 = A_2V_2[/tex]
[tex]\frac{V_2}{V_1} = \frac{A_2}{A_1}= 0.8[/tex]
Now applying the Bernoulli equation between inlet and throat section we have,
[tex]\frac{p_1}{\rho g}+ \frac{v_1^2}{2g}+z_1=\frac{p_2}{\rho g}+ \frac{v_2^2}{2g}+z_2[/tex]
Here,
[tex]z_1 = z_2[/tex]
Then for a Venturi duct, the velocity of the airplane [tex]V_1[/tex] will be
[tex]V = \sqrt{\frac{2(p_1-p_2)}{\rho[(\frac{A_1}{A_2})^2-1]}}[/tex]
Our values are,
[tex]\frac{A_2}{A_1} = 0.8[/tex]
[tex]\rho = 0.002377slug/ft^3[/tex]
[tex]p_1 = 2116lb/ft^2[/tex]
[tex]p_2 = 2100lb/ft^2[/tex]
Replacing,
[tex]V= \sqrt{\frac{2(2116-2100)}{(0.002377)[(\frac{1}{0.8})^2-1]}}[/tex]
[tex]V = 154.7ft/s[/tex]
Therefore the velocity of the airplane is 154.7ft/s
On a nonprecision approach, what is the maximum acceptable descent rate during the final stages of the approach (below 1,000 ft. AGL)?
Answer: For non-precision approaches, the maximum acceptable descent rate acceptable should be one that ensures the aircraft reaches the minimum descent altitude at a distance from the threshold that allows landing in the touch down zone. Otherwise, a decent rate greater than 1000fpm is unacceptable.
Explanation: For non-precision approaches, a descent rate should be used that ensures the aircraft reaches the minimum decent altitude at a distance from the threshold that allows landing in the touchdown zone (TDZ) . On many instrument approach procedures, this distance is annotated by a visual descent point (VDP) If no VDP is annotated, calculate a normal descent point to the TDZ. To determine the required rate of descent, subtract the TDZ elevation (TDZE) from the final approach fix (FAF) altitude and divide this by the time inbound. For illustration, if the FAF altitude is 1,000 feet mean sea level (MSL), the TDZE is 200 feet MSL and the time inbound is two minutes, an 400 fpm rate of descent should be applied.
A descent rate greater than approximately 1,000 fpm is unacceptable during the final stages of an approach (below 1,000 feet AGL). Operational experience and research shows that this is largely due to a human perceptual limitation that is independent of the airplane or helicopter type. As a result, operational practices and techniques must ensure that descent rates greater than 1,000 fpm are not permitted in either the instrument or visual portions of an approach and landing operation.
Describe briefly the main advantages offered by satellite communications. Explain what is meant by a distance-insensitive communications system.
Answer:
Advantage of satellite communication are as follow
- it help in mobile and wireless communications
- it can covers the wide area in single time
- installation of satellite communication is easy
- it can be used for any form of communication i.e. video, audio, etc.
- service can be available in uniformly
Explanation:
Advantage of satellite communication are as follow
- it helps in mobile and wireless communications
- it can cover the wide-area in a single time
- installation of satellite communication is easy
- it can be used for any form of communication i.e. video, audio, etc.
- service can be available in uniformly
Distance insensitive communication system is describe in terms of limitation discovered area. Example of this is mobile and satellite. The mobile discover area is less than satellite.
Suppose you were a heating engineer and you wished to consider a house as a dynamic system. Without a heater, the average temperature in the house would clearly vary over a 24-h period. What might you consider for inputs, outputs, and state variables for a simple dynamic model? How would you expand your model so that it would predict temperatures in several rooms of the house? How does the installation of a thermostatically controlled heater change your model?
Answer:
As a heating engineer and considering a house as a dynamic system , and that without a heater, the average temperature in the house would vary over a 24-h period.
What might you consider for inputs, outputs, and state variables for a simple dynamic model?
State variables: according to the weather conditions of the area where the house was built:
State variable # 1: minimum temperture during a day in an specific season (*4);
State variable # 2: maximum temperature during the day, in an specific season (*4) as well;
State variable # 3: average temperature during the day in an specific season (*4).
That makes 16 state variables all of them in Centigrade degrees.
Input variables:
# 1: one degree over each of the state variables given.
# 2: one degree below each of the state variables, all of them in Centigrade degrees.
Output variables:
# 1 are the temperatures reached after adding one degree to each of the input variables.
# 2 are the temperatures reached after decreasing one degree, all of them in Centigrade degrees.
How would you expand your model so that it would predict temperatures in several rooms of the house?
I would add output variables in a "Y" system to predict temperatures in several rooms of the house.
How does the installation of a thermostatically controlled heater change your model?
