To solve this problem we will use the values of the density on surface, which relates the total load and the area of the surface. From there we will get the total charge, which will allow us to find the electric flow.
Surface density is defined as,
[tex]\mu= \frac{Q_{tot}}{A_{surface}}[/tex]
Where,
[tex]A_{surface} = 4\pi r^2[/tex]
Replacing,
[tex]4*10^{-9} = \frac{Q_{tot}}{(4\pi 0.02^2)}[/tex]
[tex]Q_{tot}= 20.11*10^{-2} C[/tex]
At the same time electric flux can be defined as,
[tex]\Phi = \frac{Qtot}{\epsilon_0}[/tex]
Here,
[tex]\epsilon_0 =[/tex] Permittivity vacuum constant
Replacing,
[tex]\Phi =\frac{(20.11 * 10-12 )}{(8.85 * 10-12)}[/tex]
[tex]\Phi =2.27N \cdot m^2 \cdot C^{-1}[/tex]
Therefore the total electric flux through the concentric spherical surface is [tex]2.27Nm^2/C[/tex]
Through conc. spherical surface, the total electric flux will be:
"2.27 N.m².C⁻¹".
Electric fluxUniform surface density charge, [tex]A_{surface}[/tex] = 4.0 nC/m²
Radius, r = 2.0 cm, or
= 0.02
We know the relation,
→ Surface density, μ = [tex]\frac{Q_{tot}}{A_{surface}}[/tex]
Here, [tex]A_{surface}[/tex] = 4πr²
[tex]Q_{tot}[/tex] = 20.11 × 10⁻² C
Now,
→ Electric flux, [tex]\Phi[/tex] = [tex]\frac{Q_{tot}}{\epsilon_0}[/tex]
By substituting the values,
= [tex]\frac{20.11\times 10^{-12}}{8.85\times 10^{-12}}[/tex]
= 2.27 N.m².C⁻¹
Thus the above response is correct.
Find out more information about electric flux here:
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Which of the following method header represents Function(true, 0, 2.6)? A. void Function( bool b, double d, int c ) B. void Function( bool b, double d, double e ) C. void Function( bool b, int c, double d ) D. b and c
Answer:
(d) b and c
Explanation:
Given;
Function (true, 0, 2.6)
From the above, the function Function receives three arguments;
=>First argument (i.e true), is of type boolean
=>Second argument (i.e 0), is of type int, float or double since integers are inherently floating point numbers. Therefore 0 can be int, float or double.
=>Third argument (i.e 2.6), is a floating point number and could be of type float or double.
Therefore the method header for the function Function could be any of the following;
i). void Function (bool b, double d, double e)
ii). void Function (bool b, int c, double d)
iii). void Function (bool b, int d, float e)
iv). void Function (bool b, float d, double e)
v). void Function (bool b, double d, float e)
vi). void Function (bool b, float d, float e)
Note:
i. The ordered list above has ii and v in bold form to show that they are part of the options specified in the question.
ii. The letters b, c, d and e specified in the method declaration variations do not matter. Since they are just dummy variables, they can be any letter.
Therefore b and c are a correct representation for the method header.
2. Volcanic islands that form over mantle plumes, such as the Hawaiian chain, are home to some of Earth’s largest volcanoes. However, several volcanoes on Mars are gigantic compared to any on Earth. What does this difference tell us about the role of plate motion in shaping the Martian surface?
The theory that explains this phenomenon is linked to the convergence of tectonic plates when there is immersion over the crazy oceanic ones. This movement could be referred to as an oceanic-oceanic convergence where the immersion of two ocean slabs occurs and one descends below another plate and initiates volcanic activity by the same mechanism that operates in all subduction zones.
In this way the movement of the plate on the Martian surface could be relatively much faster than the occurrence of the movement of the plate on Earth. Giant volcanoes form because the area of the most oceanic crust converges faster.
Consider the generic state space model *(t)= Ax(t) + Bu(t) y(t) = Cx(t)+Du(t). Use the definition of Al to show that x(t)=1*[4=) Bu(t)dt is the state response under input u(t).
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]
Suppose that the cost of electrical energy is $0.15 per kilowatt hour and that your electrical bill for 30 days is $80. Assume that the power delivered is constant over the entire 30 days. What is the power in watts? If a voltage of 120 V supplies this power, what current flows? Part of your electrical load is a 60-W light that is on continuously. By what percentage can your energy consumption be reduced by turning this light off?
