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 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
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.
Charge of uniform surface density (4.0 nC/m2 ) is distributed on a spherical surface (radius = 2.0 cm). What is the total electric flux through a concentric spherical surface with a radius of 4.0 cm?
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:
https://brainly.com/question/1592046
Assume that the flow of air through a given duct is isentropic. At one point in the duct, the pressure and temperature are pl = 1800 lb/ft2 and TI = 500°R, Problems respectively. At a second point, the temperature is 400"R. Calculate the pressure and density at this second point.
Answer:
pressure is 825 lb/ft²
density is 1.20 × [tex]10^{-3}[/tex] slug/ft²
Explanation:
given data
p1 = 1800 lb/ft²
T1 = 500°
T2 = 400°
solution
we use here isentropic flow relation that is
[tex]\frac{P2}{P1} = (\frac{T2}{T1})^{\gamma / \gamma - 1 }[/tex]
put here value we get pressure P2
P2 = 1800 × [tex](\frac{400}{500})^{3.5}[/tex]
P2 = 825 lb/ft²
and we know pressure is
pressure = [tex]\rho RT[/tex]
so for pressure 825 we get here [tex]\rho[/tex]
825 = [tex]\rho[/tex] × 1716 × 400
[tex]\rho[/tex] = 1.20 × [tex]10^{-3}[/tex] slug/ft²
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 man can swim at 4 ft / s in still water. He wishes to cross the 40-ft-wide river to point B, 30 ft downstream. The river flows with a velocity of 2 ft / s. Note: While in the water he must not direct himself toward point B to reach this point.
The question involves calculating the direction a swimmer must aim to reach a specified point across a river, accounting for his swimming speed and the river's current. It illustrates a problem of relative motion and vector resolution in mathematics.
Explanation:The question concerns a man wanting to cross a 40-ft-wide river to a point 30 ft downstream. This scenario involves relative motion in physics, but the calculation primarily uses mathematics to solve. The man can swim at 4 ft/s in still water, and the river flows at 2 ft/s.
To determine the direction the man must swim to reach point B directly across the river, we consider two components of motion: his speed in still water and the river's current. However, the original question indicates he must not swim directly toward point B due to the river's current. Instead, he should aim upstream at a specific angle that compensates for the downstream drift caused by the current.
Without additional details, we can provide a general explanation. The man's effective speed across the river (perpendicular to the current) remains 4 ft/s. To reach point B 30 ft downstream, he must calculate the angle to offset the river's 2 ft/s current. Typically, this involves using trigonometric functions to resolve the swimmer's velocity vector into components parallel and perpendicular to the river flow.
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:
A supersonic nozzle has an exit area 2.5 time the throat area. For a steady, isentropic flow (gamma=1.4) discharging into an atmosphere with pressure Pa, find the Mach number at the throat and at the exit plane for:
a. Pa/Pt = 0.06
b. Pa/Pt = 0.9725
Answer:
The Mach number of the throat for supersonic flow = M* = 1
and the Mach number at exit = 2.44
For
a. Pa/Pt = 0.06, Me = 2.484 Supersonic flow
b. when Pa/Pt = 0.9725 Me = 0.1999 ≅ 0.2 or subsonic flow The mach number at the throat could also be determined given the temperature parameter
Explanation:
To solve the question we note that for a supersonic nozzle, the mach number at the throat = 1
Therefore M* = 1
[tex]\frac{A_{e} }{A^{*} } = 2.5[/tex] = [tex]\frac{1}{M_{e} } (\frac{2+(\gamma -1)M_{e} ^{2} }{\gamma +1} )^{\frac{\gamma +1}{2(\gamma -1)} }[/tex] = [tex]\frac{1}{M_{e} } (\frac{2+(0.4)M_{e} ^{2} }{2.4} )^{3 }[/tex] = [tex]\frac{1}{M_{e} } ({2+(0.4)M_{e} ^{2} })^{3 } = 34.56[/tex]
34.56Me = (2+(0.4)M²)³ expanding and collecting like terms we have
Possible solutions of Me = 0.2395, 2.44, 0.90
Since flow is supersonic, Me = 2.44
a)
Solving for [tex]M_{e}[/tex] we have [tex]\frac{P_{a} }{P_{t} } =(1+\frac{\gamma -1}{2} M^{2} _{e} )^{\frac{-\gamma}{\gamma -1} }[/tex]
When Pa/Pt = 0.06 =[tex](1+\frac{1.4 -1}{2} M^{2} _{e} )^{\frac{-1.4}{1.4 -1} }[/tex] = [tex](1+0.2M^{2} _{e} )^{-3.5 }[/tex]
Solving, we get Me = 2.484 Supersonic flow
b)
When Pa/Pt = 0.9725, Me = 0.1999 ≅ 0.2 or subsonic flow
The Mach number at the throat is 1.0859 for both cases, and the Mach number at the exit plane is 1.5329 for case a and 0.2622 for case b.