It would change on the "Y" variables as they will get a control system designed for sensors to produce from some input variables to make the system respond.
Explanation:
State-determined system models using well defined physical systems is of highly interest to engineers.
Create a program named PaintingDemo that instantiates an array of eight Room objects and demonstrates the Room methods. The Room constructor requires parameters for length, width, and height fields (all of type int); use a variety of values when constructing the objects. The Room class also contains the following fields: Area - The wall area of the Room (as an int) Gallons - The number of gallons of paint needed to paint the room (as an int)
Answer:
Explanation:
Code used will be like
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace PaintingWall
{
class Room
{
public int length, width, height,Area,Gallons;
public Room(int l,int w,int h)
{
length = l;
width = w;
height = h;
}
private int getLength()
{
return length;
}
private int getWidth()
{
return width;
}
private int getHeight()
{
return height;
}
public void WallAreaAndNumberGallons()
{
Area = getLength() * getHeight() * getWidth();
if (Area < 350)
{
Gallons = 1;
}
else if (Area > 350)
{
Gallons = 2;
}
Console.WriteLine ("The area of the Room is " + Area);
Console.WriteLine("The number of gallons paint needed to paint the Room is " + Gallons);
}
}
class PaintingDemo
{
static void Main(string[] args)
{
int l, w, h;
Room[] r = new Room[8];
for (int i = 0; i <= 7; i++)
{
Console.WriteLine("Room "+(i+1));
Console.Write("Enter Length : ");
l = Convert.ToInt32(Console.ReadLine() );
Console.Write("Enter Width : ");
w = Convert.ToInt32(Console.ReadLine());
Console.Write("Enter Height : ");
h= Convert.ToInt32(Console.ReadLine());
r[i] = new Room(l,w,h);
Console.WriteLine();
}
for (int i = 0; i <= 7; i++)
{
Console.WriteLine("Room " + (i + 1));
r[i].WallAreaAndNumberGallons();
}
Console.ReadKey();
}
}
}
A sign erected on uneven ground is guyed by cables EF and EG. If the force exerted by cable EF at E is 46 lb, determine the moment of that force about the line joining points A and D.
Answer:
M_AD = 1359.17 lb-in
Explanation:
Given:
- T_ef = 46 lb
Find:
- Moment of that force T_ef about the line joining points A and D.
Solution:
- Find the position of point E:
mag(BC) = sqrt ( 48^2 + 36^2) = 60 in
BE / BC = 45 / 60 = 0.75
Hence, E = < 0.75*48 , 96 , 36*0.75> = < 36 , 96 , 27 > in
- Find unit vector EF:
mag(EF) = sqrt ( (21-36)^2 + (96+14)^2 + (57-27)^2 ) = 115 in
vec(EF) = < -15 , -110 , 30 >
unit(EF) = (1/115) * < -15 , -110 , 30 >
- Tension T_EF = (46/115) * < -15 , -110 , 30 > = < -6 , -44 , 12 > lb
- Find unit vector AD:
mag(AD) = sqrt ( (48)^2 + (-12)^2 + (36)^2 ) = 12*sqrt(26) in
vec(AD) = < 48 , -12 , 36 >
unit(AD) = (1/12*sqrt(26)) * < 48 , -12 , 36 >
unit (AD) = <0.7845 , -0.19612 , 0.58835 >
Next:
M_AD = unit(AD) . ( E x T_EF)
[tex]M_d = \left[\begin{array}{ccc}0.7845&-0.19612&0.58835\\36&96&27\\-6&-44&12\end{array}\right][/tex]
M_AD = 1835.73 + 116.49528 - 593.0568
M_AD = 1359.17 lb-in
The moment of the force about the line joining points A and D is; 617.949 lb.in
What is the moment of the force?We are given;
Force exerted by cable EF at E; T_EF = 46 lb.
From the diagram of the guy wire, we can draw a triangle and we will have the following coordinates;
A(0, 0, 0)
D(48, -12, 36)
E(E_x, 96, E_z)
Also, we can get that;
BC² = 48² + 36²
BC = √(48² + 36²)
BC = 60 in
Also, from similar triangles, we will have the coordinate of E as;
E(36, 96, 27)
Position of Vector of EF is;
EF = {(21 - 36)i + (-14 - 96)j + (57 - 27)k} in
EF = {-15i - 110j + 30k} in
Magnitude of EF from online calculation = 115 in
Force along cable EF is;
F_EF = 46{(-15i - 110j + 30k)/115}
F_EF = {-6i - 44j + 12k} lb
Position vector of AE is {36i + 96j + 27k} in
Position vector of AD is {48i - 12j + 36k} in
Magnitude of AD = 61.188 N
Unit vector of AD; λ_ad = {48i - 12j + 36k}/61.188
λ_ad = 0.7845i - 0.1961j + 0.5883k
M_ad = λ_ad × r_ea × T_EF
M_ad = [tex]\left[\begin{array}{ccc}0.7845&-0.1961&0.5883\\36&96&27\\6&-44&12\end{array}\right][/tex]
Solving this gives;
M_ad = 617.949 lb.in
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Your organization spans multiple geographical locations. The name resolution is happening with a single DNS zone for the entire organization. Which of the following is likely to happen if you continue with the single DNS zone? [Choose all that apply.]
Name resolution traffic goes to the single zone
Granular application of policies
Centralized Management
Higher security
Administrative burden
Submit
Answer:Name resolution traffic goes to the single zone
Administrative burden
Submit
Centralized Management
Explanation:DNS(Domain name system) is a term used in the internet which Describes the conversion of alphabetical names into Numerical representations,he a large Organisation as stated which spans through different Geographical areas continue with a single Domain name system it will lead to the following.
Name resolution traffic will increase which might delay the execution of tasks
Administrative burden will be Increased as it is carrying out a wide range of activities.
Centralized management which may affect the flow of work.
What character string does the binary ASCII code 1010100 1101000 1101001 1110011 0100000 1101001 1110011 0100000 1000101 1000001 1010011 1011001 0100001?
Answer: This is EASY!
Explanation:
To make it easy, you would convert those binary numbers and to denary. And this gives:
84 104 105 115 32 105 115 32 69 65 83 89 33
Then, the denary numbers generated can be converted to ASCII code using this ASCII Table in the attachment below. And the result is: This is EASY!
Two loads connected in parallel draw a total of 2.4 kW at 0.8 pf lagging from a 120-V rms, 60-Hz line. One load absorbs 1.5 kW at a 0.707 pf lagging. Determine:
(a) the pf of the second load,
(b) the parallel element required to correct the pf to 0.9 lagging for the two loads.
Answer: a) 0.948 b) 117.5µf
Explanation:
Given the load, a total of 2.4kw and 0.8pf
V= 120V, 60 Hz
P= 2.4 kw, cos θ= 80
P= S sin θ - (p/cos θ) sin θ
= P tan θ(cos^-1 (0.8)
=2.4 tan(36.87)= 1.8KVAR
S= 2.4 + j1. 8KVA
1 load absorbs 1.5 kW at 0.707 pf lagging
P= 1.5 kW, cos θ= 0.707 and θ=45 degree
Q= Ptan θ= tan 45°
Q=P=1.5kw
S1= 1.5 +1.5j KVA
S1 + S2= S
2.4+j1.8= 1.5+1.5j + S2
S2= 0.9 + 0.3j KVA
S2= 0.949= 18.43 °
Pf= cos(18.43°) = 0.948
b.) pf to 0.9, a capacitor is needed.
Pf = 0.9
Cos θ= 0.9
θ= 25.84 °
(WC) V^2= P (tan θ1 - tan θ2)
C= 2400 ( tan (36. 87°) - tan (25.84°)) /2 πf × 120^2
f=60, π=22/7
C= 117.5µf
In this exercise we have to use the parallel plate and capacitor knowledge to find the values, so:
a) 0.948
b) 117.5µf
What is a capacitor?Capacitor is a component that stores electrical charges in an electrical field, accumulating an internal electrical charge imbalance.
Given the information that:
total of 2.4 kW at 0.8 pf absorbs 1.5 kW at a 0.707 pf V= 120V, 60 HzP= 2.4 kw, cos θ= 80Knowing that the formula is;
[tex]P= S sin \theta - (p/cos \theta) sin \theta\\= P tan \theta (cos^{-1} (0.8))\\=2.4 tan(36.87)= 1.8KVAR\\S= 2.4 + j1. 8KVA[/tex]
Continues the calculus we have:
[tex]S1= 1.5 +1.5j KVA\\S1 + S2= S\\2.4+j1.8= 1.5+1.5j + S2\\S2= 0.9 + 0.3j KVA\\S2= 0.949= 18.43 \\Pf= cos(18.43) = 0.948[/tex]
b.) pf to 0.9, a capacitor is needed.
[tex]Pf = 0.9\\Cos \theta= 0.9\\\theta = 25.84\\(WC) V^2= P (tan \theta_1 - tan \theta_2)\\2400 ( tan (36. 87) - tan (25.84)) /2 \pi f * 120^2\\[/tex]
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A horizontal rigid bar ABC is pinned at end A and supported by two cables at points B and C. A vertical load P 5 10 kN acts at end C of the bar. The two cables are made of steel with a modulus elasticity E 5 200 GPa and have the same cross-sectional area. Calculate the minimum cross-sectional area of each cable if the yield stress of the cable is 400 MPa and the factor of safety is 2.0. Consider load P only; ignore the weight of bar ABC and the cables.