Answer:
a) 740 W b) 6.2 A c) 8.1%
Explanation:
We need first to get the total energy spent during the 30 days, that can be calculated as follows:
1 month = 30 days = 720 hr
If the total cost amounts $80 (for 720 hr), and the cost per kwh is 0.15, we have:
80 $/mo = 0.15 $/Kwh*x Kwh/mo
Solving for the total energy spent in the month:
[tex]x (kwh/mo) = \frac{80}{0.15} = 533.3 kwh[/tex]
Assuming that the power delivered is constant over the entire 30 days, as power is the rate of change of energy, we can find the power as follows:
[tex]P = \frac{E}{t} = \frac{533.3 kWh}{720 h} = 0.74 kW = 740 W[/tex]
b) If the power is supplied by a voltage of 120 V, we can find the current I as follows:
[tex]I =\frac{P}{V} =\frac{740W}{120V} = 6.2A[/tex]
c) If part of the electrical load is a 60-W light, we can substract this power from the one we have just found, as follows:
P = 740 W - 60 W = 680 W
The new value of the energy spent during the entire month will be as follows:
E = 0.68 kW*24(hr/day)*30(days/mo) = 490 kWh/mo
The reduction in percentage regarding the total energy spent can be calculated as follows:
ΔE = [tex]\frac{533.3-490}{533.3} = 0.081*100= 8.1%[/tex]
⇒ ΔE(%) = 8.1%
[tex]%(E) =\frac{533.3-490}{533.3} = 0.081 * 100 = 8.1%[/tex]
What is the energy change when the temperature of 15.0 grams of solid silver is decreased from 37.3 °C to 20.5 °C ?
Answer:
Q=58.716 W
Explanation:
Given that
mass ,m = 15 g
The decrease in the temperature ,ΔT = 37.3 - 20.5 °C
ΔT = 16.8°C
We know that specific heat of silver ,Cp = 0.233 J/g°C
The energy change of the silver is given as
Q = m Cp ΔT
Now by putting the values we get
[tex]Q= 15 \times 0.233\times 16.8\ W[/tex]
Q=58.716 W
Therefore the change in the energy will be 58.716 W.
A team member who has been a good worker for many years has recently been doing poor work. You suspect that he may be tired of his job. Further, you know he is particularly sensitive to negative feedback. What is your best course of action?
Answer:
You must follow the steps below to deal with employees who have poor performance. I hope that you will be able to handle all the issues after reading this answer.
Explanation:
Be specific with facts in hand
It is important to confront your employees about their respective actions. But to convince them about their withdrawal from lack of interest, it is imperative to have a consistent record at hand as well. For example, if the employee has been constantly delayed for a period of time, specify the precise details about the frequency and intensity of absenteeism. Be sure not to overdo your statements or use hard phrases to reduce employee self-esteem. Just be direct and precise. Reiterate the guidelines accordingly.
Consider the needs of your employees
Poor performance is not always the result of an employee's carelessness. There can be multiple genuine reasons for lack of performance and it can vary from person to person. The first is to understand the reason and judge whether they are genuine or not. Even if they aren't, don't let the other person know. Focus on your concerns and provide solutions accordingly. For example, if your employees cannot focus on their work due to some personal stress, make appointments for the counseling sessions and make sure they can get back on track.
Focus on feedback
Everyone handles comments differently. Although it is always recommended to be direct and clear in your communication, there may be certain strategies you can adopt to communicate your comments effectively. If your employee has difficulties in achieving his goals, work with him and provide him with all the necessary help to improve his performance. The best way is to provide weekly or monthly comments to your employees, so they know what they must do to achieve their goals.
Provide Performance Support Technology
When faced with existing employees who do not meet expectations, it is a smart decision to offer them training (and / or training) and different resources to help them improve. As an example, you can combine a low-performance worker with someone to act as a mentor or offer you a manual with the procedures to follow. In addition, there are performance improvement tools: WalkMe is a very valuable technology that can help a worker be more efficient and precise, within the workflow (and not take them away from their daily tasks).
Offer rewards and recognition
Whenever you see that your employees have poor performance, it is always better to adopt a carrot and stick approach for instant and consistent improvement. It is a combination of rewards and punishments that you can induce for the best and worst weekly or monthly performance. This has proven to be one of the best ways to combat low performance problems in companies of all kinds for centuries.
Facing low-performance workers for the first time may not be too encouraging for managers as well, but having an adequate system to deal with them is essential, especially during the performance of change management. It is always better to deal with such situations, instead of ignoring them, to maintain the consistency of productivity and profitability in the business.
Write a static method called quadrant that takes as parameters a pair of real numbers representing an (x, y) point and that returns the quadrant number for that point. Recall that quadrants are numbered as integers from 1 to 4 with the upper-right quadrant numbered 1 and the subsequent quadrants numbered in a counter-clockwise fashion:
Full Question
Write a static method called quadrant that takes as parameters a pair of real numbers representing an (x, y) point and that returns the quadrant number for that point.
Recall that quadrants are numbered as integers from 1 to 4 with the upper-right quadrant numbered 1 and the subsequent quadrants numbered in a counter-clockwise fashion. Notice that the quadrant is determined by whether the x and y coordinates are positive or negative numbers. If a point falls on the x-axis or the y-axis, then the method should return 0.
Answer
public int quadrant(double x, double y) // Line 1
{
if(x > 0 && y > 0){ //Line 2
return 1;
}
if(x < 0 && y > 0){ // Line 3
return 2;
}
if(x < 0 && y < 0) {// Line 4
return 3;
}
if(x > 0 && y < 0) {// Line 5
return 4;
}
return 0;
}
Explanation:
Line 1 defines the static method.
The method is named quadrant and its of type integer.
Along with the method definition, two variables x and y of type double are also declared
Line 2 checks if x and y are greater than 0.
If yes then the program returns 1
Line 3 checks if x is less than 0 and y is greater than 0
If yes then the program returns 2
Line 4 checks if x and y are less than 0
If yes then the program returns 3
Line 5 checks is x is greater than 0 and y is less than 0
If yes then the program returns 4
// Program decides tuition based on several criteria: // 1 - 12 credit hours @ $150 per credit hour // 13 - 18 credit hours, flat fee $1900 // over 18 hours, $1900 plus $100 per credit hour over 18 // If year in school is 4, there is a 15% discount using System; using static System.Console; class DebugFour3 { static void Main() { int credits; year; string inputString; double tuition; const int LOWCREDITS = 12; const int HIGHCREDITS = 18; const double HOURFEE = 15000; const double DISCOUNT = 0.15; const double FLAT = 1900.00; const double RATE = 100.00; const int SENIORYEAR = 4; WriteLine("How many credits? "); inputString = ReadLine(); credits = Convert.ToInt32(inputString); WriteLine("Year in school? "); inputString = Readline(); year = Convert.ToInt32(inputString); if(credits > LOWCREDITS) tuition = HOURFEE * credits; else if(credits == HIGHCREDITS) tuition = FLAT; else tuition = FLAT + (credits - HIGHCREDITS) * RATE; if(year < SENIORYEAR) tuition = tuition - (tuition * DISCOUNT); WriteLine("For year {0}, with {1} credits", year, credits); WriteLine("Tuition is {0}", tuition.ToString("C")); } }
Answer:
using System.IO;
using System;
class Program
{
static void Main()
{
int credits, year;
string inputString;
double tuition;
const int LOWCREDITS = 12;
const int HIGHCREDITS = 18;
const double HOURFEE = 15000;
const double DISCOUNT = 0.15;
const double FLAT = 1900.00;
const double RATE = 100.00;
const int SENIORYEAR = 4;
Console.WriteLine("How many credits? ");
inputString = Console.ReadLine();
credits = Convert.ToInt32(inputString);
Console.WriteLine("Year in school? ");
inputString = Console.ReadLine();
year = Convert.ToInt32(inputString);
if(credits > LOWCREDITS)
tuition = HOURFEE * credits;
else if(credits == HIGHCREDITS)
tuition = FLAT;
else
tuition = FLAT + (credits - HIGHCREDITS) * RATE;
if(year < SENIORYEAR)
tuition = tuition - (tuition * DISCOUNT);
Console.WriteLine("For year {0}, with {1} credits",year, credits);
Console.WriteLine("Tuition is {0}", tuition.ToString("C"));
}
You haven't declared the type for the year variable in Main. There's a typo in Readline(); it should be ReadLine().
Here's the corrected version of your code:
using System;
using static System.Console;
class DebugFour3
{
static void Main()
{
int credits, year; // Declare type for year
string inputString;
double tuition;
const int LOWCREDITS = 12;
const int HIGHCREDITS = 18;
const double HOURFEE = 150.00; // Corrected constant
const double DISCOUNT = 0.15;
const double FLAT = 1900.00;
const double RATE = 100.00;
const int SENIORYEAR = 4;
WriteLine("How many credits? ");
inputString = ReadLine();
credits = Convert.ToInt32(inputString);
WriteLine("Year in school? ");
inputString = ReadLine();
year = Convert.ToInt32(inputString);
if (credits > LOWCREDITS)
tuition = HOURFEE * credits;
else if (credits <= HIGHCREDITS) // Corrected condition
tuition = FLAT;
else
tuition = FLAT + (credits - HIGHCREDITS) * RATE;
if (year < SENIORYEAR)
{
// Corrected calculation for discount
tuition = tuition - (tuition * DISCOUNT);
}
WriteLine("For year {0}, with {1} credits", year, credits);
WriteLine("Tuition is {0}", tuition.ToString("C"));
}
}
2) The magnetic field at a distance of 2 cm from a current carrying wire is 4 μT. What is the magnetic field at a distance of 4 cm from the wire?
Answer:
2 μT
Explanation:
Assuming that the diameter of the current carrying wire is very small compared with its length, we can treat it as an infinite wire.
For an infinite wire, the magnetic field, at a distance r from the wire, is given by Biot-Savart law equation, as follows:
[tex]B = \frac{\mu0*I}{2*\pi*r}[/tex]
where:
μ₀ = 4*π*10⁻⁷
I = current carried by the wire
r = distance from the wire.
So, B₁ can be calculated as follows:
[tex]B1 = \frac{\mu0*I}{2*\pi*0.02m} = 4e-6 T[/tex]
If we change the distance where we want to know the value of B, all other factors remain constant, we can write the following expression for the new value of B, B₂:
[tex]B2 = \frac{\mu0*I}{2*\pi*0.04m}[/tex]
Dividing B1 by B2, and simplifying common terms, we have:
[tex]\frac{B1}{B2} = \frac{0.04m}{0.02m} = 2[/tex]
⇒ [tex]B2 = \frac{B1}{2} = \frac{4e-6T}{2} = 2e-6 T[/tex]
⇒ B₂ = 2 μT
A quarter-circle block with a vertical rectangular end is attached to a balance beam as shown in Fig. 1. Water in the tank puts a hydrostatic pressure force on the block which causes a clockwise moment about the pivot point. This moment is balanced by the counterclockwise moment produced by the weight placed at the end of the balance beam. The purpose of this experiment is to determine the weight, W, needed to balance the beam as a function of the water depth, h.
