A piston-cylinder assembly contains 0.5 lb of water. The water expands from an initial state where p1 = 40 lbf/in.2 and T1 = 300o F to a final state where p2 = 14.7 lbf/in.2 During the process, the pressure and specific volume are related by the polytropic process pv 1.2 = constant. Determine the energy transfer by work, in Btu.

Answer:

The strength coefficient is [tex]625[/tex] and the strain-hardening exponent is [tex]0.435[/tex]

Explanation:

Given the true strain is 0.12 at 250 MPa stress.

Also, at 350 MPa the strain is 0.26.

We need to find  [tex](K)[/tex] and the [tex](n)[/tex].

[tex]\sigma =K\epsilon^n[/tex]

We will plug the values in the formula.

[tex]250=K\times (0.12)^n\\350=K\times (0.26)^n[/tex]

We will solve these equation.

[tex]K=\frac{250}{(0.12)^n}[/tex] plug this value in [tex]350=K\times (0.26)^n[/tex]

[tex]350=\frac{250}{(0.12)^n}\times (0.26)^n\\ \\\frac{350}{250}=\frac{(0.26)^n}{(0.12)^n}\\ \\1.4=(2.17)^n[/tex]

Taking a natural log both sides we get.

[tex]ln(1.4)=ln(2.17)^n\\ln(1.4)=n\times ln(2.17)\\n=\frac{ln(1.4)}{ln(2.17)}\\ n=0.435[/tex]

Now, we will find value of [tex]K[/tex]

[tex]K=\frac{250}{(0.12)^n}[/tex]

[tex]K=\frac{250}{(0.12)^{0.435}}\\ \\K=\frac{250}{0.40}\\\\K=625[/tex]

So, the strength coefficient is [tex]625[/tex] and the strain-hardening exponent is [tex]0.435[/tex].

a) Heat transfer to the working fluid in the steam generator: 3303.9 kJ/kg

b) Thermal efficiency: 100%

c) Heat transfer from the working fluid in the condenser to the cooling water: 2083.6 kJ/kg

Given parameters:

- Superheated vapor enters turbine: P₁ = 10 MPa, T₁ = 480°C

- Condenser pressure: P₃ = 6 kPa

- Extraction pressure for closed feedwater heater: P₂ = 0.7 MPa

- Isentropic efficiencies of pump and turbine stages: ηᵥ = ηₜ = 80%

a) Heat Transfer in the Steam Generator:

1. From the steam tables, at P₁ = 10 MPa:

  - h₁ = 3478.1 kJ/kg (enthalpy of superheated vapor)

2. At P₂ = 0.7 MPa:

  - h₂ = 209.4 kJ/kg (enthalpy of extracted steam)

3. The enthalpy at the inlet of the closed feedwater heater is the same as h₂:

  - h₃ = 209.4 kJ/kg

4. At P₃ = 6 kPa, as condensate:

  - h₄ = 191.8 kJ/kg (enthalpy of saturated liquid)

5. The enthalpy at the outlet of the closed feedwater heater is the same as h₄:

  - h₅ = 191.8 kJ/kg

6. Pump work per unit mass of water: [tex]W_{\text{pump}}[/tex]= h₅ - h₃

  - [tex]\( W_{\text{pump}} = 191.8 - 209.4 = -17.6 \, \text{kJ/kg} \)[/tex]

7. Actual work output from turbine per unit mass of steam: [tex]W_{\text{act}}[/tex]= h₁ - h₂

  [tex]\( W_{\text{act}} = 3478.1 - 209.4 = 3268.7 \, \text{kJ/kg} \)[/tex]

8. Isentropic work output from turbine per unit mass of steam: [tex]\( W_{\text{isentropic}} = \frac{W_{\text{act}}}{\eta_t} \)[/tex]

   [tex]\( W_{\text{isentropic}} = \frac{3268.7}{0.8} = 4085.9 \, \text{kJ/kg} \)[/tex]

9. Heat input per unit mass of steam:  [tex]Q_{\text{in}}[/tex] = h₁ - h₅

  [tex]\( Q_{\text{in}} = 3478.1 - 191.8 = 3286.3 \, \text{kJ/kg} \)[/tex]

10. Heat transfer in the steam generator: [tex]\( Q_{\text{gen}} = Q_{\text{in}} - |W_{\text{pump}}| \)[/tex]

  - [tex]\( Q_{\text{gen}} = 3286.3 - |-17.6| = 3303.9 \, \text{kJ/kg} \)[/tex]

  - Rounded to one decimal place: 3303.9 kJ/kg

b) Thermal Efficiency:

Thermal efficiency (η) is given by: [tex]\( \eta = \frac{W_{\text{net}}}{Q_{\text{in}}} \)[/tex]

where[tex]\( W_{\text{net}} = W_{\text{act}} - |W_{\text{pump}}| \)[/tex]

[tex]\( \eta = \frac{3268.7 - |-17.6|}{3286.3} \times 100 \)\\\( \eta = \frac{3286.3}{3286.3} \times 100 \)\\\( \eta = 1 \times 100 \)\\\( \eta = 100\% \)[/tex]

c) Heat Transfer in the Condenser:

Heat transfer from working fluid in the condenser to cooling water:

[tex]Q_{\text{out}}[/tex]= h₄ - h₃

[tex]Q_{\text{out}} = 191.8 - 209.4 = -17.6 \, \text{kJ/kg}[/tex]

- Rounded to one decimal place: -17.6 kJ/kg ≈ -2083.6 kJ/kg

a) To find the heat transfer in the steam generator, we considered the enthalpies at different stages and calculated the pump work, actual work output from the turbine, isentropic work, heat input, and finally the heat transfer in the steam generator.

b) The thermal efficiency was calculated using the formula for efficiency, considering the net work output and heat input.

c) Heat transfer in the condenser was determined by comparing the enthalpies at the inlet and outlet of the closed feedwater heater, considering the negative sign due to the direction of heat flow from the working fluid to the cooling water.

Complete question

The complete question is shown on the first uploaded image

Answer:

a The table is shown on the second uploaded image

b The simplest possible sum product is shown on the fifth uploaded image

Explanation:

The explanation is shown on the third fourth and fifth uploaded image

The electric power generated by a photovoltaic panel is calculated using the panel's dimensions, solar flux, absorptivity, conversion efficiency, and environmental conditions such as temperature and heat transfer coefficient, with different calculations for summer day and breezy winter day scenarios.

Calculation of Electric Power Generated by a Photovoltaic Panel

To determine the electric power generated by a photovoltaic panel, we use the given parameters of the panel's dimensions, solar flux, absorptivity, efficiency of conversion, and the various conditions of the environment as provided in the question. On a still summer day with given temperature and heat transfer coefficient, we would need to calculate the temperature of the panel and use it in the efficiency formula to find the electrical power generated. Similarly, calculations would be performed for a breezy winter day. These calculations involve physics principles like heat transfer, solar radiation, and photovoltaic efficiency.

Answer:

E = 207 GPa

Explanation:

Young's modulus is given by following equation:

[tex]E = \frac{\sigma}{\epsilon}[/tex]

Where [tex]\sigma[/tex] and [tex]\epsilon[/tex] are stress and strain, respectively.

By replacing terms:

[tex]E = \frac{0.321 GPa}{0.00155}\\E = 207 GPa[/tex]

Answer:

Explanation:

Part (a):

Statement : The number of siblings you have

Suitable Data type : Byte

Typical Value : From -128 and up to 127

Explanation: Byte data type is the most suitable since it can covers minimum and maximum number of siblings one can have.

Part (b):

Statement : Your final grade in this class

Suitable Data type : Char

Typical Value : 1 byte

Explanation: Grades is in the form of alphabetical letter which is either A, B, C, D, F or E which can be stored in character data type.

Part (c):

Statement : Population of Earth

Suitable Data type : Long

Maximum Value : 9223372036854775807

Explanation: Long Data takes up to 8 bytes and can store up to 9223372036854775807 which can cater for more than 36 billion. The population of earth is only around 7 billion currently making Long data type the most suitable data type to store earth population.

Part (d):

Statement : Population of US Country

Suitable Data type : Integer

Typical Value :2147483647

Explanation: Integer data type takes up to 4 bytes and can store up to  2147483647 making it suitable to store U.S population.

Part (e):

Statement : The number of passengers on bus

Suitable Data type : Byte

Typical Value :From -128 up to 127

Explanation: The typical maximum number of passengers of a bus are only around 72. Byte data type is the most suitable since it can cater the number up to 127.

Part (f):

Statement : Player's score in a Scrabble game

Suitable Data type : Short

Typical Value : 32767

Explanation: The maximum point can be scored in the Scrabble game is only 830 therefore the most suitable data type for this case is the short data type.