To determine the minimum cross-sectional area of the steel cables, calculate the allowable stress and use it along with the provided load. The result is a minimum area of 50 mm² for each cable.
Explanation:The question involves calculating the minimum cross-sectional area of each steel cable designed to support a load with a safety factor, given the yield stress of the material. First, determine the allowable stress by dividing the yield stress by the factor of safety. In this case, the allowable stress is 200 MPa (400 MPa / 2.0). To find the minimum cross-sectional area (A), use the formula A = P / σ, where P = 10 kN = 10,000 N (the load) and σ (sigma) is the allowable stress in N/m². Convert 200 MPa to N/m² to get 200,000 N/m². Therefore, the minimum cross-sectional area required for each cable is 50 mm² (10,000 N / 200,000 N/m²).
The net potential energy EN between two adjacent ions, is sometimes represented by the expression
EN = -C/r + Dexp (-r/p)
in which r is the interionic separation and C, D, and rho are constants whose values depend on the specific material.
Derive an expression for the bonding energy E0 in terms of the equilibrium interionic separation r0 and the constants D and rho using the following procedure:
1. Differentiate EN with respect to r, and then set the resulting expression equal to zero.
2. Solve for C in terms of D, rho and r0.
3. Determine the expression for E0 by substitution for C in the equation above.
What is the equation that represents the correct expression for E0?
The expression for the bonding energy E₀ in terms of the equilibrium interionic separation r₀ and the given constants is;
E₀ = rD[(exp(-r₀/ρ) + exp(-r/ρ)] - EN/r₀
Bonding Energy and Net Potential Energy
We are given the expression;
EN = -C/r + Dexp(-r/ρ)
where;
EN is net potential energy
r is the interionic separation
C, D, and rho(ρ) are constants whose values depend on the specific material.
The formula for the bonding energy is usually;
E₀ = -C/r₀ + D exp(-r₀/ρ)
Step 1; We are to differentiate EN with respect to r. Thus, we have;
dEN/dr = C/r² - D exp(-r/p)
Step 2; We are to solve for C in terms of D, rho(ρ) and r₀. Thus;
E₀ + (C/r₀) = -D*exp(-r₀/ρ)
⇒ C/r₀ = -Dexp(-r₀/ρ) - E₀
Thus, multiplying both sides by r₀ gives;
C = -r₀(Dexp(-r₀/ρ) + E₀)
Step 3; We are to determine the expression for E₀ by substitution for C in the equation given. This gives us;
EN = -r₀(Dexp(-r₀/ρ) + E₀)/r + Dexp(-r/ρ)
EN = r₀*D*exp(-r₀/ρ) - (r₀E₀/r) + D*exp(-r/ρ)
EN + (r₀E₀/r) = r₀*D*exp(-r₀/ρ) + D*exp(-r/ρ)
r₀E₀/r = D[(exp(-r₀/ρ) + exp(-r/ρ)] - EN
Thus; E₀ = rD[(exp(-r₀/ρ) + exp(-r/ρ)] - EN/r₀
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5. A driver is traveling at 90 km/h on a wet road. An object is spotted on the road 140m ahead and the driver is able to come to a stop just before hitting the object. Assuming standard reaction time and using the practical-stopping distance equation, determine the grade of the road.
Answer: Check the attached
Explanation:
You are traveling along an interstate highway at 32.0 m/s (about 72 mph) when a truck stops suddenly in front of you. You immediately apply your brakes and cut your speed in half after 6.0 s.(a) What was your acceleration, assuming it was constant?
Answer:
a= - 2.6 m/s².
Explanation:
u = 32 m/s
The speed after 6 s is half of u
[tex]v= \dfrac{32}{2}=16\ m/s[/tex]
t= 6 s
The average acceleration = a
We know v = u +at
v=final velocity
u=initial velocity
Now by putting the values in the above equation
16= 32 + a x 6
[tex]a=\dfrac{16-32}{6}\ m/s^2[/tex]
[tex]a=-2.6\ m/s^2[/tex]
Therefore the acceleration will be - 2.6 m/s².
a= - 2.6 m/s².
Negative indicates that velocity and acceleration is is opposite direction.
Current density is given in cylindrical coordinates as J = −106z1.5az A/m2 in the region 0 ≤ rho ≤ 20 µm; for rho ≥ 20 µm, J = 0. (a) Find the total current crossing the surface z = 0.1 m in the az direction. (b) If the charge velocity is 2 × 106 m/s at z = 0.1 m, find rhoν there. (c) If the volume charge density at z = 0.15 m is −2000 C/m3, find the charge velocity there.