For a given water depth, h, determine the hydrostatic pressure force, FR = γhcA, on the vertical end of the block. Also determine the point of action of this force, a distance yR – yc below the centroid of the area. Note that the equations for FR and yR – yc are different when the water level is below the end of the block (h < R2 – R1) (partially submerged) than when it is above the end of the block (h > R1 – R2) (fully submerged).
(a) Sketch the problem for a fully immersed block. Find W = f1 (h) symbolically. Then substitute for the constants. Obtain a linear relation for W as a function of h.
(b) Sketch the problem for a partially immersed block. Find W = f2 (h) symbolically. Then substitute for the constants. Obtain a cubic relation for W as a function of h.
Answer:
See attachment below
Explanation:
A water tank filled with water to a depth of 16 ft has in inspection cover (1 in. 3 1 in.) at its base, held in place by a plastic bracket. The bracket can hold a load of 9 lbf. Is the bracket strong enough? If it is, what would the water depth have to be to cause the bracket to break?
Answer:
[tex]F=6.88 [lbf][/tex]
The bracket is strong enough.
[tex]h=20.91 [ft][/tex]
Explanation:
Let's recall that the variation of the pressure respect to displacement in a liquid incomprehensible and static will be:
[tex]\frac{dP}{dy}=-\rho g[/tex]
If we take ρ (density) as a constant and solving this differential equation, we will have:
[tex]\Delta P=\rho gh[/tex]
P is the total pressureh is the heightNow, the pressure at the base will be:
[tex]P_{base}=\rho gh[/tex]
We use this equation knowing that we have atmospheric pressure on the outside of the tank.
The force on the inspection cover will be (A=1 in²):
[tex]F=P_{base}A=\rho ghA= 62.4 [lb/ft^{3}]*32.2 [ft/s^{2}]*16 [ft]*0.00689 [ft^{2}]=221.47 [\frac{lb*ft}{s^{2}}][/tex]
We know that 1 lbf = 32.17 (lb*ft)/s², so:
[tex]F=6.88 [lbf][/tex]
The statement says that the bracket can hold a load of 9 lbf, therefore the bracket is strong enough.
We can use the equation of the force to find the depth.
[tex]F=\rho ghA[/tex]
If we solve it for h we will have:
[tex]h=\frac{F}{\rho gA}[/tex]
F is the force that bracket can hold (9 lbf or 289.53 (lb*ft)/s²)A is the area (A=0.00689 ft²)[tex]h=\frac{289.53}{62.4*32.2*0.00689}=20.91 [ft][/tex]
I hope it helps you!
A submarine dives at a constant speed pf 3 m/s. It is know that the water pressure increase at a rate of one atmospheric pressure (105 Pa) per 10 meter of depth. Find the time rate of pressure change measured by the submarine.
Answer:
dP/dt = 0.3003003003
Explanation:
speed of submarine = 3m/s
rate of change of pressure with depth = (105 Pa) per 10 meter of depth
to get time taken by the submarine to cover 10m ; t = 10/3 = 3.33s
time rate of change of pressure = dP/dt = 105 Pa/3.33
= 30030.03003or 0.3003003003
You buy a 75-W lightbulb in Europe, where electricity is delivered to homes at 240V. If you use the lightbulb in the United States at 120 V (assume its resistance does not change), how bright will it be relative to 75-W 129-V bulbs?
Answer:
The bulb will be [tex]\frac{1}{4}[/tex] times as bright as it is in Europe.
Explanation:
Data provided in the question:
Power of bulb in Europe = 75 W
Voltage provided in Europe = 240 V
Voltage provided in United Stated = 120 V
Now,
We know,
Power = Voltage² ÷ Resistance
Therefore,
75 = 240² ÷ Resistance
or
Resistance = 240² ÷ 75
or
Resistance = 768 Ω
Therefore,
Power in United States = Voltage² ÷ Resistance
= 120² ÷ 768
= 18.75 W
Therefore,
Ratio of powers
[tex]\frac{\text{Power in united states}}{\text{Power in Europe}}=\frac{18.75}{75}[/tex]
= [tex]\frac{1}{4}[/tex]
Hence,
The bulb will be [tex]\frac{1}{4}[/tex] times as bright as it is in Europe.
a) the pre-exponential factor and activation energy for the hydrolysis of t-butyl chloride are 2.1xE16 sec-1 and 102 kJ mol-1, respectively. calculate the values of delta S and delta H at 286 K for the reaction.
b)the pre-exponential factor and activation energy for the gas-phase cycloaddition of maleic anyhyrdride and cycopentadiene are 5.9xE7 M-1S-1 and 51 kJ mol-1 respectively. calculate values of delta S and delta H at 293 K for the reaction.
Answer:
Delta S = 356.64J/mol and Delta H = 99620J/mol
Explanation:
The detailed steps and appropriate use of the arrehenius equation and necessary susbtitution were made as shown in the attached file.
Define a function pyramid_volume with parameters base_length, base_width, and pyramid_height, that returns the volume of a pyramid with a rectangular base. Sample output with inputs: 4.5 2.1 3.0 Volume for 4.5, 2.1, 3.0 is: 9.45 Relevant geometry equations: Volume
Hi, you haven't provided the programing language in which you need the code, I'll just explain how to do it using Python, and you can apply a similar method for any programming language.
