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 ?

Answers

Answer 1

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.


Related Questions

In a conduit with a diameter of 4.5 ft, the depth of flow is 4.0 ft. (a) Determine the hydraulic radius, hydraulic depth, and section factors for critical and normal flows. (b) Determine the alternate depth of flow that will carry the same discharge.

Answers

Answer:

(a) 1.125 ft, Section factor = 22.78

(b) 42.75 ft

Explanation:

Hydraulic radius is given by [tex]R_{H} = \frac{A}{P}[/tex] Where

A = Cross sectional area of flow and

P = Perimeter  h

Since the cross section is a circle  then at depth 4 of 4.5 the perimeter

[tex]=2 \pi r-\frac{\theta }{360} *2 \pi r[/tex]  where r = 2.25 and θ = 102.1 °

perimeter = 10.1 ft and the  area = [tex]=\pi r^2-\frac{\theta }{360} * \pi r^2[/tex] =  11.39 ft²

Therefore [tex]R_{H} = \frac{11.39}{10.1} = 1.125 ft[/tex]

Section factor is given by for critical flow = Z = A×√D

= 11.39 ft² × √(4 ft) = 22.78

for normal flow Z =[tex]Z_{} ^{2} = \frac{A^{3}}{T}[/tex] = 22.78

(b) The alternate depth of flow is given by

for a given flow rate, we have from chart for flow in circular pipes

Alternative depth = 0.9×45 = 42.75 ft

Your organization spans multiple geographical locations. The name resolution is happening with a single DNS zone for the entire organization. Which of the following is likely to happen if you continue with the single DNS zone? [Choose all that apply.]

Name resolution traffic goes to the single zone

Granular application of policies

Centralized Management

Higher security

Administrative burden
Submit

Answers

Answer:Name resolution traffic goes to the single zone

Administrative burden

Submit

Centralized Management

Explanation:DNS(Domain name system) is a term used in the internet which Describes the conversion of alphabetical names into Numerical representations,he a large Organisation as stated which spans through different Geographical areas continue with a single Domain name system it will lead to the following.

Name resolution traffic will increase which might delay the execution of tasks

Administrative burden will be Increased as it is carrying out a wide range of activities.

Centralized management which may affect the flow of work.

A Scotch-yoke mechanism is used to convert rotary motion into reciprocating motion. As the disk rotates at the constant angular rate , a pin A slides in a vertical slot causing the slotted member to displace horizontally according to x = r sin(t) relative to the fi xed disk center O. Determine the expressions for the velocity and acceleration of a point P

Answers

Answer:

The question continues ; Determine the expressions for the velocity and acceleration of a point P as a function of time t, and determine the maximum velocity of point P during one cycle. Use the values r = 75mm and w = pie-rads/s

Explanation:

The diagram and the detailed step by step explanation is as shown in the attachment

Final answer:

In the Scotch-yoke mechanism, if the displacement is given by x = r sin(t), the velocity v = r cos(t) and acceleration a = -r sin(t). The negative sign in acceleration indicates that it is in the opposite direction to displacement.

Explanation:

The Scotch-yoke mechanism which converts rotary motion into reciprocating motion can be analyzed using principles of kinematics. If we have the displacement given by x = r sin(t), the velocity and acceleration can be derived from this displacement equation.

The velocity (v) is the time derivative of the displacement function, i.e., v = dx/dt = r cos(t).

The acceleration (a) is the time derivative of the velocity function, so a = dv/dt = -r sin(t).

The negative sign signifies that acceleration is in the opposite direction to displacement.

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A pipe produces successive harmonics at 300 Hz and 350 Hz. Calculate the length of the pipe and state whether it is closed at one end or not. Assume the speed of sound to be 340 m/s.

Answers

Answer:

The pipe is open ended and the length of pipe is 3.4 m.

Explanation:

For identification of the type of pipe checking the successive frequencies in both the open pipe and closed pipe as below

Equation for nth frequency for open end pipe is given as

[tex]f_n=\frac{nv}{2L}[/tex]

For (n+1)th value the frequency is

[tex]f_{n+1}=\frac{(n+1)v}{2L}[/tex]

Taking a ratio of both equation and solving for n such that the value of n is a whole number

[tex]\frac{f_{n+1}}{f_n}=\frac{\frac{(n+1)v}{2L}}{\frac{nv}{2L}}\\\frac{350}{300}=\frac{(n+1)}{n}\\350n =300n+300\\50n =300\\n =6\\[/tex]

So n is a whole number this means that the pipe is open ended.

For confirmation the  nth frequency for a closed ended pipe is given as

[tex]f_n=\frac{(2n+1)v}{4L}[/tex]

For (n+1)th value the frequency is

[tex]f_{n+1}=\frac{(2n+3)v}{4L}[/tex]

Taking a ratio of both equation and solving for n such that the value of n is a whole number

[tex]\frac{f_{n+1}}{f_n}=\frac{\frac{(2n+3)v}{2L}}{\frac{(2n+1)v}{2L}}\\\frac{350}{300}=\frac{(2n+3)}{(2n+1)}\\700n+350 =600n+900\\100n =550\\n =5.5\\[/tex]

As n is not a whole number so this is further confirmed that the pipe is open ended.

Now from the equation of, with n=6, v=340 m/s and f=300 Hz

[tex]f_n=\frac{nv}{2L}\\300=\frac{6 \times 340}{2L}\\L=\frac{2040}{600}\\L=3.4 m[/tex]

The value of length is 3.4m.

The space shuttle fleet was designed with two booster stages. If the first stage provides a thrust of 686.68 Mega-newtons(MN) and the space shuttle has a mass of 5,470,0005, 470,000 pound-mass, what is the acceleration of the space craft in miles per hour squared?

Answers

Answer:

6.30 miles/hour

Explanation:

Newton's second law applies here. In simple terms:

[tex]F = ma[/tex]

where F = Force (Thrust) in N

           a = acceleration (m/s²)

The acceleration can be given by rearranging the  formula to give:

[tex]a = \frac{F}{N}[/tex]

  = [tex]\frac{(686.68*10^{6} )}{24811505120150.2656} \\= 0.0000277 m/s\\= 6.03 miles/hr[/tex]

A cylindrical tank has a thin barrier and it carries two fluids, one of the fluids has specific gravity of 2.0 and the other fluid has a specific weight of 100 lbf/ft3. The mass of the tank is 20lb-mass. Determine the magnitude of the vertical force required to give the tank a downward acceleration of 10 ft/s2.

Answers

Answer:

attached below

Explanation:

A manometer containing a fluid with a density of 60 lbm/ft3 is attached to a tank filled with air. If the gage pressure of the air in the tank is 9.4 psig and the atmospheric pressure is 12.5 psia, the fluid-level difference between the two columns, h, in feet is

Answers

Answer:

The fluid level difference in the manometer arm = 22.56 ft.

Explanation:

Assumption: The fluid in the manometer is incompressible, that is, its density is constant.

The fluid level difference between the two arms of the manometer gives the gage pressure of the air in the tank.

And P(gage) = ρgh

ρ = density of the manometer fluid = 60 lbm/ft³

g = acceleration due to gravity = 32.2 ft/s²

ρg = 60 × 32.2 = 1932 lbm/ft²s²

ρg = 1932 lbm/ft²s² × 1lbf.s²/32.2lbm.ft = 60 lbf/ft³

h = fluid level difference between the two arms of the manometer = ?

P(gage) = 9.4 psig = 9.4 × 144 = 1353.6 lbf/ft²

1353.6 = ρg × h = 60 lbf/ft³ × h

h = 1353.6/60 = 22.56 ft

A diagrammatic representation of this setup is presented in the attached image.

Hope this helps!

A 15 Watt desk-type fluorescent lamp has an effective resistance of 200 ohms when operating (note: the 15 Watts is only associated with the lamp). It is in series with a ballast that has a resistance of 80 ohms and an inductance of .9H. The lamp and ballast are operated at 120V, 60Hz. Draw the circuit

Answers

The question is incomplete! Complete question along with answers and explanation is provided below.

Question:

A 15 Watt desk-type fluorescent lamp has an effective resistance of 200 ohms when operating (note: the 15 Watts is only associated with the lamp). It is in series with a ballast that has a resistance of 80 ohms and an inductance of .9H. The lamp and ballast are operated at 120V, 60Hz.

a) Draw the circuit

b) Calculate the power drawn by the lamp

c) Calculate the apparent power

d) Calculate the power factor

e) Calculate the reactive power

f) Calculate the size of the capacitor necessary to provide unity power factor correction

Explanation:

a) draw the circuit

Refer to the attached image.

As you can see in the attached drawing, it is a series circuit containing  two resistors and one inductor.

In a series circuit, current remains same throughout the circuit

The circuit is powered by an AC voltage source having voltage of 120 V and frequency 60 Hz.