Question:
Current density is given in cylindrical coordinates as J = −10^6z^1.5az A/m² in the region 0 ≤ ρ ≤ 20 µm; for ρ ≥ 20 µm, J = 0.
(a) Find the total current crossing the surface z = 0.1 m in the az direction.
(b) If the charge velocity is 2 × 10^6 m/s at z = 0.1 m, find ρν there.
(c) If the volume charge density at z = 0.15 m is −2000 C/m3, find the charge velocity there.
Answer:
a. -39.8μA
b. -15.81mC/m³
c. 29.05m/s
Explanation:
Given
Density = J = −10^6z^1.5az A/m²
Region: 0 ≤ ρ ≤ 20 µm
ρ ≥ 20 µm
J = 0.
a. Total current is calculated by.
J * ½((ρ1)² - (ρ0)²) * 2 π * φdza.
Where J = Density = -10^6 * z^1.5
ρ1 = Upper bound of ρ = 20
ρ0 = Lower bound of ρ = 0
π = 22/7
φdza = 10^-6
z = 0.1
Total current
= -10^6 * z^1.5 * ½(20² - 0²) * 2 * 22/7 * 10^-6
= 10^6 * 0.1^1.5 * ½(20² - 0²) * 2 * 22/7 * 10^-6
= −39.7543477278310
= -39.8μA
b. Calculating velocity charge density at (ρv)
Density (J) = ρv * V
Where J = Density = -10^6 * z^1.5
V = 2 * 10^6
z = 0.1
Substitute the above values
-10^6 * 0.1 ^1.5 = ρv * 2 * 10^6
ρv = (-10^6 * 0.1^1.5)/(2 * 10^6)
ρv = -0.1^1.5/(2)
ρv = -0.015811388300841
ρv = -0.01581 --------- Approximated
ρv = -15.81mC/m³
c. Calculating Velocity
Velocity = J/V
Where Velocity Charge Density = -2000 C/m3
Where J = -10^6 * z^1.5
z = 0.15
J = -10^6 * 0.15^1.5
J = -58094.75019311125
Velocity = -58094.75019311125/-2000
Velocity = 29.047375096555625m/s
Velocity = 29.05m/s
A) The total current crossing the surface z = 0.1 m in the z^ direction is; I_tot ≈ -39.8μA
B) If the charge velocity is 2 × 10⁶ m/s, then ρv is; -15.81 mC/m³
C) If the volume charge density at z = 0.15 m is −2000 C/m³, the Charge velocity is; 29.05m/s
We are given;
Current Density; J = −10⁶z^(1.5) (z^) A/m²
Region: 0 ≤ ρ ≤ 20 µm
At ρ ≥ 20µm, J = 0.
A) Total current is gotten from the formula;
I_tot = J × ½((ρ1)² - (ρ0)²) × 2π × φdza.
Where;
J is current Density = −10⁶z^(1.5) A/m²
ρ1 is Upper bound of ρ = 20 µm
ρ0 is Lower bound of ρ = 0 µm
φdza = 10⁻⁶
z = 0.1
Thus plugging in the relevant values, we have;
Total current;
I_tot = -10⁶ × 0.1^(1.5) * ½(20² - 0²) × 2 × π × 10⁻⁶
I_tot ≈ -39.8μA
B) Formula to find ρv is;
ρv = J/V
where;
J is current density = −10⁶z^(1.5) A/m²
V is charge velocity = 2 × 10⁶ m/s
z = 0.1
Thus;
ρv = ( −10⁶ × 0.1^(1.5))/(2 × 10^6)
ρv = -¹/₂(0.1^(1.5))
ρv ≈ -0.01581 C/m³
Thus;
ρv = -15.81 mC/m³
C) Formula for the charge velocity is;
Charge Velocity = J/V
Where;
J is current density = −10⁶z^(1.5) A/m²
V is volume charge density = -2000 C/m³
At z = 0.15;
Charge velocity = (−10⁶ × 0.15^(1.5))/(-2000)
Charge velocity ≈ 29.05m/s
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the correlation between a car's engine size and its fuel economy is r = -0.774. what fraction of the variability in fuel economy is accounted for by the engine size?
Answer:
59.9%
Explanation:
R^2 =(-0.774)^2 = 0.599
59.9% of fuel is accounted for
For a certain mechanical element the constitutive relation is Y = AV^3, where Y is a system variable, a is a constant, and V is velocity. Give each answer below in terms of V. (a) If Y is the force, i.e. (Y = F), find an expression for the power in the element? (b) If Y is the linear momentum, i.e. (Y = p), find an expression for the energy stored in the element?
Answer:
a) Power = Av⁴
b) Energy = Av⁴/2 or (A²v⁶)/2m
Explanation:
Given Y = Av³
A = a constant, v = Velocity, Y = system variable.
a) If Y = Force, F, Find Power in the element.