Answer:
1. def pyramid_volume(base_length, base_width, pyramid_height):
2. volume = base_length*base_width*pyramid_height/3
3. return(volume)
Explanation step by step:
In the first line of code, we define the function pyramid_volume and it's input parametersIn the second line, we perform operations with the input values to get the volume of the pyramid with a rectangular base, the formula is V = l*w*h/3In the last line of code, we return the volumeIn the image below you can see the result of calling the function with input 4.5, 2.1, 3.0.
The function 'pyramid_volume' can be used to compute the volume of a pyramid with a rectangular base. Its calculation uses the equation: V = (base_area * height) / 3, where base_area = base_length * base_width. Given the inputs 4.5, 2.1, and 3.0 for base_length, base_width, and pyramid_height respectively, the volume would be 9.45
Explanation:The function pyramid_volume is best defined in the context of Mathematics, particularly, geometric formulas. To compute the volume of a pyramid with a rectangular base, you can utilize the formula: V = (base_area * height) / 3. Here, base_area is equal to base_length times base_width. Hence, the function pyramid_volume could be defined as follows:
Volume = lambda base_length, base_width, pyramid_height: (base_length * base_width * pyramid_height) / 3.
Thus, if you input 4.5, 2.1, and 3.0 as your base_length, base_width, pyramid_height respectively into the function, it will return the volume of the pyramid, which in this case is 9.45.
Learn more about Volume Computation here:https://brainly.com/question/33636745
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A ½ in diameter rod of 5 in length is being considered as part of a mechanical
linkage in which it can experience a tensile loading and unloading during application. The
material being considered is a standard aluminum bronze with a Young’s modulus of 15.5 Msi.
What is the maximum load the rod can take before it starts to permanently elongate?
Considering the rod is designed so that it does not yield, what is the maximum energy that the
rod would store? [Hint: You can look up some properties of aluminum bronze in your machine
design book].
Answer:
The maximum load the rod can take before it starts to permanently elongate = = 6086.84 lbf or 27075.59 N
The maximum energy that the rod would store =2.536 ft·lbf or 3.439 J
Explanation:
We list out the variables to the question as follows
Rod diameter = 0.5 in
Length of rod = 5 in
Young's modulus of elasticity for the material = 15.5 Msi
The maximum load rthe rd can take before it starts to permanently elongate is the yield stress and it can vbe calclulated by applyng the 0.2% offset rule by taking the yeild strain to be equal to 2%
Thus Young's Modulus at yield point =[tex]\frac{Stress}{Strain}[/tex] so that the yield stess is
Yield stress = strain × Young's Modulus
= 2/100 × E = 0.002 × 15.5 Msi = 0.031 Msi = 31 ksi
The maximum load that the rod would take before it starts to permanently elongate = F = Stress×Area
where area = π·D²/4 = 0.196 in²
F = 31 kSi × 0.196 in² = 6086.84 lbf or 27075.59 N
To calculate the strain energy stored in the rod we apply the strain energy formula thus
U = V×σ²/2·E
To calculate the volume we have V = L ×π·D²/4 =5 in × 0.196 in² = 0.98 in³
then U = 0.98 × (31000)²/(2·15.5×10⁶)
= 30.43 in³·psi = 2.536 ft·lbf
or 3.439 J
The maximum energy that the rod would store = 2.536 ft·lbf
The passageway between the riser and the main cavity must have a small cross-sectional area to minimize waste of casting metal: True or False
Answer:
True
Explanation:
The passageway between the riser and the main cavity must have a small cross-sectional area to reduce waste of material in metal casting.
Maintaining an optimistic attitude, building self-efficacy, strengthening stress inoculation, experiencing secure personal relationships, and practicing spirituality or religiosity are factors to help individuals thrive in the face of _____.
Optimistic attitude, self-efficacy, stress inoculation, secure relationships, and spirituality are factors that help individuals cope with adversity. These elements foster hardiness and stress-related growth, allowing individuals to manage chronic and acute stressors more effectively and pursue personal development.
Explanation:Maintaining an optimistic attitude, building self-efficacy, strengthening stress inoculation, experiencing secure personal relationships, and practicing spirituality or religiosity are factors to help individuals thrive in the face of adversity.
Adversity can include both chronic stressors, such as long-term unemployment, and acute stressors, like illnesses or loss. These stressors require coping mechanisms and personal strengths for an individual to manage effectively. Optimism is a general tendency to expect positive outcomes and it contributes to less stress and happier dispositions. Self-efficacy, defined by Albert Bandura, is the belief in one's abilities to reach goals, which empowers a proactive response to stressors. The concept of hardiness, which relates to optimism and self-efficacy, describes those who are less affected by life's stressors and use effective coping strategies.
Furthermore, stress-related growth or thriving describes the ability of individuals to exceed previous performances and show increased strength in the face of adversity. The nurturing of secure personal relationships provides social support that can aid in increasing positive affect and reducing the number of physical symptoms during stressful times.
the car travels around the circular track having a radius of r=300 m such that when it is at point A it has a velocity of 5m/s which is increasing at the rate of v =(0.06t) m/s^2, where t is in seconds. Determine the magnitudes of its velocity and accerlation when it has traveled one third of the way around the track
The magnitude of the velocity and the acceleration is 0.06 m/s^2., the velocity is 11 m/s
How to solve for the velocity and the accelerationGiven the information, we can calculate the velocity and acceleration of the car as it moves one-third of the way around the circular track.
First, we need to find the time t when the car has traveled one-third of the way around the track. The distance traveled by the car in one-third of the way around the track is d = r/3, where r is the radius of the track. The velocity of the car at time t is v = 5 + 0.06t. The distance traveled by the car can be found using the equation d = vt. Setting these two equations equal to each other, we have:
d = r/3 = 5t + 0.03t^2
Solving for t, we find that t = 20 seconds.
Next, we can find the velocity of the car at time t = 20 seconds using the equation v = 5 + 0.06t = 5 + 0.06 * 20 = 11 m/s.
Finally, the acceleration of the car can be found using the equation a = dv/dt = 0.06 m/s^2.
So, when the car has traveled one-third of the way around the track, its velocity is 11 m/s and its acceleration is 0.06 m/s^2.
Read more on velocity and acceleration here:https://brainly.com/question/460763
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Peter the postman became bored one night and, to break the monotony of the night shift, he carried out the following experiment with a row of mailboxes in the post office. These mailboxes were numbered 1 through 150, and beginning with mailbox 2, he opened the doors of all the even-numbered mailboxes, leaving the others closed. Next, beginning with mailbox 3, he went to every third mail box, opening its door if it were closed, and closing it if it were open. Then he repeated this procedure with every fourth mailbox, then every fifth mailbox, and so on. When he finished, he was surprised at the distribution of closed mailboxes. Write a program to determine which mailboxes these were.
Answer:
//This Program is written in C++
// Comments are used for explanatory purpose
#include <iostream>
using namespace std;
enum mailbox{open, close};
int box[149];
void closeAllBoxes();
void OpenClose();
void printAll();
int main()
{
closeAllBoxes();
OpenClose();
printAll();
return 0;
}
void closeAllBoxes()
{
for (int i = 0; i < 150; i++) //Iterate through from 0 to 149 which literarily means 1 to 150
{
box[i] = close; //Close all boxes
}
}
void OpenClose()
{
for(int i = 2; i < 150; i++) {
for(int j = i; j < 150; j += i) {
if (box[j] == close) //Open box if box is closed
box[j] = open;
else
box[j] = close; // Close box if box is opened
}
}
// At the end of this test, all boxes would be closed
}
void printAll()
{
for (int x = 0; x < 150; x++) //use this to test
{
if (box[x] = 1)
{
cout << "Mailbox #" << x+1 << " is closed" << endl;
// Print all close boxes
}
}
}
Explanation:
The lattice constant of a body centered cubic unit cell is a=4.85 oA. Determine the volume density of atoms. (Hints: Volume density=number of unit cell atoms/unit cell volume).
Answer:
the volume density of atoms = 1.75 x 10^22 /cm3
Explanation:
The detailed steps is as shown in the attached file.
A 6-pack of canned drinks is to be cooled from 23°C to 3°C. The mass of each canned drink is 0.355 kg. The drinks can be treated as water, and the energy stored in the aluminum can itself is negligible. The amount of heat transfer from the six canned drinks is _____ kJ.
Answer:
178 kJ
Explanation:
Assuming no heat transfer out of the cooling device, and if we can neglect the energy stored in the aluminum can, the energy transferred by the canned drinks, would be equal to the change in the internal energy of the canned drinks, as follows:
ΔU = -Q = -c*m*ΔT (1)
where c= specific heat of water = 4180 J/kg*ºC
m= total mass = 6*0.355 Kg = 2.13 kg
ΔT = difference between final and initial temperatures = 20ºC
Replacing by these values in (1), we can solve for Q as follows:
Q = 4180 J/kg*ºC * 2.13 kg * -20 ºC = -178 kJ
So, the amount of heat transfer from the six canned drinks is 178 kJ.
-178.92 kJ
Explanation:The amount or quantity (Q) of heat transferred during a chemical process such as cooling is the product of the mass (m) of the substance involved, the specific heat capacity (c) of the substance and the change in temperature (ΔT) of the substance. i.e
Q = m x c x Δ T ----------------------(i)
From the question;
m = total mass of the canned drinks = 6 x 0.355kg = 2.13kg
ΔT = final temperature - initial temperature = 3°C - 23°C = -20°C
Known constant;
c = specific heat capacity of water = 4200J/kg°C
Substitute all these values into equation (i) as follows;
Q = 2.13 x 4200 x (-20)
Q = -178920 J
Divide the result by 1000 to convert it to kJ as follows;
Q = (-178920 / 1000) kJ
Q = -178.92 kJ
Therefore, the quantity of heat transferred from the six canned drinks is -178.92 kJ
Note;
The -ve sign shows that the heat transferred is actually the heat lost in cooling the drinks from 23°C to 3°C
g Let the charges start infinitely far away and infinitely far apart. They are placed at (6 cm, 0) and (0, 3 cm), respectively, from their initial infinite positions. A. Determine the electric potential due to the two charges at the origin. B. Determine the electric potential energy associated with the system, relative to their infinite initial positions.