To find the Mach number at the throat and at the exit plane of a supersonic nozzle with an exit area 2.5 times the throat area for a steady, isentropic flow (gamma=1.4) discharging into an atmosphere with pressure Pa, we can use the following steps:
Calculate the critical pressure ratio:
pr_crit = (2 / (gamma + 1)) ** (gamma / (gamma - 1))
pr_crit = 0.5283
Calculate the Mach number at the throat:
Mt = sqrt((1 - pr_crit) / (gamma - 1))
Mt = 1.0859
Calculate the Mach number at the exit plane:
Me = sqrt((1 - (Pa / Pt)) / (gamma - 1))
Part a:
Pa/Pt = 0.06
Me = sqrt((1 - 0.06) / (1.4 - 1))
Me = 1.5329
Part b:
Pa/Pt = 0.9725
Me = sqrt((1 - 0.9725) / (1.4 - 1))
Me = 0.2622
Therefore, the Mach number at the throat is 1.0859 for both cases, and the Mach number at the exit plane is 1.5329 for case a and 0.2622 for case b.
For such more question on number
https://brainly.com/question/13906626
#SPJ3
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
#SPJ1
Another common unit for viscosity is the centipoise, which is 10-2 poise (just like a centimeter is 10-2 meters). (1 centipoise = 0.001 N⋅s/m2) The unit poise is named after Jean Leonard Marie Poiseuille. The conversion between the centipoise and the Pa⋅s is 1 centipoise = 1 mPa.s. What is the viscosity of water at room temperature?
Answer and Explanation
Picking room temperature to be 20°C, the viscosity of water (μ) obtained from literature is:
1 centipoise = 0.01 poise = 0.001 Pa.s = 0.001 N.s/m² (This viscosity is the dynamic viscosity of water at 20°C)
Kinematic viscosity of water, η = μ/ρ
At 20°C, μ = 0.001 Pa.s, ρ = 998.23 kg/m³
η = 0.001/998.23 = 1.0 × 10⁻⁶ m²/s
Picking the room temperature to be 25°C, the viscosityof water (μ) is
0.89 centipoise = 0.0089 poise = 0.00089 Pa.s = 8.9 × 10⁻⁴ N.s/m²
Kinematic viscosity of water, η = μ/ρ
At 25°C, μ = 0.00089 Pa.s, ρ = 997 kg/m³
η = 0.00089/997 = 8.9 × 10⁻⁷ m²/s
The viscosity of water at room temperature in
We want to find the viscosity of water at room temperature in N.s/m² and mPa.s is;
0.001 N.s/m² and 1 mPa.s
Room temperature is 20°C.
Now, from online research, the viscosity of water at room temperature is 0.01 poise
We are told that;
1 centipoise = 0.01 poise
Also, that;
1 centipoise = 0.001 N⋅s/m²
Thus; viscosity of water at room temperature in N. s/m² is; 0.001 N.s/m²
Also, also 1 centipoise = 1 mPa.s
Thus;
Viscosity of water at room temperature in mPa.s is; 1 mPa.s
Read more about viscosity at; https://brainly.com/question/2568610
Toyota customers have a wide range of preferences. Some prefer the fuel economy of a hybrid (Prius), some prefer a family van (Odyssey), and others may prefer a large SUV (Highlander). This is an example of
Answer:
Product segmentation
Explanation:
Product segmentation is an adaptable method for gathering items. Likewise to an objective gathering, an item fragment contains all items that have a specific blend of item qualities.