Power is the dot product of Force and velocity and it's a scalar quantity.
i.e. P = F.v = F.v (cos θ) where θ is the angle between the force and velocity vector.
But in this case, average power is simply given by Fv.
P(avg) = Fv = Yv = (Av³) × v = Av⁴
b) If Y = linear momentum, p, Find the energy stored in the element.
Energy is related to linear momentum by the relationship between kinetic energy and linear momentum.
p = mv and E = mv²/2 = (mv)(v)/2, so,
E = pv/2
For this question, p = Y = Av³
E = Yv/2 = (Av³)v/2 = Av⁴/2
Kinetic energy is often related to momentum through this expression too,
p = mv; p² = m²v²
E = mv²/2; E = (mv²/2) × (m/m) = m²v²/2m = p²/2m
Therefore, E = Y²/2m = (Av³)²/2m = (A²v⁶)/2m
Hope this helps!
Air is compressed slowly in a piston–cylinder assembly from an initial state where p1 = 1.4 bar, V1 = 4.25 m3 , to a final state where p2 = 6.8 bar. During the process, the relation between pressure and volume follows pV = constan
The work done by the gas is -940 kJ
Explanation:
In this process, we are told that the product of pressure and volume remains constant:
[tex]pV=const.[/tex]
so we can write
[tex]p_1 V_1 = p_2 V_2[/tex]
where
[tex]p_1 = 1.4 bar[/tex] is the initial pressure
[tex]p_2 = 6.8 bar[/tex] is the final pressure
[tex]V_1=4.25 m^2[/tex] is the initial volume
Solving for [tex]V_2[/tex], we find the final volume:
[tex]V_2=\frac{p_1V_1}{p_2}=\frac{(1.4)(4.25)}{6.8}=0.875 m^3[/tex]
Now by looking at the equation of state of an ideal gas:
[tex]pV=nRT[/tex] (1)
we notice that since [tex]pV=const.[/tex], this means that also the absolute temperature of the gas T remains constant (because the number of moles n does not change). Therefore this is an isothermal process: the work done in an isothermal process is given by
[tex]W=nRTln(\frac{V_2}{V_1})[/tex]
And by looking again at (1), we can substitute (nRT) with (pV), so we get
[tex]W=p_1 V_1 ln (\frac{V_2}{V_1})[/tex]
Converting the pressure into SI units,
[tex]p_1 = 1.4 bar = 1.4\cdot 10^5 Pa[/tex]
So the work done is
[tex]W=(1.4\cdot 10^5)(4.25)ln(\frac{0.875}{4.25})=-9.4\cdot 10^5 J[/tex]
Which means -940 kJ. This value is negative since the work is done by the surroundings on the gas (because the gas is compressed).
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A 15 Watt desk-type fluorescent lamp has an effective resistance of 200 ohms when operating (note: the 15 Watts is only associated with the lamp). It is in series with a ballast that has a resistance of 80 ohms and an inductance of .9H. The lamp and ballast are operated at 120V, 60Hz. Draw the circuit
The question is incomplete! Complete question along with answers and explanation is provided below.
Question:
A 15 Watt desk-type fluorescent lamp has an effective resistance of 200 ohms when operating (note: the 15 Watts is only associated with the lamp). It is in series with a ballast that has a resistance of 80 ohms and an inductance of .9H. The lamp and ballast are operated at 120V, 60Hz.
a) Draw the circuit
b) Calculate the power drawn by the lamp
c) Calculate the apparent power
d) Calculate the power factor
e) Calculate the reactive power
f) Calculate the size of the capacitor necessary to provide unity power factor correction
Explanation:
a) draw the circuit
Refer to the attached image.
As you can see in the attached drawing, it is a series circuit containing two resistors and one inductor.
In a series circuit, current remains same throughout the circuit
The circuit is powered by an AC voltage source having voltage of 120 V and frequency 60 Hz.
The current flowing in the circuit can be found by ohm's law
I = V/Z
where V is the voltage and Z is the total impedance of the circuit
Z = R + XL
where XL is the inductive reactance
XL = j2 π f L
XL = j2*π*60*0.9
XL = j339.29Ω
Total resistance is
R =200 + 80 = 280 Ω
Total impedance is
Z = 280 + j339.29 Ω
b) Calculate the power drawn by the lamp
First calculate the current
I = V/Z
I = 120/(280 + j339.29)
I = 0.272<-50.46° A (complex notation)
P = I²R
P = (0.272)²200
P ≈ 15 W
Power drawn by the circuit
P=V*I*cos(50.46°)
P=20.77 W
c) Calculate the apparent power
A = VI*
A = 120*0.272<50.46°
A = 32.64<50.46° VA
d) Calculate the power factor
PF = cos(50.46)
PF = 0.63
e) Calculate the reactive power
Q = VIsin(50.46)
Q = 120*0.272<-50.46*sin(50.46)
Q = 25.13<-50.46 VAR
f) Calculate the size of the capacitor necessary to provide unity power factor correction
The required reactive compensation power is
Qc = P (tan(old) - tan(new))
Qc = 20.77 (tan(50.46) - tan(0))
Qc = 25.16 VAR
C = Qc/2πfV²
C = 25.16/2*π*60*120²
C = 4.63 uF
Hence adding a capacitor of 4.63 uF parallel to the load will improve the PF from 0.63 to 1.