The current flowing in the circuit can be found by ohm's law

 I = V/Z

where V is the voltage and Z is the total impedance of the circuit

 Z = R + XL

where  XL is the inductive reactance

XL = j2 π f L

XL = j2*π*60*0.9

XL = j339.29Ω

Total resistance is

R =200 + 80 = 280 Ω

Total impedance is

Z = 280 + j339.29 Ω

b) Calculate the power drawn by the lamp

First calculate the current

I = V/Z

I = 120/(280 + j339.29)

I = 0.272<-50.46° A  (complex notation)

P = I²R

P = (0.272)²200

P ≈ 15 W

Power drawn by the circuit

P=V*I*cos(50.46°)

P=20.77 W

c) Calculate the apparent power

A = VI*

A = 120*0.272<50.46°

A = 32.64<50.46° VA

d) Calculate the power factor

PF = cos(50.46)

PF = 0.63

e) Calculate the reactive power

Q = VIsin(50.46)

Q = 120*0.272<-50.46*sin(50.46)

Q = 25.13<-50.46  VAR

f) Calculate the size of the capacitor necessary to provide unity power factor correction

The required reactive compensation power is

Qc = P (tan(old) - tan(new))

Qc = 20.77 (tan(50.46) - tan(0))

Qc = 25.16 VAR

C = Qc/2πfV²

C = 25.16/2*π*60*120²

C = 4.63 uF

Hence adding a capacitor of 4.63 uF parallel to the load will improve the PF from 0.63 to 1.

Code a Boolean expression that tests if a decimal variable named currentSales is greater than or equal to 1000 or a Boolean variable named newCustomer is equal to true. Code this statement so both conditions will always be tested, and code it in the shortest way possible]

Answers

Answer:

Given

Decimal variable: currentSales

Decimal test value: 1000

Boolean variable: newCustomer

Boolean default value: true

The following code segment is written in Java

if(currentSales == 1000 || newCustomer)

{

//Some statements

}

The Program above tests two conditions using one of statement

The first condition is to check if currentSales is 1000

== Sign is s a relational operator used for comparison (it's different from=)

|| represents OR

The second condition is newCustomer, which is a Boolean variable

If one or both of the conditions is true, the statements within the {} will be executed

Meaning that, both conditions doesn't have to be true;

At least 1 condition must be true for the statement within the curly braces to be executed

A cylinder fitted with a frictionless piston contains 2 kg of R-134a at 3.5 bar and 100 C. The cylinder is now cooled so that the R-134a is kept at constant pressure until a final state is reached with a quality of 25%. Calculate the heat transfer in the process.

Answers

Answer:

The answer to the question is

The heat transferred in the process is -274.645 kJ

Explanation:

To solve the question, we list out the variables thus

R-134a = Tetrafluoroethane

Intitial Temperaturte t₁ = 100 °C

Initial pressure = 3.5 bar = 350 kPa

For closed system we have m₁ = m₂ = m

ΔU = m×(u₂ - u₁) = ₁Q₂ -₁W₂

For constant pressure process we have

Work done = W = [tex]\int\limits^a_b P \, dV[/tex]  = P×ΔV = P × (V₂ - V₁) = P×m×(v₂ - v₁)

From the tables we have

State 1 we have h₁ = (490.48 +489.52)/2 = 490 kJ/kg

State 2 gives h₂ = 206.75 + 0.75 × 194.57= 352.6775 kJ/kg

Therefore Q₁₂ = m×(u₂ - u₁) + W₁₂ = m × (u₂ - u₁) + P×m×(v₂ - v₁)

= m×(h₂ - h₁) = 2.0 kg × (352.6775 kJ/kg - 490 kJ/kg) =-274.645 kJ

An engineering student claims that a country road can be safely negotiated at 65 mi/h in rainy weather. Because of the winding nature of the road, one stretch of level pavement has a sight distance of only 510 ft. Assuming practical stopping distance, comment on the student

Answers

Answer:

Negotiated speed should be lower. Perception/reaction time is too less than design values.

Explanation:

Given:

- The claimed safe speed V_1 = 65 mi/h

- Sight distance D = 510 ft

- The practical deceleration a = 11.2 ft/s   ... according to standards

Find:

Assuming practical stopping distance, comment on the student whether the claim is correct or not

Solution:

- Calculate the practical stopping distance:

                      d = V_1^2 / 2*a

                      d = ( 65 * 1.46 )^2 / 2*11.2 = 402.054 ft

- Solve for reaction distance d_r is as follows:

                     d_r = D - d = 510 - 402.054 = 107.945 ft

- The perception/time reaction is:

                   t_r = d_r/V_1 = 107.945 / 94.9

                  t_r = 1.17 sec

Answer: The perception/reaction time t_r = 1.17 s is well below the t = 2.3 s.

Hence, the safe speed should be lower.