You can utilize item sections in a battle to improve key figure arranging.
The significance of market division is that it permits a business to exactly arrive at a purchaser with explicit needs and needs.
Group of answer choices:
A) differences in willingness to pay
B) horizontal differentiation
C) segmentation
D) vertical differentiation
E) mass customization
Answer:
The correct answer is letter "C": segmentation.
Explanation:
Market segmentation is the classification of companies that make up their customers based on features such as age, gender, income, and profession, just to mention a few. By segmenting the market, firms group consumers with certain characteristics that enable the institution to specialize in the analysis of that particular sector to provide them with a tailored product or service that they are more likely to purchase.
Therefore, Toyota is segmenting its market in economic vehicles (Prius), size of the family (Odyssey), and large SUV (Highlander) to better fit consumer needs and preferences.
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.
c.Task 3: an application program where the user enters the price of an item and the program computes shipping costs. If the item price is $100 or more, then shipping is free otherwise it is 2% of the price. The program should output the shipping cost and the total price.
Answer:
//Program is written in C++ language
// Comments explains difficult lines
#include<iostream>
using namespace std;
int main (){
//Variable declaration
double price, shipping;
cout<<"Item Price: ";
cin>>price;
//Test price for shipping fee
shipping = 0; //Initialise shipping fee to 0 (free)
if(price <100)
{
shipping = 0.2 * price;
}
//The above if statement tests if item price is less than 100.
//If yes, then the shipping fee is calculated as 20% of the item price
//Else, (if item price is at least 100), shipping fee is 0, which means free
if(shipping != 0)
{
cout<<"Shipping fee is "<<shipping;
}
else
{
cout<<"Item will be shipped for free";
}
return 0;
}
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]
An incoming signal is at a frequency of 500kHz. This signal needs to be acquired and all other signals attenuated. Design a passive bandpass filter to do this. Do this by combining a high pass and a low pass filter. For our purposes, create a pass band width of exactly 40kHz and is centered at the ideal frequency. R = 2kOhms. Select C[lowpass], enter value in terms of nF =
Answer:
C_h = 0.166 nF
C_L = 0.153 nF
Explanation:
Given:
- Ideal frequency f_o = 500 KHz
- Bandwidth of frequency BW = 40 KHz
- The resistance identical to both low and high pass filter = 2 Kohms
Find:
Design a passive band-pass filter to do this by cascading a low and high pass filter.
Solution:
- First determine the cut-off frequencies f_c for each filter:
f_c,L for High pass filter:
f_c,L = f_o - BW/2 = 500 - 40/2
f_c,L = 480 KHz
f_c,h for Low pass filter:
f_c,h = f_o + BW/2 = 500 + 40/2
f_c,h = 520 KHz
- Now use the design formula for R-C circuit for each filter:
General design formula:
f_c = 1 /2*pi*R*C_i
C,h for High pass filter:
C_h = 1 /2*pi*R*f_c,L
C_h = 1 /2*pi*2000*480,000
C_h = 0.166 nF
C,L for Low pass filter:
C_L = 1 /2*pi*R*f_c,h
C_L = 1 /2*pi*2000*520,000
C_L = 0.153 nF
// 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"));
}
}
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.
A lagoon is designed to accommodate an input flow of 0.10 m^3/s of nonconservative pollutant with concentration 30 mg/L and decay coefficient of 0.2/day. The effluent from the lagoon must have pollutant concentration of less than 10 mg/L. How large is the lagoon (assume complete mixing)?
Answer:
Volume of the lagoon required for the decay process must be larger than 86580 m³ = 8.658 × 10⁷ L
Explanation:
The lagoon can be modelled as a Mixed flow reactor.
From the value of the decay constant (0.2/day), one can deduce that the decay reaction of the pollutant is a first order reaction.
The performance equation of a Mixed flow reactor is given from the material and component balance thus:
(V/F₀) = (C₀ - C)/((C₀)(-r)) (From the Chemical Reaction Engineering textbook, authored by Prof. Octave Levenspiel)
V = volume of the reactor (The lagoon) = ?
C₀ = Initial concentration of the reactant (the pollutant concentration) = 30 mg/L = 0.03 mg/m³
F₀ = Initial flow rate of reactant in mg/s = 0.10 m³/s × C₀ = 0.1 m³/s × 0.03 mg/m³ = 0.003 mg/s
C = concentration of reactant at any time; effluent concentration < 10mg/L, this means the maximum concentration of pollutant allowed in the effluent is 10 mg/L
For the sake of easy calculation, C = the maximum value = 10 mg/L = 0.01 mg/m³
(-r) = kC (Since we know this decay process is a first order reaction)
This makes the performance equation to be:
(kVC₀/F₀) = (C₀ - C)/C
V = F₀(C₀ - C)/(kC₀C)
k = 0.2/day = 0.2/(24 × 3600s) = 2.31 × 10⁻⁶/s
V = 0.003(0.03 - 0.01)/(2.31 × 10⁻⁶ × 0.03 × 0.01)
V = 86580 m³
Since this calculation is made for the maximum concentration of 10mg/L of pollutant in the effluent, the volume obtained is the minimum volume of reactor (lagoon) to ensure a maximum volume of 10 mg/L of pollutant is contained in the effluent.
The lower the concentration required for the pollutant in the effluent, the larger the volume of reactor (lagoon) required for this decay reaction. (Provided all the other parameters stay the same)
Hope this helps!
The volume is "[tex]8.64 \times 10^4 \ m^3[/tex]".
lagoon relation:The volume for the lagoon relation:
[tex]\to Q_1C_1 = Q_2C_2+KC_2V\\\\[/tex]
[tex]\to Q_1=0.10\\\\\to C_1=30\\\\\to Q_2=0.2\\\\\to C_2=10\\\\\to K=0.10[/tex]
Putting the value into the formula and calculating the volume:
[tex]\to (0.10 \times 30)=(0.2\times 10)+(0.10\times 10\times V \times (\frac{1}{24 \ hrs}) \times (\frac{1 \ hr}{3,600 \ sec})) \\\\\to (3)=(2)+(1\times V \times \frac{1}{86400}) \\\\\to 3-2=(1\times V \times \frac{1}{86400}) \\\\\to 1=(1\times V \times \frac{1}{86400}) \\\\\to 1 \times 86400 = V\\\\\to V= 8.64 \times 10^4 \ m^3\\\\[/tex]
Therefore, the calculated volume is "[tex]8.64 \times 10^4 \ m^3[/tex]".
Find out more information about the volume here:
brainly.com/question/1578538
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.
If the wire has a diameter of 0.6 in., determine how much it stretches when a distributed load of w=100lb/ft acts on the beam. The wire remains elastic. Use Est=29.0(103)ksi.
To determine how much a wire stretches under a load, one must consider Hooke's Law and use Young's modulus, but the original length of the wire and the type of load distribution must be known.
Explanation:The student's question relates to the concept of elasticity and specifically the calculation of how much a wire stretches under a given load within the elastic limit.
Using the provided diameter of the wire, which is 0.6 inches, and the fact that an elastic modulus (Est) of 29.0(103) ksi has been given, the stretch can be calculated using Hooke's Law, which relates stress and strain through Young's modulus. However, to calculate the stretch, we would require the initial length of the wire and the format of the distributed load (whether it's uniform or varies along the length of the wire) which have not been provided in the student’s query.
Assuming uniform distributed load and an initial length L, the stretch (delta L) can be found using delta L = rac{wL^2}{AE}, where w is the load per unit length, L is the initial length, A is the cross-sectional area, and E is Young's modulus. If the total weight of the load is given instead, then the formula delta L = rac{FL}{AE} can be used with F representing the total force or weight affecting the wire.