A gas contained in a piston cylinder assembly undergoes a process from state 1 to state 2 defined by the following relationship and given properties. Determine the final volume (V2) of the gas. P*V= constant P1 = 445 kPa V1 = 2.6 m^3 P2 = 140 kPa ans is 8.3 but how do i get it?
Answer:
V2 = final volume = 8.3m^3
Explanation:
Given P1 = 445 kPa, V1 = 2.6 m^3, P2 = 140 kPa
From PV = constant; P1V1 =P2V2 , where V2 = final volume
V2 = P1V1/P2
Substituting in the equation ;
V2 = 445 x 2.6 / 140
V2 = final volume = 8.3m^3
A water jet that leaves a nozzle at 95 m/s at a flow rate of 120 kg/s is to be used to generate power by striking the buckets located on the perimeter of a wheel. Determine the power generation potential of this water jet. The power generation potential of the water jet is kW.
Answer:
P= 541.5 kW.
Explanation:
Given that
velocity of water after leaving the nozzle ,v= 95 m/s
The mass flow rate of the water , m= 120 kg/s
The power generated P is given as
[tex]P=\dfrac{1}{2}mv^2[/tex]
Now by putting the values in the above equation we get
[tex]P=\dfrac{1}{2}\times 120\times 95^2\ W[/tex]
P=541500 W
The power in kW will be 541.5 kW.
Therefore the answer will be 541.5 kW
P= 541.5 kW.
The power generation potential of the water jet is 542.25 kW.
Explanation:To determine the power generation potential of the water jet, we need to calculate the kinetic energy of the water jet and then convert it to power. The kinetic energy of the water jet can be calculated using the formula KE = 0.5 * m * v^2, where m is the mass flow rate of the water and v is the velocity of the water jet. Given that the flow rate is 120 kg/s and the velocity is 95 m/s, we can calculate the kinetic energy to be KE = 0.5 * 120 * 95^2 = 0.5 * 120 * 9025 = 542,250 J/s.
To convert the kinetic energy to power, we divide by the time taken to deliver the energy. Since the flow rate is given in kg/s, we can assume the time taken is 1 second. Therefore, the power generation potential of the water jet is 542,250 J/s, or 542.25 kW.
A gas is compressed from an initial volume of 0.42 m3 to a final volume of 0.12 m3. During the quasi-equilibrium process, the pressure changes with volume according to the relation P = aV + b, where a = –1200 kPa/m3 and b = 500 kPa. Calculate the work done during this process (a) by plotting the process on a P-V diagram and finding the area under the process curve and (b) by performing the necessary integrations.
Answer:
[tex]W=-52 800\ \text{J}=-52.8\ \text{kJ}[/tex]
Explanation:
First I sketched the compression of the gas with the help of the given pressure change process relation. That is your pressure change due to change in volume.
To find the area underneath the curve (the same as saying to find the work done) you should integrate the given relation for pressure change:
[tex]W=\int_{0.42}^{0.12}-1200V+500dV=-52.8\ \text{kJ}[/tex]
The flexural strength or MOR of a ceramic is 310 MPa. A block of the ceramic, which is 20 mm wide, 15 mm high, and 300 mm long, is supported between two rods 150 mm apart. Determine the force required to fracture the material, assuming no plastic deformation occurs.
Answer:
[tex]F=6200\ \text{N}\\[/tex]
Explanation:
In this problem you need to define the force that acts upon a beam in a 3 point bending problem. I put a picture of the problem taken from Wikipedia:
In this problem the flexural strength is defined with the following formula:
[tex]\sigma=\cfrac{3FL}{2bd^2}[/tex]
where F is the force applied, L the length between the two rods, b the width of the ceramic block and d it's height.
The force is then defined as:
[tex]F=\cfrac{2\sigma bd^2}{3L}=6200\ \text{N}[/tex]
For flow over a plate, the variation of velocity with vertical distance y from the plate is given as u(y) = ay − by2 where a and b are constants. Choose the correct relation for the wall shear stress in terms of a, b, and μ.