Answer:
a) V =10¹¹*(1.5q₁ + 3q₂)
b) U = 1.34*10¹¹q₁q₂
Explanation:
Given
x₁ = 6 cm
y₁ = 0 cm
x₂ = 0 cm
y₂ = 3 cm
q₁ = unknown value in Coulomb
q₂ = unknown value in Coulomb
A) V₁ = Kq₁/r₁
where r₁ = √((6-0)²+(0-0)²)cm = 6 cm = 0.06 m
V₁ = 9*10⁹q₁/(0.06) = 1.5*10¹¹q₁
V₂ = Kq₂/r₂
where r₂ = √((0-0)²+(3-0)²)cm = 3 cm = 0.03 m
V₂ = 9*10⁹q₂/(0.03) = 3*10¹¹q₂
The electric potential due to the two charges at the origin is
V = ∑Vi = V₁ + V₂ = 1.5*10¹¹q₁ + 3*10¹¹q₂ = 10¹¹*(1.5q₁ + 3q₂)
B) The electric potential energy associated with the system, relative to their infinite initial positions, can be obtained as follows
U = Kq₁q₂/r₁₂
where
r₁₂ = √((0-6)²+(3-0)²)cm = √45 cm = 3√5 cm = (3√5/100) m
then
U = 9*10⁹q₁q₂/(3√5/100)
⇒ U = 1.34*10¹¹q₁q₂
Compute the estimated actual endurance limit for SAE 4130 WQT 1300 steel bar with a rectangular cross section of 20.0 mm by 60 mm. It is to be machined and subjected to repeated and reversed bending stress. A reliability of 99% is desired
Answer:
estimated actual endurance strength = 183.22 MPa
Explanation:
given data
actual endurance limit for SAE 4130 WQT 1300
cross section = 20.0 mm by 60 mm
reliability = 99%
solution
we use here table for get some value for AISI 4130 WQT 1300 steel
Sut = 676 MPa
Sn = 260 MPa
and here
stress factor Cst = 1
and wrought steel Cm = 1
and reliability factor Cr is = 0.81
so here equivalent diameter will be
equivalent diameter = 0.808 [tex]\sqrt{bh}[/tex] ...........1
equivalent diameter = 0. 808 [tex]\sqrt{20*60}[/tex]
equivalent diameter = 28 mm
so here size factor will be = 0.87 by the table
so now we can get estimated actual endurance strength will be
actual endurance strength = Sn × Cm × Cst × Cr × Cs ...............2
put here value
actual endurance strength = 260 × 1 × 1 × 0.81 × 0.87
actual endurance strength = 183.22 MPa
Consider the following two well-mixed, isothermal gas-phase batch reactors for the elementary and irreversible decomposition of A to B given by the following chemical equation: A k −−→ 2 B (1) In reactor 1, the reactor volume is held constant (reactor pressure therefore changes). In reactor 2, the reactor pressure is held constant (reactor volume therefore changes). Both reactors are charged with pure A at 2.0 atm, A and B are considered ideal gases, and k = 0.35 min−1 .
(a) What is the fractional decrease in the concentration of A in reactors 1 and 2 after five minutes?
(b) What is the total molar conversion of A in reactors 1 and 2 after five minutes?
Answer:
attached below
Explanation:
How many electrons move past a fixed reference point every t = 2.55 ps if the current is i = 7.3 μA ? Express your answer as an integer.
The number of electrons that move past a fixed reference point in 2.55 ps when the current is 7.3 extmu A is approximately 116, after calculating the total charge and then dividing it by the charge of an electron.
Explanation:The number of electrons moving past a fixed reference point in a given time interval when the electric current is known can be calculated by first determining the total charge that passes through the reference point and then dividing that by the charge per electron. The total charge ( extit{Q}) that moves through a fixed point is the product of the current ( extit{I}) and the time interval ( extit{t}), which is expressed by the formula extit{Q} = extit{I} imes extit{t}. Given that the charge of an electron is approximately -1.60 imes 10^{-19} C (coulombs), one can calculate the number of electrons by dividing the total charge by the electronic charge.
In this case, the current is 7.3 extmu A (microamperes), which is equal to 7.3 imes 10^{-6} A (amperes), and the time interval is 2.55 ps (picoseconds), or 2.55 imes 10^{-12} s (seconds). Using the formula extit{Q} = extit{I} imes extit{t}, the total charge moved past the reference point is calculated as follows:
Q = 7.3 imes 10^{-6} A imes 2.55 imes 10^{-12} s = 1.8615 imes 10^{-17} C
Next, to find the number of electrons that move past the reference point, we divide the total charge by the charge of an electron:
Number of electrons = rac{Q}{e} = rac{1.8615 imes 10^{-17} C}{1.60 imes 10^{-19} C/e^-}
This yields approximately 116.34 electrons. Since the question asks for an integer value, we round this to 116 electrons as the number of electrons that move past a fixed reference point every 2.55 ps if the current is 7.3 extmu A.
To find the number of electrons moving past a fixed reference point with a given current, you can use the formula: Current (i) = Charge (q) / Time (t). By substituting the known values and calculating, you can determine the number of electrons passing the point at a specific time.
To find the number of electrons moving past a fixed reference point with given current:
Given: Current (i) = 7.3 μA = 7.3 x 10^-6 A
Formula: Current (i) = Charge (q) / Time (t)
Calculation: Number of electrons = (Current x Time) / Charge of an electron
Substitute values and calculate.
Define the following concepts in your own words: (a) stiffness, (b) strength, (c) ductility, (d) yielding, (e) toughness, and (f) strain hardening.
Answer:
Please find detail answer given below
Explanation:
(a) Stiffness
It is defined as the property of a material that is rigid and difficult to bend. The example of stiffness is the rubber band. If an elastic band is stretched with two fingers, the stiffness is less and the flexibility is greater. Similarly, if we use the rubber band assembly and stretch it with two fingers, the stiffness will be more rigid and the flexibility will be less.
(b) Strength
Force measures how much stress can be applied to an element before it is permanently deformed or fractured.
(c) ductility
Ductility is a measure of a metal's ability to resist tensile stress: any force that separates the two ends of an object. The tug of war game provides a good example of the tension of tension that is applied to a rope. Ductility is the plastic deformation that occurs in the metal as a result of such types of deformation. The term "ductile" literally means that a metallic substance is capable of stretching on a thin wire without weakening or becoming more fragile in the process.
(d) Yielding
The performance of a material is explained as the tension at which a material begins to deform irreversibly. Before the elastic limit, the material will deform elastically, which means that it will return to its original shape when the applied tension is eliminated (that is, without permanent and visible change in the shape of the material).
(e) toughness
The ability of a metal to deform plastically and to absorb energy in the process before the fracture is called resistance. The emphasis of this definition should be placed on the ability to absorb energy before the fracture.
(f) strain hardening
The process of strain hardening is at what time when metal is strained beyond the yield point. An intensifying pressure is required to produce extra plastic deformation and the metal apparently becomes stronger and more difficult to deform.
A signal is band-limited between 2 kHz and 2.1 kHz. If the signal is sampled at 1750 Hz, the baseband signal will lie between?a. 00 and 650 Hzb. 750 and 2000 Hzc. 100 and 110 Hzd. 250 and 350 Hz
Answer:
Explanation: 100 and 110 Hz
Here fL=2kHz, fH=2.1kHz, B=0.1kHz, fL=2kHz,
n≤fH/(fH-fL)
n≤(2.1*1000)/((2.1-2)*1000)
n≤20
2fH/n ≤ fs
(2*2.1*1000)/20 ≤ fs
210≤ fs
The following questions present a twist on the scenario above to test your understanding.
Suppose another stone is thrown horizontally from the same building. If it strikes the ground 67 m away, find the following values.
(a) time of flight
s
(b) initial speed
m/s
(c) speed and angle with respect to the horizontal of the velocity vector at impact
m/s
Answer:
a) t = √(2*h/g)
b) v₀ = 67*√(g/(2*h))
c) ∅ = tan⁻¹(2*h/67)
Explanation:
Suppose that the height of the building is known (h), then we have
a) y = h = g*t²/2
then the time of flight is
t = √(2*h/g)
b) The initial speed (v₀) can be obtained as follows
x = v₀*t ⇒ v₀ = x / t
where
x = xmax = 67 m and t = √(2*h/g)
So, we have
v₀ = 67 / √(2*h/g) = 67*√(g/(2*h))
c) In order to get the speed and the angle with respect to the horizontal of the velocity vector at impact, we obtain vx and vy as follows
vx = v₀ = 67*√(g/(2*h))
and
vy = gt = g*√(2*h/g) ⇒ vy = √(2*h*g)
then we apply
v = √(vx² + vy²) = √((67*√(g/(2*h)))² + (√(2*h*g))²)
⇒ v = √(g*(4489 + 4*h²)/(2*h))
The angle is obtained as follows
tan ∅ = vy / vx ⇒ tan ∅ = √(2*h*g) / 67*√(g/(2*h))
⇒ tan ∅ = 2*h / 67 ⇒ ∅ = tan⁻¹(2*h/67)
Python Codio Question:
This is a tricky challenge for you.
We will pass in a value N. N can be positive or negative.
If N is positive then output all values from N down to and excluding 0.
If N is negative, then output every value from N up to and excluding 0.
# Get N from the command line
import sys
N = int(sys.argv[1])
# Your code goes here
Answer:
# -*- coding: utf-8 -*-
# Get N from the command line
import sys
N = int(sys.argv[1])
if N > 0: #N is positive
positve = list(range(N,0,-1))
print(positve)
elif N < 0: #N is negative
negative = list(range(N,0,1))
print(negative)
else:
print("Invalid in input")
Explanation:
First, you need to identify if the number entered is positive, negative or none of them, for that we use one if, one elif and one else statement:
If the number entered (N) is greater than zero (is positive) we print a list in the range N to 0 in steps of minus one, the zero is not printed because the function range by default omits the last valueIf the previous statement was False and the number entered is smaller than zero (is negative) we print a list in the range N to 0 in steps of one, the zero is not printed because the function range by default omits the last valueFinally, if the two previous events were false print invalid input