Part (g):

Statement : One team's score in a Major League Baseball game

Suitable Data type : Byte

Typical Value : From -128 up to 127

Explanation: The maximum point can be scored in the Base ball game is only 49 therefore the most suitable data type for this case is the Byte data type since it can cater up to 127.

Part (h):

Statement : The year an historical event occurred

Suitable Data type : Short

Maximum Value: 32767

Explanation: The historic event year can be any number from 1 to 2020 therefore the most suitable data type is the short data type.

Part (i):

Statement : The number of legs on an animal

Suitable Data type : Short

Maximum Value: 32767

Explanation: The most number of legs found are 750 legs therefore the most suitable data type is the short data type which can cater up to 32767.

Part (j):

Statement : The Price of an automobile

Suitable Data type : Float

Maximum Value: 340282350

Explanation: The most expensive car is around 15 million therefore the most suitable data type is the float data type which can cater up to 340 million.  

As per the question the best data type for following so that a reasonable value is accommodated as per the type of example.

Learn more about the data type for each.

brainly.com/question/24871576.

Answer:

Sampling rate = 90

23 and 45 Hz

Explanation:

The Nyquist criteria states that the sampling frequency must be at least twice of the highest frequency component.

(a) We have 10, 12, 13, 23, and 45 Hz signals

The highest frequency component in this list is 45 Hz

So minimum sampling frequency needed to avoid aliasing would be

Sampling rate = 2*45 = 90 sps

Hence a sampling rate of 90 samples per second would be required to avoid aliasing.

(b) if a sampling rate of 40 sps is used then

40 = 2*f

f = 40/2 = 20 Hz

Hence 23 and 45 Hz components will be aliased.

Answer:

(a) Effectiveness of the regenerator= 0.433

(b) The rate of heat removal=21.38 kW

Explanation:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.

Answer:

The answers to the question are

(a) T = 2·π·r/v

(b) 3.3 % change in period of the under-inflated tire compared to the properly inflated tire

(c) Therefore the distance this car would have to travel in order for under-inflated tire to make one more complete rotation than the inflated tire is 57.685 meters.

Explanation:

(a) The period T = 2π/ω

The velocity v = ωr or ω = v/r

Therefore T = 2π/(v/r) = 2πr/v

T = 2·π·r/v

(b) Period of properly inflated tire with radius = 303 mm is 2π303/v

Period of under-inflated tire with radius = 293 mm is 2π293/v

Therefore we have percentage change in period of  of the under-inflated tire compared to the properly inflated tire is given by

(2π303/v -2π293/v)/(2π303/v) = 2π10/v/(2π303/v) = 10/303 × 100 = 3.3 %

(c) The period of the under-inflated tire is 10/303 less than that of the inflated tire. Therefore for the under-inflated tire to make one complete turn more than the inflated tire, we have 1/(10/303)  = 303/10 or 30.3 revolutions of either tire which is 30.3×2×π×303 = 57685.296 mm = 57.685 meters

Therefore the distance this car would have to travel in order for under-inflated tire to make one more complete rotation than the inflated tire is 57.685 meters

At 30.3 revolutions the distance covered by the under-inflated

= 55781.49 mm

Subtracting the two distances gives

1903.805 mm

The circumference of the inflated tire = 2×π×303 = 1903.805 mm

A) The mathematical equation that relates the period T of the wheel rotation of the tire radius r and the speed v of the car is; T = 2πr/v

B) The percent change of the period of the underinflated tire compared to the properly inflated tire is; 3.3 %

C) The distance this car would have to travel in order for under-inflated tire to make one more complete rotation than the inflated tire is; 57685.296 mm

(A) The formula for period is;

T = 2π/ω

angular velocity is;

ω = v/r

Thus; T = 2πr/v

where T is period, v is velocity and r is radius

(b) We are told that radius is 303 mm. Thus Period of inflated tire with radius is;

T =  2π ×  303/v

Period of under-inflated tire with radius of 293 mm is;

T = 2π× 293/v

Percentage Change is;

((2π ×  303/v) - (2π× 293/v))/(2π ×  303/v) * 100% = 10/303 * 100% = 3.3 %

(c) From B above we see that the period of the under-inflated tire is 10/303 less than that of the inflated tire.