For flow over a plate, the variation of velocity with vertical distance y from the plate, the correct relation for the wall shear stress is [tex]\( \tau = \mu (a - 2by) \)[/tex]. The correct option is B.
The wall shear stress ([tex]\( \tau \)[/tex]) for flow over a plate is given by the following relation, assuming the fluid has constant viscosity ([tex]\( \mu \)[/tex]):
[tex]\[ \tau = \mu \frac{du}{dy} \][/tex]
Where:
[tex]\( \mu \)[/tex] = dynamic viscosity of the fluid,
[tex]\( \dfrac{du}{dy} \)[/tex] = rate of change of velocity with respect to the vertical distance (y) from the plate.
In your case, the velocity profile [tex]\( u(y) = ay - by^2 \)[/tex] is given. To find the rate of change of velocity with respect to y, we differentiate [tex]\( u(y) \)[/tex] with respect to y:
[tex]\[ \frac{du}{dy} = a - 2by \][/tex]
Now, substitute this into the formula for wall shear stress:
[tex]\[ \tau = \mu \left( a - 2by \right) \][/tex]
So, the correct relation for the wall shear stress ([tex]\( \tau \)[/tex]) in terms of a, b, and [tex]\( \mu \)[/tex] is:
[tex]\[ \tau = \mu \left( a - 2by \right) \][/tex]
Thus, the correct option is B.
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Your question seems incomplete, the probable complete question is:
Question:
For flow over a flat plate, the velocity variation with vertical distance y from the plate is given by [tex]\( u(y) = ay - by^2 \)[/tex], where a and b are constants. What is the correct expression for the wall shear stress ([tex]\( \tau \)[/tex]) in terms of a b, and dynamic viscosity ([tex]\( \mu \)[/tex])?
A) [tex]\( \tau = \mu (a - by) \)[/tex]
B) [tex]\( \tau = \mu (a - 2by) \)[/tex]
C) [tex]\( \tau = \mu (a + by) \)[/tex]
D) [tex]\( \tau = \mu (a + 2by) \)[/tex]
A driver is traveling at 52 mi/h on a wet road. an object is spotted on the road 415 ft ahead and the driver is able to come to a stop just before hitting the object. assuming standard perception/ reaction time and practical stopping distance, determine the grade of the road.
Answer:
G = 0.424
Explanation:
Ds = ( 0.278tr * V ) + (0.278 * V²)/ ( 19.6* ( f ± G))
Where Ds = stopping sight distance = 415miles = 126.5m
G = absolute grade road
V = velocity of vehicle = 52miles/hr
f = friction = 0 because the road is wet
tr = standard perception / reaction time = 2.5s
So therefore:
Substituting to get G
We have
2479.4G = 705.6G + 751.72
1773.8G = 751.72
G = 751.72/1773.8
G = 0.424
Based on the calculations, the grade of the road is equal to 2.43 %.
Given the following data:
Velocity = 52 mi/h to km/h = 83.6859 km/h.Assumed data:
Acceleration = 3.5 [tex]m/s^2[/tex]Acceleration due to gravity = 9.8 [tex]m/s^2[/tex]Standard perception/reaction time = 2.5 secondsStopping distance = 135 meters.The formula for stopping distance.Mathematically, the stopping distance for an inclined surface with a coefficient of friction is given by this formula:
[tex]SD = 0.278Vt + \frac{V^2}{254} (f \pm 0.01G)[/tex]
Where:
V is the velocity.t is the perception/reaction timeG is the grade of a surface.f is the coefficient of friction.Substituting the given parameters into the formula, we have;
[tex]135 = 0.278 \times 83.6859 \times 2.5 + [\frac{83.69^2}{254} \times (\frac{3.5}{9.8} + 0.01G)]\\\\135 = 58.14 + [27.58 \times (0.36 + 0.01G)]\\\\135 = 58.14 +9.93+0.2758G\\\\0.2758G=135 - 58.14 -9.93\\\\0.2757G=66.9923\\\\G=\frac{66.93}{0.2757}[/tex]
G = 242.76 ≈ 243
G = 2.43 %
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A lab technician is ordered to take a sample of your blood. As she inserts the needle, she says, "My, you have tough skin!" What would be an equivalent translation of this statement?
Answer:
Your stratified squamous epithelium is difficult to penetrate!
Explanation:
The epithelium is the tissue formed by one or several layers of cells attached to each other that cover all the free surfaces of the organism, and constitute the internal lining of the cavities, organs, hollows, ducts of the body and skin and also form the Mucous and glands. The epithelia also forms the parenchyma of many organs, such as the liver.
Flat or squamous epithelia: Formed by flat cells, with much less height than width and a flattened nucleus. It is one of the most external being part of the epidermis and generating some inconveniences when penetrating with needles to perform blood extractions when you have certain characteristics of hardness.