Air is compressed slowly in a piston–cylinder assembly from an initial state where p1 = 1.4 bar, V1 = 4.25 m3 , to a final state where p2 = 6.8 bar. During the process, the relation between pressure and volume follows pV = constan

Answers

The work done by the gas is -940 kJ

Explanation:

In this process, we are told that the product of pressure and volume remains constant:

[tex]pV=const.[/tex]

so we can write

[tex]p_1 V_1 = p_2 V_2[/tex]

where

[tex]p_1 = 1.4 bar[/tex] is the initial pressure

[tex]p_2 = 6.8 bar[/tex] is the final pressure

[tex]V_1=4.25 m^2[/tex] is the initial volume

Solving for [tex]V_2[/tex], we find the final volume:

[tex]V_2=\frac{p_1V_1}{p_2}=\frac{(1.4)(4.25)}{6.8}=0.875 m^3[/tex]

Now by looking at the equation of state of an ideal gas:

[tex]pV=nRT[/tex] (1)

we notice that since [tex]pV=const.[/tex], this means that also the absolute temperature of the gas T remains constant (because the number of moles n does not change). Therefore this is an isothermal process: the work done in an isothermal process is given by

[tex]W=nRTln(\frac{V_2}{V_1})[/tex]

And by looking again at (1), we  can substitute (nRT) with (pV), so we get

[tex]W=p_1 V_1 ln (\frac{V_2}{V_1})[/tex]

Converting the pressure into SI units,

[tex]p_1 = 1.4 bar = 1.4\cdot 10^5 Pa[/tex]

So the work done is

[tex]W=(1.4\cdot 10^5)(4.25)ln(\frac{0.875}{4.25})=-9.4\cdot 10^5 J[/tex]

Which means -940 kJ. This value is negative since the work is done by the surroundings on the gas (because the gas is compressed).

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Suppose you were a heating engineer and you wished to consider a house as a dynamic system. Without a heater, the average temperature in the house would clearly vary over a 24-h period. What might you consider for inputs, outputs, and state variables for a simple dynamic model? How would you expand your model so that it would predict temperatures in several rooms of the house? How does the installation of a thermostatically controlled heater change your model?

Answers

Answer:

As a heating engineer and considering a house as a dynamic system , and that without a heater, the average temperature in the house would vary over a 24-h period.

What might you consider for inputs, outputs, and state variables for a simple dynamic model?

State variables: according to the weather conditions of the area where the house was built:

State variable # 1: minimum temperture during a day in an specific season (*4);

State variable # 2: maximum temperature during the day, in an specific season (*4) as well;

State variable # 3: average temperature during the day in an specific season (*4).

That makes 16 state variables all of them in Centigrade degrees.

Input variables:

# 1: one degree over each of the state variables given.

# 2: one degree below each of the state variables, all of them in Centigrade           degrees.

Output variables:

# 1 are the temperatures reached after adding one degree to each of the input variables.

# 2 are the temperatures reached after decreasing one degree, all of them in Centigrade degrees.

How would you expand your model so that it would predict temperatures in several rooms of the house?

I would add output variables in a "Y" system to predict temperatures in several rooms of the house.

How does the installation of a thermostatically controlled heater change your model?

It would change on the "Y" variables as they will get a control system  designed for sensors to produce from some input variables to make the system respond.

Explanation:

State-determined system models using well defined physical systems is of highly interest to engineers.

Create a program named PaintingDemo that instantiates an array of eight Room objects and demonstrates the Room methods. The Room constructor requires parameters for length, width, and height fields (all of type int); use a variety of values when constructing the objects. The Room class also contains the following fields: Area - The wall area of the Room (as an int) Gallons - The number of gallons of paint needed to paint the room (as an int)

Answers

Answer:

Explanation:

Code used will be like

using System;

using System.Collections.Generic;

using System.Linq;

using System.Text;

using System.Threading.Tasks;

namespace PaintingWall

{

class Room

{

public int length, width, height,Area,Gallons;

public Room(int l,int w,int h)

{

length = l;

width = w;

height = h;  

}

private int getLength()

{

return length;

}

private int getWidth()

{

return width;

}

private int getHeight()

{

return height;

}

public void WallAreaAndNumberGallons()

{

Area = getLength() * getHeight() * getWidth();

if (Area < 350)

{

Gallons = 1;

}

else if (Area > 350)

{

Gallons = 2;

}    

Console.WriteLine ("The area of the Room is " + Area);

Console.WriteLine("The number of gallons paint needed to paint the Room is " + Gallons);

}

 

}

class PaintingDemo

{

static void Main(string[] args)

{

int l, w, h;

Room[] r = new Room[8];

for (int i = 0; i <= 7; i++)

{

Console.WriteLine("Room "+(i+1));

Console.Write("Enter Length : ");

l = Convert.ToInt32(Console.ReadLine() );