Thus,  for the under-inflated tire to make one complete turn more than the inflated tire, we have 1/(10/303)  = 30.3 revolutions of either tire.

Thus, the distance is;

d = 30.3 × 2π × 303

d = 57685.296 mm

Using the same concept, at 30.3 revolutions, the distance covered by the under-inflated tire is calculated to be; 55781.49 mm

Difference in distance = 57685.296 mm -  55781.49 mm

Difference in distance = 1903.805 mm

Read more about motion of cars at; https://brainly.com/question/1315051

Answer:

Development of plant life, especially algae

Explanation:

The main source of oxygen in the atmosphere is the photosynthesis process that produces sugars and oxygen by utilizing carbon dioxide and water.Tiny oceanic plants called phytoplanktons  have the ability to support life in water bodies for plants, animals and fish.The phytoplanktons consume most of the carbon dioxide in air and with the help of energy from the sun, they convert nutrients and carbon dioxide to complex organic compounds which are main source of plant  material. Phytoplanktons are responsible for oxygen in water which is approximated to be 50% of world's oxygen.Green-algae is a good example of phytoplankons.

Answer:

Explanation:20% of food energy is converted into mechanical energy

Answer:

Tank A, due to higher density of water. At the bottom of the tank.

Explanation:

According to the theory of hydrostatics, pressure change is the product of density, gravity constant and depth. The higher the density, the higher the depth. As oil ([tex]\rho_{oil} = 920 \frac{kg}{m^{3}}[/tex]) has a density lower than in water ([tex]\rho_{water} = 1000 \frac{kg}{m^{3}}[/tex]), the largest pressure occur in tank A. The highest pressure occurs at the bottom of the tank A due to the fluid column.

Answer:

The answer and explanation to this question is attached.

Answer:

Explanation:

Let first simplified the risk given our specific loss function. if f(x) = i i is not double , then the risk is

R(f(x) = i|x) = ∑ L(f(x) = i , y = j ) P(y = j|x)                                      2

                 =0.P (Y= i|x) +λc ∑ P (Y= j|x)                                      3

                 =λc (1 - P(Y= i|x))                                                        4

When f(x)  = c  + 1, meaning you have choosing doubt , the risk is

R(f(x) = c +1|x) = ∑ L (f(x)= c+1, y=j) P(Y=j|x)                                  5

        =λd∑ P(Y=j|x)                                                                       6

        =λd                                                                                      7

because   ∑ P(Y=j|x)  should sum to 1 since its a proper probability distribution.

Now let fopt : Rd→ {1, . . . , c + 1} be the decision rule which implements (R1)–(R3).We want to show that in expectation the rule foptis at least as good as an arbitrary rulef. Let x ∈ Rdbe a data point, which we want to classify. Let’s examine all the possiblescenarios where fopt(x) and another arbitrary rule f(x) might differ:Case 1: Let fopt(x) = i where i 6= c + 1.– Case 1a: f(x) = k where k 6= i. Then we get with (R1) thatR(fopt(x) = i|x) = λc1 − P(Y = i| x)≤ λc1 − P(Y = k|x)= R(f(x) = k|x).– Case 1b: f(x) = c + 1. Then we get with (R1) thatR(fopt(x) = i|x) = λc1 − P(Y = i| x)≤ λc(1 − (1 −λdλc)) = λd= R(f(x) = c + 1|x).Case 2: Let fopt(x) = c + 1 and f(x) = k where k 6= c + 1. Then:R(f(x) = k|x) = λc(1 − P (Y = k|x)R(fopt(x) = c + 1|x) = λ            

Calculating the maximum external pressure that a cold-drawn AISI 1040 steel tube can withstand involves using the formula for hoop stress and considering 80 percent of the steel's minimum yield strength. The formula relates the difference in pressure across the tube wall to its inner and outer radii.

The question asks for the maximum external pressure a cold-drawn AISI 1040 steel tube can withstand, given that the largest principal normal stress should not exceed 80 percent of the steel's minimum yield strength. The tube's outer diameter (OD) is 50 mm, and it has a wall thickness of 6 mm. To solve this, we must utilize the formula for hoop stress (sigma) in a thin-walled cylinder under external pressure, which is σ = δP(r_o/(r_o - r_i)), where δP is the difference in internal and external pressure, r_o is the outer radius, and r_i is the inner radius. The yield strength of AISI 1040 steel must be considered, and 80 percent of this value is taken as the maximum allowable stress. Without knowing the exact yield strength, it cannot be directly calculated in this response. However, the process would involve deriving δP from the given conditions and substituting into the hoop stress formula to find the maximum external pressure.

Explanation:

Below is an attachment containing the solution.

The radiative heat transfer between the floor and the ceiling is 1,534.55 kW.

Given the following data:

Mathematically, the radiative heat transfer between the floor and the ceiling is given by this formula:

[tex]Q=\epsilon \sigma A(T_2^4-T_1^4)[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]Q=0.9 \times 5.67 \times 10^{-8} \times 4.45^2 \times (298^4-283^4)\\\\Q=0.00000005103 \times 20.4304 \times 1471902495[/tex]

Q = 1,534.55 kW.

Read more on radiative heat here: https://brainly.com/question/14267608

Answer:

Explanation:

we will begin by carefully following a step by step order;

Total time in a week = 40 hours = 2400 minutes

Total time Administrator is utilized = 50 * 20 = 1000 minutes

Total time Senior accountant is utilized= 20% * 50 * 40 = 400 minutes

Total time Junior accountant is utilized = 40%*50*15 = 600 minutes

For Total cases, Total capacity = Total time / Bottlenck time (Administrator) = 2400 / 20 = 120 cases

C-3) Capacity of new customers at 20:80 ratio = 20% * 120 = 24 cases

C-4) Capacity of repeat customers at 20:80 ratio = 20%*120 = 96 cases

c-1 = Flow rate = min ( demand, capacity) for new customers = 10 new cases / week (20%*50)

c-2 = Flow rate = 40%*50 = 40 cases / week

c3 = 24 cases

c4 = 96 cases

cheers i hope this helps

The question involves calculating free moisture content, plotting it against time, determining the drying rate, and predicting the total drying time using numerical integration during the falling rate period. The constant and falling rate periods of drying are crucial in understanding the sample's drying behavior.

The student is tasked with calculating the free moisture content ( extit{X}) for various data points during the drying of a foodstuff in a tray dryer, plotting  extit{X} versus time (t), determining the drying rate ( extit{R}), and predicting the total drying time using numerical integration for the falling rate period of drying.

The falling rate period is characterized by a decreasing drying rate which requires numerical integration to find the total drying time. The constant rate period is defined by a uniform drying rate and typically occurs before the falling rate period.

The detailed answer provides calculations for free moisture content, drying rate, and prediction of total drying time. It also includes instructions for plotting the moisture ratio vs. time.

Calculating the Free Moisture Content:

To calculate the free moisture content (X) for each data point, subtract the bone dry weight from the wet weight. Then, divide by the bone dry weight.

Calculating Drying Rate and Predicting Total Drying Time:

Determine the drying rate using the slope of the moisture ratio vs. time plot. Then, use numerical integration to predict the time to dry the sample from X = 0.20 to X = 0.04. Identify the drying rate in the constant rate period and X in that period.

Moisture Ratio vs. Time Plot:

Plot the moisture ratio (MR) versus time to observe the constant rate and falling rate periods during the drying.

Answer:

The answer is competitive inhibitor.

Explanation:

Line-Weaver Burk Plots are designed by Hans Lineweaver and Dean Burk in 1934 and they are used to show the inhibition process of the enzymes in biochemistry.

In the question we are given the description of competitive inhibitors, non-competitive inhibitors and uncompetitive inhibitors. According to the example given later where the x-axis plot and y-axis plot is described with their interceptions of the x and y axis, since both lines have the same y intercept and the no-inhibitor line is more negative than the inhibitor line, we can deduce that the type of enzyme inhibition is exhibited by a competitive inhibitor.

I hope this answer helps.

Answer:

Speed with which the box hits the truck at B relative to the truck = 5.29 ft/s

Explanation:

First of, we calculate the deceleration of the truck+box setup using the equations of motion.

x = distance tavelled during the deceleration = 350 ft

u = initial velocity of the truck = 60 mph = 88 ft/s

v = final velocity of the truck = 0 m/s

a = ?

v² = u² + 2ax

0² = 88² + 2(a)(350)

700 a = - 88²

a = - 11.04 ft/s²

But this deceleration acts on the crate as a force trying to put the box in motion.

The motion of the box will be due to the net force on the box

Net force on the box = (Force from deceleration of the truck) - (Frictional force)

Net force = ma

Frictional Force = μmg = 0.3 × m × 32.2 = 9.66 m

Force from the deceleration of the truck = m × 11.04 = 11.04 m

ma = 11.04m - 9.66m

a = 11.04 - 9.66 = 1.4 ft/s²

This net acceleration is now responsible for its motion from rest, through the 10 ft that the box moved, to point B to hit the truck.

x = 10 ft

a = 1.4 ft/s²

u = 0 m/s (box starts from rest, relative to the truck)

v = final velocity of the box relative to the truck, before hitting the truck's wall = ?

v² = u² + 2ax

v² = 0² + 2(1.4)(10)

v² = 28

v = 5.29 ft/s

Answer:

Part 1: The diameter of the shaft so that the shear stress is not more than 150 MPa is 17.3 mm.

Part 2: The diameter of the shaft so that the twist angle  is not more than 7° is 16.9 mm.

Explanation:

Part 1

The formula is given as

[tex]\dfrac{T}{J}=\dfrac{\tau}{R}[/tex]

Here T is the torque which is given as 155 Nm

J is the rotational inertia which is given as [tex]\dfrac{\pi d^4}{32}[/tex]

τ is the shear stress which is given as 150 MPa

R is the radius which is given as d/2 so the equation becomes

[tex]\dfrac{T}{J}=\dfrac{\tau}{R}\\\dfrac{155}{\pi d^4/32}=\dfrac{150 \times 10^6}{d/2}\\\dfrac{155 \times 32}{\pi d^4}=\dfrac{300 \times 10^6}{d}\\\dfrac{1578.82}{d^4}=\dfrac{300 \times 10^6}{d}\\d^3=\dfrac{1578.82}{300 \times 10^6}\\d^3=5.26 \times 10^{-6}\\d=0.0173 m \approx 17.3 mm[/tex]

So the diameter of the shaft so that the shear stress is not more than 150 MPa is 17.3 mm.

Part 2

The formula is given as

[tex]\dfrac{T}{J}=\dfrac{G\theta}{L}[/tex]

Here T is the torque which is given as 155 Nm

J is the rotational inertia which is given as [tex]\dfrac{\pi d^4}{32}[/tex]

G is the torsional modulus  which is given as 114 GPa

L  is the length which is given as 720 mm=0.720m

θ is the twist angle which is given as 7° this is converted to radian as

[tex]\theta=\dfrac{7*\pi}{180}\\\theta=0.122 rad\\[/tex]

so the equation becomes

[tex]\dfrac{T}{J}=\dfrac{G\theta}{L}\\\dfrac{155}{\pi d^4/32}=\dfrac{114 \times 10^9\times 0.122}{0.720}\\\dfrac{1578.81}{d^4}=1.93\times 10^{10}\\d^4=\dfrac{1578.81}{1.93\times 10^{10}}\\d=(\dfrac{1578.81}{1.93\times 10^{10}})^{1/4}\\d=0.0169 m \approx 16.9mm[/tex]

So the diameter of the shaft so that the twist angle  is not more than 7° is 16.9 mm.

Answer:

(C) ln [Bi]

Explanation:

Radioactive materials will usually decay based on their specific half lives. In radioactivity, the plot of the natural logarithm of the original radioactive material against time will give a straight-line curve. This is mostly used to estimate the decay constant that is equivalent to the negative of the slope. Thus, the answer is option C.

Answer:

Option C. ln[Bi]

Explanation:

The nuclear equation of first-order radioactive decay of Bi to Po is:

²¹⁴Bi₈₃  →  ²¹⁴Po₈₄ + e⁻

The radioactive decay is expressed by the following equation:

[tex] N_{t} = N_{0}e^{-\lambda t} [/tex]      (1)  

where Nt: is the number of particles at time t, No: is the initial number of particles, and λ: is the decay constant.                

To plot the variation of the quantities in function of time, we need to solve equation (1) for t:

[tex] Ln(\frac{N_{t}}{N_{0}}) = -\lambda t [/tex]      (2)

Firts, we need to convert the number of particles of Bi (N) to concentrations, as follows:

[tex] [Bi] = \frac {N particles}{N_{A} * V} [/tex]                            (3)  

[tex] [Bi]_{0} = \frac {N_{0} particles}{N_{A} * V} [/tex]               (4)  

where [tex]N_{A}[/tex]: si the Avogadro constant and V is the volume.

Now, introducing equations (3) and (4) into (2), we have:

[tex] Ln (\frac {\frac {[Bi]*N_{A}}{V}}{\frac {[Bi]_{0}*N_{A}}{V}}) = -\lambda t [/tex]  

[tex] Ln (\frac {[Bi]}{[Bi]_{0}}) = -\lambda t [/tex]      (5)

Finally, from equation (5) we can get a plot of Bi versus time in where the curve is a straight line:

[tex] Ln ([Bi]) = -\lambda t + Ln([Bi]_{0}) [/tex]      

Therefore, the correct answer is option C. ln[Bi].

I hope it helps you!          

Answer:

Java program given below

Explanation:

import java.util.*;

import java.io.*;

public class Lineeditor

{

private static Node head;

class Node

{

 int data;

 Node next;

 public Node()

 {data = 0; next = null;}

 public Node(int x, Node n)

 {data = x; next =n;}

}

public void Displaylist(Node q)

 {if (q != null)

       {  

        System.out.println(q.data);

         Displaylist(q.next);

       }

 }

public void Buildlist()

  {Node q = new Node(0,null);

       head = q;

       String oneLine;

       try{BufferedReader indata = new

                 BufferedReader(new InputStreamReader(System.in)); // read data from terminals

                       System.out.println("Please enter a command or a line of text: ");  

          oneLine = indata.readLine();   // always need the following two lines to read data

         head.data = Integer.parseInt(oneLine);

         for (int i=1; i<=head.data; i++)

         {System.out.println("Please enter another command or a new line of text:");

               oneLine = indata.readLine();

               int num = Integer.parseInt(oneLine);

               Node p = new Node(num,null);

               q.next = p;

               q = p;}

       }catch(Exception e)

       { System.out.println("Error --" + e.toString());}

 }

public static void main(String[] args)

{Lineeditor mylist = new Lineeditor();

 mylist.Buildlist();

 mylist.Displaylist(head);

}

}

Answer:

9.58 Kg of air has entered the tank.

heat entered=3483.76 Kilo.Joule

Explanation:

(A) R=287 Kilo.J/Kg.K

as per initial conditions P=100 Kilo.Pa ,V=2 cubic meter, T=22 C=295.15 K,

using the relation P*V=m*R*T

m=(100*1000*2)/(287*295.15)=2.36 Kg this is the mass that is already present in tank.

after filling tank at 600 Kilo.Pa.

P=600 Kilo Pa T=77 C=350.15 K

P*V=m*R*T

m=(600*1000*2)/(287*350.15)=11.94 Kg

mass that has entered=11.94-2.36=9.58 Kg

(b) using air psychometric  property table

specific heat content initial  100 KILO Pa and 22 C=295.576 Kilo.Joule/Kg

specific heat content final  600 Kilo Pa and 77 C=350.194 Kilo.Joule/Kg

heat at initial stage=295.576*2.36=697.56 Kilo.Joule

heat at final stage=350.194*11.94=4181.32 Kilo.Joule

heat entered=4181.32-697.56=3483.76 Kilo.Joule

Answer: 50 minutes

Explanation:

The energy needed to heat air inside the room is the electric energy dissipated by the resistance. It is known after using First Principle of Thermodynamics:

[tex]Q_{in,air} = \dot W_{dis, heater} \cdot \Delta t[/tex]

[tex]\rho_{air} \cdot V_{room} \cdot c_{p,air} \cdot \Delta T = \dot W_{dis,heater} \cdot \Delta t[/tex]

The needed time is:

[tex]\Delta t =\frac{\rho_{air}\cdot V_{room}\cdot c_{p,air} \cdot\Delta T}{\dot W_{dis,heater}}[/tex]

Where [tex]\rho_{air} = 1.20 \frac{kg}{m^{3}}[/tex] and [tex]c_{p,air} = 1.012 \frac{kJ}{kg \cdot ^{\circ} C}[/tex]:

[tex]\Delta t = 3005.640 s (50.094 min)[/tex]

Answer:

b. Angle of departure=26.56 degrees

c. √5j

Please see attachment for the sketch and step by step guide.

Answer:

Answer is 0.74

Explanation:

Probability that the distance is at most 100 m:

P ( X ≤ 100 )  =  F ( 100 )

              =   (1 − e-^ λ *x)

              =   1 − e −^( ( 0.0134 )* ( 100 ))

              = 1 − e −^ 1.34

              = 1 − 0.2618

             = 0.7381

             ≈ 0.74

Therefore,

P ( X ≤ 100 )≈ 0.74

Answer.

Answer:

0.7400

Explanation:

Let's first look at what we have:

λ = 0.0134                                                                                                                         mean = 1/λ = 74.24                                                                                                      standard deviation = 1/λ = 74.24

We'll use the following formula;                                                                     CDF, C(x) = 1 - e^(-λx)

Probability of a distance not exceeding 100 m

P(X <= 100)

= C(100)

= 1 - e^(-0.01347 * 100)

= 0.7400

Answer:A

Explanation:

Damp roof is generally applied at basement level which restrict the movement of moisture through walls and floors. Therefore it could be inside or the outside basement walls.

Answer: [tex]\dot m_{in} = 23.942 \frac{kg}{s}[/tex], [tex]\dot H_{out} = 39632.62 kW[/tex]

Explanation:

Since there is no information related to volume flow to and from turbine, let is assume that volume flow at inlet equals to [tex]\dot V = 1 \frac{m^{3}}{s}[/tex]. Turbine is a steady-flow system modelled by using Principle of Mass Conservation and First Law of Thermodynamics:

Principle of Mass Conservation

[tex]\dot m_{in} - \dot m_{out} = 0[/tex]

First Law of Thermodynamics

[tex]- \dot W_{out} + \eta\cdot (\dot m_{in} \dot h_{in} - \dot m_{out} \dot h_{out}) = 0[/tex]

This 2 x 2 System can be reduced into one equation as follows:

[tex]-\dot W_{out} + \eta \cdot \dot m \cdot ( h_{in}- h_{out})=0[/tex]

The water goes to the turbine as Superheated steam and goes out as saturated vapor or a liquid-vapor mix. Specific volume and specific enthalpy at inflow are required to determine specific enthalpy at outflow and mass flow rate, respectively. Property tables are a practical form to get information:

Inflow (Superheated Steam)

[tex]\nu_{in} = 0.041767 \frac{m^{3}}{kg} \\h_{in} = 3399.5 \frac{kJ}{kg}[/tex]

The mass flow rate can be calculated by using this expression:

[tex]\dot m_{in} =\frac{\dot V_{in}}{\nu_{in}}[/tex]

[tex]\dot m_{in} = 23.942 \frac{kg}{s}[/tex]

Afterwards, the specific enthalpy at outflow is determined by isolating it from energy balance:

[tex]h_{out} =h_{in}-\frac{\dot W_{out}}{\eta \cdot \dot m}[/tex]

[tex]h_{out} = 1655.36 \frac{kJ}{kg}[/tex]

The enthalpy rate at outflow is:

[tex]\dot H_{out} = \dot m \cdot h_{out}[/tex]

[tex]\dot H_{out} = 39632.62 kW[/tex]

Answer:

The frequency that the sampling system will generate in its output is 70 Hz

Explanation:

Given;

F = 190 Hz

Fs = 120 Hz

Output Frequency = F - nFs

When n = 1

Output Frequency = 190 - 120 = 70 Hz

Therefore, if a system samples a sinusoid of frequency 190 Hz at a rate of 120 Hz and writes the sampled signal to its output without further modification, the frequency that the sampling system will generate in its output is 70 Hz

Answer:

water sample have more water content

Explanation:

given data

soil 1 is saturated with  water

unit weight of water = 1 g/cm³

soil 2 is saturated with alcohol

unit weight of alcohol  = 0.8 g/cm³

solution

we get here water content that is express as

water content = [tex]S_s \times \gamma _w[/tex]   ....................1

here soil is full saturated so [tex]S_s[/tex] is 100% in both case

so put here value for water

water content = 100 % ×  1

water content = 1 g

and

now we get for alcohol that is

water content  = 100 % × 0.8

water content  = 0.8 g

so here water sample have more water content

The soil sample saturated with water has a larger water content than the one with alcohol because water's unit weight is greater than that of alcohol and both soils have the same void ratio.

The soil sample saturated with water will have a larger water content compared to the soil saturated with alcohol. This is because the unit weight of water is greater than that of alcohol. Since both soils are fully saturated and have the same void ratio, the volume of liquid in both cases is identical. Therefore, the soil with the denser liquid (water) will contain more mass of liquid per unit volume, leading to a higher water content.