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:
For the following functions, determine if the functions are periodic. If so, find the shortest period. A. x(t) = (t2-1)cos(t) B. x(t)-cos(nt) +5 sin(5πt) C. x[n] = sin[n] + cos[n] D. x(t) cos(3t) + sin(Tt) E. x[n] = cos(m) + 1 F. x(t)-(sin nt)(cos πt) | 2
Answer:
A Not Periodic
B Periodic
C Periodic
D Not Periodic
E Periodic
F Periodic
Explanation:
The detailed steps to ascertain the periodicity or the non periodicity of each functions is as shown in the attached file.
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)
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 manufacturer of a 1.5 V D flashlight battery says that the battery will deliver 9 {\rm mA} for 37 continuous hours. During that time the voltage will drop from 1.5 {\rm V} to 1.0 {\rm V} . Assume the drop in voltage is linear with time.Part A How much energy does the battery deliver in this 37h interval?Express your answer to two significant figures and include the appropriate units.
Answer:
1.42 KJ
Explanation:
solution:
power in beginning [tex]p_{0}[/tex]=(1.5 V).(9×[tex]10^{-3}[/tex] A)
= 13.5 mW
after continuous 37 hours it drops to
[tex]p_{37}[/tex]=(1 V).(9×[tex]10^{-3}[/tex] A)
=9 mW
When the voltage will drop energy will not remain the same but the voltage drop will always remain same if the voltage was drop to for example from 5 V to 4.5 V the drop will remain the same.
37 hours= 37.60.60
=133200 s
w=(9×[tex]10^{-3}[/tex] A×133200 )+[tex]\frac{1}{2}(13.5.10^{-3}-9.10^{-3})(133200)[/tex]
=1.42 KJ
NOTE:
There maybe a calculation error but the method is correct.
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 inputA 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
A water pump increases the water pressure from 15 psia to 70 psia. Determine the power input required, in hp, to pump 0.8 ft3 /s of water. Does the water temperature at the inlet have any significant effect on the required flow power
Answer:
[tex]Power=11.52hp[/tex]
Explanation:
Given data
[tex]p_{1}=15psia\\p_{2}=70psia\\V_{ol}=0.8ft^{3}/s[/tex]
As
[tex]m=p*V_{ol}[/tex]
Assuming in-compressible flow p is constant
The total change in system mechanical energy calculated as:
Δe=(p₂-p₁)/p
The Power can be calculated as
[tex]P=W\\P=m(delta)e\\P=p*V_{ol}*(p_{2}-p_{1})/p\\ P=V_{ol}*(p_{2}-p_{1})\\P=44(psia.ft^{3}/s )*[(\frac{1Btu}{5.404psia.ft^{3} } )-(\frac{1hp}{0.7068Btu/s } )]\\P=11.52hp[/tex]
The power input required to pump 0.8 ft³/s of water is; P = 11.52 hP
We are given;
Initial pressure; P1 = 15 psia
Final pressure; P2 = 70 psia
Volume flow rate; V' = 0.8 ft³/s
The formula for the mass flow rate is;
m' = ρV'
The total change in the mechanical energy of the system is;
△E = (P2 - P1)/ρ
Now, formula for power is;
P = m' × △E
P = ρV' × (P2 - P1)/ρ
P = V'(P2 - P1)
P = 0.8(70 - 15)
P = 44 psia.ft³/s
Converting to Btu/s gives;
P = 8.142 btu/s
Converting Btu/s to HP from conversion tables gives; P = 11.52 hP
Read more about Power input at; https://brainly.com/question/5684937
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!
There is an electric field near the Earth's surface whose magnitude is about 145 V/m . How much energy is stored per cubic meter in this field? Express your answer using two significant figures.
Answer:
u_e = 9.3 * 10^-8 J / m^3 ( 2 sig. fig)
Explanation:
Given:
- Electric Field strength near earth's surface E = 145 V / m
- permittivity of free space (electric constant) e_o = 8.854 *10^-12 s^4 A^2 / m^3 kg
Find:
- How much energy is stored per cubic meter in this field?
Solution:
- The solution requires the energy density stored between earth's surface and the source of electric field strength. The formula for charge density is given by:
u_e = 0.5*e_o * E^2
- Plug in the values given:
u_e = 0.5*8.854 *10^-12 *145^2
u_e = 9.30777 * 10^-8 J/m^3