Console.Write("Enter Width : ");

w = Convert.ToInt32(Console.ReadLine());

Console.Write("Enter Height : ");

h= Convert.ToInt32(Console.ReadLine());

r[i] = new Room(l,w,h);

Console.WriteLine();

}

for (int i = 0; i <= 7; i++)

{

Console.WriteLine("Room " + (i + 1));

r[i].WallAreaAndNumberGallons();

}

Console.ReadKey();  

}

}

}

A horizontal rigid bar ABC is pinned at end A and supported by two cables at points B and C. A vertical load P 5 10 kN acts at end C of the bar. The two cables are made of steel with a modulus elasticity E 5 200 GPa and have the same cross-sectional area. Calculate the minimum cross-sectional area of each cable if the yield stress of the cable is 400 MPa and the factor of safety is 2.0. Consider load P only; ignore the weight of bar ABC and the cables.

Answers

Final answer:

To determine the minimum cross-sectional area of the steel cables, calculate the allowable stress and use it along with the provided load. The result is a minimum area of 50 mm² for each cable.

Explanation:

The question involves calculating the minimum cross-sectional area of each steel cable designed to support a load with a safety factor, given the yield stress of the material. First, determine the allowable stress by dividing the yield stress by the factor of safety. In this case, the allowable stress is 200 MPa (400 MPa / 2.0). To find the minimum cross-sectional area (A), use the formula A = P / σ, where P = 10 kN = 10,000 N (the load) and σ (sigma) is the allowable stress in N/m². Convert 200 MPa to N/m² to get 200,000 N/m². Therefore, the minimum cross-sectional area required for each cable is 50 mm² (10,000 N / 200,000 N/m²).

You are traveling along an interstate highway at 32.0 m/s (about 72 mph) when a truck stops suddenly in front of you. You immediately apply your brakes and cut your speed in half after 6.0 s.(a) What was your acceleration, assuming it was constant?

Answers

Answer:

a= - 2.6 m/s².

Explanation:

u = 32 m/s

The speed after 6 s is half of u

[tex]v= \dfrac{32}{2}=16\ m/s[/tex]

t= 6 s

The average acceleration = a

We know v = u +at

v=final velocity

u=initial velocity

Now by putting the values in the above equation

16= 32 + a x 6

[tex]a=\dfrac{16-32}{6}\ m/s^2[/tex]

[tex]a=-2.6\ m/s^2[/tex]

Therefore the acceleration will be - 2.6 m/s².

a= - 2.6 m/s².

Negative indicates that velocity and acceleration is is opposite direction.

The flexural strength or MOR of a ceramic is 310 MPa. A block of the ceramic, which is 20 mm wide, 15 mm high, and 300 mm long, is supported between two rods 150 mm apart. Determine the force required to fracture the material, assuming no plastic deformation occurs.

Answers

Answer:

[tex]F=6200\ \text{N}\\[/tex]

Explanation:

In this problem you need to define the force that acts upon a beam in a 3 point bending problem. I put a picture of the problem taken from Wikipedia:

In this problem the flexural strength is defined with the following formula:

[tex]\sigma=\cfrac{3FL}{2bd^2}[/tex]

where F is the force applied, L the length between the two rods, b the width of the ceramic block and d it's height.

The force is then defined as:

[tex]F=\cfrac{2\sigma bd^2}{3L}=6200\ \text{N}[/tex]

the correlation between a car's engine size and its fuel economy is r = -0.774. what fraction of the variability in fuel economy is accounted for by the engine size?

Answers

Answer:

59.9%

Explanation:

R^2 =(-0.774)^2  = 0.599

59.9% of fuel is accounted for

For the following conditions determine whether a CMFR or a PFR is more efficient in removing a reactive compound from the waste stream under steady-state conditions with a first-order reaction: reactor volume = 280 m3, flow rate = 14 m3 · day−1, and reaction rate coefficient = 0.05 day−1.

Answers

Answer:

The PFR is more efficient in the removal of the reactive compound as it has the higher conversion ratio.

Xₚբᵣ = 0.632

X꜀ₘբᵣ = 0.5

Xₚբᵣ > X꜀ₘբᵣ

Explanation:

From the reaction rate coefficient, it is evident the reaction is a first order reaction

Performance equation for a CMFR for a first order reaction is

kτ = (X)/(1 - X)

k = reaction rate constant = 0.05 /day

τ = Time constant or holding time = V/F₀

V = volume of reactor = 280 m³

F₀ = Flowrate into the reactor = 14 m³/day

X = conversion

k(V/F₀) = (X)/(1 - X)

0.05 × (280/14) = X/(1 - X)

1 = X/(1 - X)

X = 1 - X

2X = 1

X = 1/2 = 0.5

For the PFR

Performance equation for a first order reaction is given by

kτ = In [1/(1 - X)]

The parameters are the same as above,

0.05 × (280/14) = In (1/(1-X)

1 = In (1/(1-X))

e = 1/(1 - X)

2.718 = 1/(1 - X)

1 - X = 1/2.718

1 - X = 0.3679

X = 1 - 0.3679

X = 0.632

The PFR is evidently more efficient in the removal of the reactive compound as it has the higher conversion ratio.

Final answer:

To determine whether a CMFR or a PFR is more efficient in removing a reactive compound from the waste stream, we compare their volumes and flow rates. For a first-order reaction, the reaction rate is given by the equation r = kC. In a CMFR, the volume is constant, while in a PFR, the volume varies. Therefore, a PFR may be more efficient depending on the reactor design.

Explanation:

To determine whether a CMFR (Continuous Mixed Flow Reactor) or a PFR (Plug Flow Reactor) is more efficient in removing a reactive compound from the waste stream, we need to compare their volumes and flow rates. For a first-order reaction, the reaction rate is given by the equation: r = kC, where r is the reaction rate, k is the reaction rate coefficient, and C is the concentration of the reactive compound.

In a CMFR, the volume is constant, so the reactor volume (280 m3) is equal to the product of the flow rate (14 m3·day−1) and the residence time (t), which is the time it takes for the fluid to pass through the reactor. Therefore, t = V/Q = 280/14 = 20 days.

In a PFR, the volume varies along the length of the reactor, and the residence time is defined as the integral of the volume divided by the flow rate. Using the equation t = ∫V/Q, we can calculate the residence time for a PFR.

Since the residence time for a CMFR is fixed at 20 days, and the residence time for a PFR can be longer or shorter depending on the reactor design, a PFR may be more efficient in removing the reactive compound from the waste stream under steady-state conditions with a first-order reaction.

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A water jet that leaves a nozzle at 95 m/s at a flow rate of 120 kg/s is to be used to generate power by striking the buckets located on the perimeter of a wheel. Determine the power generation potential of this water jet. The power generation potential of the water jet is kW.

Answers

Answer:

P= 541.5 kW.

Explanation:

Given that

velocity of water after leaving the nozzle ,v= 95 m/s

The mass flow rate of the water , m= 120 kg/s

The power generated P is given as

[tex]P=\dfrac{1}{2}mv^2[/tex]

Now by putting the values in the above equation we get

[tex]P=\dfrac{1}{2}\times 120\times 95^2\ W[/tex]

P=541500  W

The  power in kW will be 541.5 kW.

Therefore the answer will be 541.5 kW

P= 541.5 kW.

Final answer:

The power generation potential of the water jet is 542.25 kW.

Explanation:

To determine the power generation potential of the water jet, we need to calculate the kinetic energy of the water jet and then convert it to power. The kinetic energy of the water jet can be calculated using the formula KE = 0.5 * m * v^2, where m is the mass flow rate of the water and v is the velocity of the water jet. Given that the flow rate is 120 kg/s and the velocity is 95 m/s, we can calculate the kinetic energy to be KE = 0.5 * 120 * 95^2 = 0.5 * 120 * 9025 = 542,250 J/s.

To convert the kinetic energy to power, we divide by the time taken to deliver the energy. Since the flow rate is given in kg/s, we can assume the time taken is 1 second. Therefore, the power generation potential of the water jet is 542,250 J/s, or 542.25 kW.

The terms batten seam, standing seam, and flat seam all describe types of:

(A) architectural sheet metal roofing.
(B) methods for glazing large windows.
(C) types of EPDM roofing membranes.
(D) framing methods for hollow steel doors.
(E) built-up roofing systems.

Answers

Answer:

(A) architectural sheet metal roofing

Explanation:

 By the name itself we can judge that the 'Architectural sheet metal roofing' is a kind of metal roofing.

And these type of metal roofing is primarily used for small and big houses, small buildings and as well as in a building that is for commercial use they can be totally flat as well as little bit sloped.  

And the words similarly like batten and standing seam, and flat seam all tells us that these are the types of architectural sheet metal roofing.

It is usually easy to minimize errors due to the input bias current of an opamp by adding a resistor in the input terminal, but this still leaves a small error due to the input offset current. Select one: True False

Answers

Answer:True

Explanation:

Answer:

True

Explanation:

Input bias current:

It is a small current that flows in parallel with the input terminals of op-amp to bias the input transistors. This current gets converted into voltage and amplified which results in incorrect output results. This bias current Ib+ and Ib- flows in the positive and negative input terminals of the op-amp.  

Ib+ and Ib- create errors of opposite polarity. Therefore, bias current can be minimized by carefully adding a resistor in the positive input terminal.

Input offset current:

Unfortunately, a small error still remains due to the mismatch between input currents Ib+ and Ib-.

This input offset current error can be adjusted by adding a potentiometer and resistor in the negative input terminal.

5. A driver is traveling at 90 km/h on a wet road. An object is spotted on the road 140m ahead and the driver is able to come to a stop just before hitting the object. Assuming standard reaction time and using the practical-stopping distance equation, determine the grade of the road.

Answers

Answer: Check the attached

Explanation:

The calorie is a unit of energy defined as the amount of energy needed to raise 1 g of water by 1°C. a. How many calories are required to bring a pot of water at 1°C to a boil? The pot is full to the brim, with diameter 20 cm and depth 20 cm. The density of water is 1000 kg/m3. b. If we consider that D for the pot is 20 cm, approximately how much more energy is needed to heat a hot tub with D = 2 m? How many calories is that?

Answers

Answer:

a. Calories required = 622710 calories

b. Energy = 1000 times much energy

Calories = 622710000 calories

Explanation:

Given:

h = Depth of pot = 20cm = 0.2m

Diameter of pot = 20cm = 0.2m

r = ½ *diameter = ½ * 0.2

r = 0.1m

Density = 1000kg/m³

Water temperature = 1°C

a.

First, we calculate the volume of the water(pot)

V = Volume = πr²h

V = 22/7 * 0.1² * 0.2

V = 0.044/7

V = 0.00629m³

M = Mass of water = Volume * Density

M = 0.00629m³ * 1000kg/m³

M = 6.29kg

M = 6.29 * 1000 grams

M = 6290g

The water is at 1°C, so it needs to gain 99°C to reach boiling point

So, Calories = 99 * 6290

Calories required = 622710 calories

b.

If we consider that D for the pot is 20 cm, approximately how much more energy is needed to heat a hot tub with D = 2 m? How many calories is that?

Depth of pot = 20cm

Depth of pot = 0.2m

Depth of hot tube = 2m

Energy is directly proportional to D³

Since the depth of hot the is 10 times greater than that of the pot

It'll require 10³ much more energy

Energy = 10³

Energy = 1000 times much energy

Calories required = 622710 * 1000

Calories = 622710000 calories

A liquid phase reaction, A → B, is to be carried out in an isothermal, well mixed batch reactor with a volume of 1L. Initially there are 6 moles of A. The rate of destruction of A is given by –rA =k1CA/ (1+k2CA), where k1=4, k2 =5. The unit of time in the rate constants is hours. Calculate the time, in hours, that the reaction must proceed in the reactor in order to result in 3 moles of A remaining in the reactor.

Answers

Answer:

the time, in hours = 4.07hrs

Explanation:

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

Consider a constant volume process involving heat addition to a closed system consisting of an ideal gas with no changes in kinetic or potential energy. Is the required heat transfer for raising the temperature from 295 to 305 K the same as the heat transfer required from 345 to 355 K?

Answers

Answer:

Yes and no

Explanation:

The thermodynamic equation for the heat transfer in a constant volume process is the following:

[tex]Q=\Delta U=mC_V\Delta T[/tex]

where Q is the required heat, U is the internal energy, m the mass of the gas, C_V the heat capacity assuming consant volume and [tex]\Delta T[/tex] is the change in temperature.

If you assume the heat capacity doesn't change with temperature at which the gas is currently at then the heat transfer depends solely on the change in temperature. With this assumption the transfered heat would be the same in both cases.

In reality the heat capacity does change with respect to temperature. Depending on the type of gas. In reality there would be difference in heat transfered between 295/205 K and 245/255 . Only then you wouldn't use the [tex]\Delta T[/tex] expression since the integral would be different depending on the heat capacity in relation to temperature.

A lab technician is ordered to take a sample of your blood. As she inserts the needle, she says, "My, you have tough skin!" What would be an equivalent translation of this statement?

Answers

Answer:

Your stratified squamous epithelium is difficult to penetrate!

Explanation:

The epithelium is the tissue formed by one or several layers of cells attached to each other that cover all the free surfaces of the organism, and constitute the internal lining of the cavities, organs, hollows, ducts of the body and skin and also form the Mucous and glands. The epithelia also forms the parenchyma of many organs, such as the liver.

Flat or squamous epithelia: Formed by flat cells, with much less height than width and a flattened nucleus. It is one of the most external being part of the epidermis and generating some inconveniences when penetrating with needles to perform blood extractions when you have certain characteristics of hardness.

An airplane starts from rest, travels 5000ft down a runway, and after uniform acceleration, takes off with a speed of 162 mi/h. it then climbs in a straight line with a uniform acceleration of 3 ft/s^s until it reaches a constant speed of 220 mi/h. draw the st, vt, and at graphs that describe the motion.

Answers

Answer:

Explanation:

Given

Take off speed [tex]v=162\ mph\approx 237.6\ ft/s[/tex]

distance traveled in runway [tex]d=5000 ft[/tex]

using motion of equation

[tex]v^2-u^2=2as[/tex]

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

[tex](237.6)^2=2\times a\times 5000[/tex]

[tex]a=5.64\ ft/s^2[/tex]

Acceleration after take off [tex]a_2=3\ ft/s^2[/tex]

time taken to reach [tex]237.6 ft/s[/tex]

[tex]v=u+at[/tex]

[tex]237.6=0+5.64\times t[/tex]

[tex]t=42.127\ s [/tex]

after take off it take [tex]t_2[/tex] time to reach [tex]220 mph\approx 322.67[/tex]

[tex]322.67=237.6+3\times t_2[/tex]

[tex]t_2=28.35\ s[/tex]

total time taken [tex]t_0=t+t_1[/tex]

[tex]t_0=70.48\ s[/tex]

Given the following materials and their corresponding thermal conductivity values, list them in order from most conductive to least conductive.Sheet Rock: k = 0.43 W/(m*K)Masonite: k = 0.047 W/(m*K)Glass: k = 0.72 W/(m*K)Lexan: k = 0.19 W/(m*K)b) Given the following information, calculate the thermal conductivity using Fourier's Equation.q = 100 WA = 8 m^2ATΔT= 10L = 7 m

Answers

Answer:

1) Glass

2) Rock sheet

3) Lexan

4) Masonite

b) k = 8.75 W/m.K

Explanation:

Given:

The thermal conductivity of certain materials as follows:

-Sheet Rock: k = 0.43 W/(m*K)

-Masonite: k = 0.047 W/(m*K)

-Glass: k = 0.72 W/(m*K)

-Lexan: k = 0.19 W/(m*K)

Data Given:

- Q = 100 W

- A = 8 m^2

- dT = 10 C

- L = 7 m

Find:

a) list the materials in order from most conductive to least conductive

b) calculate the thermal conductivity using Fourier's Equation

Solution:

- We know from Fourier's Law the relation between Heat transfer and thermal conductivity as follows:

                                   Q = k*A*dT / L

- From the relation above we can see that rate of heat transfer is directly proportional to thermal conductivity k.

- Hence, the list in order of decreasing conductivity is as follows:

- The list of materials in the decreasing order of thermal conductivity k is:

           1) Glass                 k = 0.72 W/m.K        

           2) Rock sheet      k = 0.43 W/m.K

           3) Lexan               k = 0.19 W/m.K

           4) Masonite          k = 0.047 W/m.K

- Use the relation given above we can compute the thermal conductivity k with the given data:

                                 k = Q*L / (A*dT)

                                 k = (100 W * 7 m) / (8 m^2*10 C)

                                 k = 8.75 W/m.K

For a certain mechanical element the constitutive relation is Y = AV^3, where Y is a system variable, a is a constant, and V is velocity. Give each answer below in terms of V. (a) If Y is the force, i.e. (Y = F), find an expression for the power in the element? (b) If Y is the linear momentum, i.e. (Y = p), find an expression for the energy stored in the element?

Answers

Answer:

a) Power = Av⁴

b) Energy = Av⁴/2 or (A²v⁶)/2m

Explanation:

Given Y = Av³

A = a constant, v = Velocity, Y = system variable.

a) If Y = Force, F, Find Power in the element.

Power is the dot product of Force and velocity and it's a scalar quantity.

i.e. P = F.v = F.v (cos θ) where θ is the angle between the force and velocity vector.

But in this case, average power is simply given by Fv.

P(avg) = Fv = Yv = (Av³) × v = Av⁴

b) If Y = linear momentum, p, Find the energy stored in the element.

Energy is related to linear momentum by the relationship between kinetic energy and linear momentum.

p = mv and E = mv²/2 = (mv)(v)/2, so,

E = pv/2

For this question, p = Y = Av³

E = Yv/2 = (Av³)v/2 = Av⁴/2

Kinetic energy is often related to momentum through this expression too,

p = mv; p² = m²v²

E = mv²/2; E = (mv²/2) × (m/m) = m²v²/2m = p²/2m

Therefore, E = Y²/2m = (Av³)²/2m = (A²v⁶)/2m

Hope this helps!

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