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?

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.

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.

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]                            

Now, 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]

[tex]h=\frac{289.53}{62.4*32.2*0.00689}=20.91 [ft][/tex]

I hope it helps you!

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;

}

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

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

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.

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.

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.

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.

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.

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]

Water is likely to leak out of the hole in the pipe when the average velocity is 5.0 m/s due to high dynamic pressure. With a slower velocity of 0.5 m/s, air might enter the pipe if the static pressure at the hole is less than atmospheric, but this requires additional details to confirm.

The question addresses the behavior of water within a pipe system under different flow conditions, involving principles of fluid dynamics. Specifically, it asks whether water will leak out of a small hole in a vertical pipe or if air will enter into the pipe through the hole given two different average velocities of water flow.

Case 1: Average velocity of 5.0 m/s

With an average velocity of 5.0 m/s, the dynamic pressure of the flowing water is considerable, and thus, water is likely to leak out of the hole due to the higher pressure inside the pipe compared to atmospheric pressure.

Case 2: Average velocity of 0.5 m/s

With a decreased velocity of 0.5 m/s, the dynamic pressure is significantly lower. If the static pressure at the hole's location is less than the atmospheric pressure, air might enter the pipe; however, if it is still higher than atmospheric, water would continue to leak out. The determination requires additional information, such as the height of the water column above the hole and any applied pressures at the water's source.

Generally, the behavior can be predicted using Bernoulli's principle and the continuity equation for incompressible flow, which together relate the velocities, pressures, and cross-sectional areas in different sections of a pipe.

The magnitude of the velocity and the acceleration is 0.06 m/s^2., the velocity is 11 m/s

Given 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

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"));

   }

}

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".

Uniform 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

Answer:

a) -k* dT / dx = q_o

b) T(x) = -280*x + 80

c) T(L) = -4 C

Explanation:

Given:

- large plane wall of thickness L = 0.3 m

- thermal conductivity k = 2.5 W/m · °C

- surface area A =12 m2.

- left side of the wall at net heat flux q_o = 700 W/m2 @ x = 0

- temperature at that surface is measured to be T1 =80°C.

Find:

- Express the differential equation and the boundary conditions for steady one dimensional heat conduction through the wall.

- Obtain a relation for the variation of temperature in the wall by solving the differential equation

- Evaluate the temperature of the right surface of the wall at x = L.

Solution:

- The mathematical formulation of Rate of change of temperature is as follows:

                                    d^2T / dx^2 = 0

- Using energy balance:

                                    E_out = E_in

                                   -k* dT / dx = q_o

- Integrate the ODE with respect to x:

                                     T(x) = - (q_o / k)*x + C

- Use the boundary conditions, T(0) = T_1 = 80C

                                     80 = - (q_o / k)*0 + C

                                      C = 80 C

-Hence the Temperature distribution in the wall along the thickness is:

                                    T(x) = - (q_o / k)*x + 80

                                    T(x) = -(700/2.5)*x + 80

                                    T(x) = -280*x + 80

- Use the above relation and compute T(L):

                                     T(L) = -280*0.3 + 80

                                     T(L) = -84 + 80 = -4 C

Differential equation: [tex]\(\frac{{d}}{{dx}} \left( k \frac{{dT}}{{dx}} \right) = 0\).[/tex]Temperature variation: [tex]\(T(x) = T_1\).[/tex]Temperature at x = L is [tex]\(T(L) = T_1\)[/tex].

a. The differential equation for steady one-dimensional heat conduction through the wall is given by Fourier's law:

[tex]\[ \frac{{d}}{{dx}} \left( k \frac{{dT}}{{dx}} \right) = 0 \][/tex]

This equation states that the rate of change of heat flux with respect to distance ( x ) is constant and equal to zero in steady-state conditions.

The boundary conditions are:

1. At  x = 0 : [tex]\( q = q_0 \)[/tex], [tex]\( T = T_1 \)[/tex]

2. At  x = L : [tex]\( \frac{{dT}}{{dx}} = 0 \)[/tex], as there is no heat flux across the right surface of the wall.

b. To solve the differential equation, integrate it twice:

[tex]\[ k \frac{{dT}}{{dx}} = C_1 \][/tex]

[tex]\[ \frac{{dT}}{{dx}} = \frac{{C_1}}{{k}} \][/tex]

[tex]\[ T = \frac{{C_1}}{{k}} x + C_2 \][/tex]

Apply the boundary conditions:

At  x = 0 : [tex]\( T = T_1 \)[/tex]

[tex]\[ C_2 = T_1 \][/tex]

At  x = L : [tex]\( \frac{{dT}}{{dx}} = 0 \)[/tex]

[tex]\[ \frac{{C_1}}{{k}} = 0 \][/tex]

[tex]\[ C_1 = 0 \][/tex]

Therefore, the temperature variation in the wall is given by:

[tex]\[ T(x) = T_1 \][/tex]

c. The temperature of the right surface of the wall at  x = L  is equal to [tex]( T(L) = T_1 \)[/tex], as there is no variation in temperature along the wall according to the solution obtained in part b.

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

Answer:

the volume density of atoms = 1.75 x 10^22 /cm3

Explanation:

The detailed steps is as shown in the attached file.

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

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

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]".

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

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

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

Answer:

the time, in hours = 4.07hrs

Explanation:

The detailed step by step derivation and appropriate integration is as shown in the attached files.

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

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)

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:

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.

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:

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          

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.

Answer:

See attachment below

Explanation: