Before entering an underground utility vault to do repairs, a work crew analyzed the gas in the vault and found that it contained 29 mg/m3of hydrogen sulfide. Because the allowable exposure level is 14 mg/m3, the work crew began ventilating the vault with a blower. If the volume of the vault is 160 m3and the flow rate of contaminant-free air is 10 m3/min, how long will it take to lower the hydrogen sulfide level to a level that will allow the work crew to enter? Assume the manhole behaves as a CMFR and that hydrogen sulfide is nonreactive in the time period considered.

Answer:

a) 44.4%

b) 72 mm

Explanation:

See attached pictures.

a. The field relative density is 44.44%.

b. A 3 m thick stratum of this sand will settle by 0.111 m when densified to a relative density of 65%.

Let's solve the problem step-by-step.

Given Data

- Wet density of sand [tex](\( \rho_{wet} \)) = 1.9 Mg/m\(^3\)[/tex]

- Field water content ( w ) = 10% = 0.10

- Density of solids [tex](\( \rho_s \))[/tex] = 2.66 [tex]Mg/m\(^3\)[/tex]

- Maximum void ratio [tex](\( e_{max} \))[/tex] = 0.62

- Minimum void ratio [tex](\( e_{min} \))[/tex] = 0.44

a. Calculation of Field Relative Density

First, we need to calculate the dry density of the sand:

[tex]\[ \rho_{dry} = \frac{\rho_{wet}}{1 + w} = \frac{1.9}{1 + 0.10} = \frac{1.9}{1.10} = 1.727 \text{ Mg/m}^3 \][/tex]

Now, we use the dry density to find the void ratio  e :

[tex]\[ e = \frac{\rho_s}{\rho_{dry}} - 1 = \frac{2.66}{1.727} - 1 = 1.54 - 1 = 0.54 \][/tex]

Relative density [tex](\( D_r \))[/tex] is given by the formula:

[tex]\[ D_r = \frac{e_{max} - e}{e_{max} - e_{min}} \times 100\% \][/tex]

Substituting the values:

[tex]\[ D_r = \frac{0.62 - 0.54}{0.62 - 0.44} \times 100\% = \frac{0.08}{0.18} \times 100\% = 44.44\% \][/tex]

b. Settlement Calculation

To find the settlement of a 3 m thick stratum of sand when densified to a relative density of 65%, we need to determine the void ratio corresponding to 65% relative density.

[tex]\[ D_r = 65\% = 0.65 \][/tex]

Using the relative density formula again, solve for  e :

[tex]\[ 0.65 = \frac{e_{max} - e_{new}}{e_{max} - e_{min}} \][/tex]

[tex]\[ 0.65 = \frac{0.62 - e_{new}}{0.62 - 0.44} \][/tex]

[tex]\[ 0.65 \times (0.62 - 0.44) = 0.62 - e_{new} \][/tex]

[tex]\[ 0.65 \times 0.18 = 0.62 - e_{new} \][/tex]

[tex]\[ 0.117 = 0.62 - e_{new} \][/tex]

[tex]\[ e_{new} = 0.62 - 0.117 = 0.503 \][/tex]

Now calculate the initial and final volumes of voids:

Initial void ratio [tex]\( e_{initial} = 0.54 \)[/tex]

Final void ratio [tex]\( e_{new} = 0.503 \)[/tex]

Initial volume of voids [tex]\( V_{v_initial} \):[/tex]

[tex]\[ V_{v_initial} = e_{initial} \times V_s \][/tex]

Final volume of voids [tex]\( V_{v_final} \):[/tex]

[tex]\[ V_{v_final} = e_{new} \times V_s \][/tex]

The change in void volume:

[tex]\[ \Delta V_v = V_{v_initial} - V_{v_final} = (e_{initial} - e_{new}) \times V_s \][/tex]

For the 3 m thick stratum:

[tex]\[ \Delta H = \Delta V_v \][/tex]

[tex]\[ \Delta H = (e_{initial} - e_{new}) \times H \][/tex]

[tex]\[ \Delta H = (0.54 - 0.503) \times 3 \text{ m} \][/tex]

[tex]\[ \Delta H = 0.037 \times 3 \text{ m} \][/tex]

[tex]\[ \Delta H = 0.111 \text{ m} \][/tex]

So, the sand stratum will settle by 0.111 m when densified to a relative density of 65%.

Answer:

Rate of heat transfer is 0.56592 kg/hour

Explanation:

Q = kA(T2 - T1)/t

Q is rate of heat transfer in Watts or Joules per second

k is thermal conductivity of the styrofoam = 0.035 W/(mK)

A is area of the cubical picnic chest = 6L^2 = 6(0.5)^2 = 6×0.25 = 1.5 m^2

T1 is initial temperature of ice = 0 °C = 0+273 = 273 K

T2 is temperature of the styrofoam = 25 °C = 25+273 = 298 K

t is thickness of styrofoam = 0.025 m

Q = 0.035×1.5(298-273)/0.025 = 1.3125/0.025 = 52.5 W = 52.5 J/s

Mass flow rate = rate of heat transfer ÷ latent heat of melting of ice = 52.5 J/s ÷ 3.34×10^ 5 J/kg = 1.572×10^-4 kg/s = 1.572×10^-4 kg/s × 3600 s/1 hr = 0.56592 kg/hr

Your question is incomplete, please let me assume this to be your complete question;

ORIGINAL SOURCE:

When instructors are creating discussion board activities for online courses, at least two questions must be answered. First, what is the objective of the discussions? Different objectives might be to create a "social presence" among students so that they do not feel isolated, to ask questions regarding assignments or topics, or to determine if students understand a topic by having them analyze and evaluate contextual situations. Based on the response to this question, different rules might be implemented to focus on the quality of the interaction more so than the quantity. The second question is, how important is online discussions in comparison to the other activities that students will perform? This question alludes to the amount of participation that instructors expect from students in online discussions along with the other required activities for the course. If a small percentage of student effort is designated for class participation, our results show that it can affect the quality and quantity of interactions.

References:

Moore, J. L., & Marra, R. M. (2005) A comparative analysis of online discussion participation protocols.Journal of Research on Technology in Education, 38(2), 191-212.

STUDENT VERSION:

According to Moore and Marra's (2005) case study, which observed two online courses, students in the first course implemented a constructive argumentation approach while students in second course had less structure for their postings. As they stated, when instructors create online discussion board activities, they must answer at least two questions. These questions are: "What is the objective of the discussions?" And "How important are online discussions in comparison to the other activities that students will perform?". According to their findings, the discussion activities that were designed based on the answers to these questions can influence the quality and quantity of interactions (Moore & Marra, 2005).

References:

Moore, J. L., & Marra, R. M. (2005) A comparative analysis of online discussion participation protocols.Journal of Research on Technology in Education, 38(2), 191-212.

Which of the following is true for the students work;

Word-for-word plagiarism

Paraphrasing plagiarism

Not Plagiarism

ANSWER: IT IS NOT PLAGIARISM

Explanation: plagiarism is the act of extracting knowledge from someone's literature work, without acknowledging the literature work. In other words, this can be called a theft of knowledge, because when you failed to acknowledge the literature source that helped you to produce your paper work, it means you have claimed to be the original owner of that knowledge.

This is not a Plagiarism because the student has acknowledged the source of the knowledge in it's literature work. The original source and the student has cited the same literature work, this why their work looks similar but not exactly the same. So the student has not committed Plagiarism

Answer:

E = 7333.33 mm

Explanation:

The annual evapotranspiration (E) amount can be calculated using the water budget equation:

[tex] P*A + Q_{in}*\Delta t = E*A + \Delta S + Q_{out}*\Delta t [/tex]   (1)

Where:

P: is the precipitation = 2500 mm,

Q(in): is the water flow into the river of the farmland = 5 m³/s,

ΔS: is the change in water storage = 2.5x10⁶ m³,  

Q(out): is the water flow out of the river of the farmland = 4 m³/s.

Δt: is the time interval = 1 year = 3.15x10⁷ s

A: is the surface area of the farmland = 6.0x10⁶ m²  

Solving equation (1) for ET we have:

[tex] E = \frac{P*A + Q_{in}*\Delta t - \Delta S - Q_{out}*\Delta t}{A} [/tex]

[tex] E = \frac{2.5 m \cdot 6.0 \cdot 10^{6} m^{2} + 5 m^{3}/s \cdot 3.15 \cdot 10^{7} s - 2.5 \cdot 10^{6} m^{3} - 4 m^{3}/s \cdot 3.15 \cdot 10^{7} s}{6.0\cdot 10^{6} m^{2}} [/tex]                                  

[tex] E = 7333.33 mm [/tex]

Therefore, the annual evapotranspiration amount is 7333.33 mm.

I hope it helps you!  

The annual evapotranspiration amount can be calculated using the hydrologic budget equation, by arranging known values of rainfall, inflow and outflow rates, and increase in water storage. This gives us the evapotranspiration amount in cubic meters which can be converted into depth (in mm) by dividing by the total land area.

The annual evapotranspiration amount can be calculated using the concept of the hydrologic budget equation, which states that the change in storage equals the sum of inputs minus the sum of outputs. In this scenario, rainfall and river inflow are the water inputs while evapotranspiration and river outflow are the water outputs. Given that the increase in water storage, rainfall, and river flow rates are known, we can rearrange the equation to find the evapotranspiration.

It results in:
Evapotranspiration (in m3) = Rainfall + Inflow - Outflow - Increase in Storage
Substituting the given values:
Evapotranspiration (in m3) = (2500 mm * 600 ha * 10,000 m2/ha * 1m/1000mm) + (5 m3/s * 31,536,000 s) - (4 m3/s * 31,536,000 s) - 2.5 * 106 m3;

To convert evapotranspiration volume to depth (in mm), we divide by the total land area:
Evapotranspiration (in mm) = Evapotranspiration (in m3) / (600ha * 10,000 m2/ha * 1mm/1m)

After computing the above equations, we arrive at the annual evapotranspiration amount in mm.

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Answer:

a) 2.42 [tex]kg/s[/tex]

b) 37.20 m/s

c) 120.56 kW

Explanation:

Given that:

The fluid in the Refrigerant = R-134a

Diameter (d) = 0.1 m

In the Inlet:

Temperature [tex]T_1 = 40^0C[/tex]

Pressure [tex]P_1= 300kPa[/tex]

Velocity [tex]V_1[/tex] = 25 m/s

At the exit:

Temperature [tex]T_2 = 90^0C[/tex]

Pressure [tex]P_2 = 240 kPa[/tex]

From the  Table A-12 for Refrigerant R-134a at [tex]T_1 = 40^0C[/tex] and [tex]P_1= 300kPa[/tex]

Specific Volume [tex]v_1 = 0.0809 m^3/kg[/tex]

From the  Table A-12 for Refrigerant R-134a at [tex]T_2 = 90^0C[/tex] and [tex]P_2 = 240 kPa[/tex]

Specific Volume [tex]v_2 = 0.12038 kJ/kg[/tex]

Their corresponding Enthalpy [tex]h_1[/tex] and [tex]h_2[/tex] are as follows:

Enthalpy [tex]h_1[/tex]  =284.05 kJ/kg

Enthalpy [tex]h_2[/tex] = 333 kJ/kg

a) The mass flow rate of the refrigerant can be calculated as :

[tex]m_1 = \frac{AV_1}{v_1}[/tex]

[tex]m_1 = \frac{\frac{\pi (0.1)^2}{4}*25}{0.08089}[/tex]

[tex]m_1 = 2.42 kg/s[/tex]

b) The velocity at the exit point:

we knew that:

[tex]m=m_1 =m_2[/tex]

∴

[tex]\frac{AV_1}{v_1} =\frac{AV_2}{v_2}[/tex]

[tex]V_2 = \frac{v_2}{v_1} V_1[/tex]

[tex]V_2 = \frac{0.12038}{0.08089} *25[/tex]

[tex]V_2 = 37.20 m/s[/tex]

c) Expression for calculating heat transfer (as long as there is no work that is said to be done and the pipe is horizontal) can be represented as:

[tex]Q_{cv}= m[(h_2-h_1)+\frac{1}{2}(V_2^2-V_1^2)][/tex]

[tex]Q_{cv}= 2.42*[(333.49-284.05)+\frac{1}{2}(37.20^2-25^2)][/tex]

[tex]Q_{cv}= 2.42*[49.44+379.42][/tex]

[tex]Q_{cv}= 119.6448kW+918.19W(\frac{1kW}{1000W} )[/tex]

[tex]Q_{cv}= 119.6448kW+0.92 kW[/tex]

[tex]Q_{cv} = 120.56 kW[/tex]

Following are the solution to the given points:

Obtain the following property at pressure [tex]300\ kPa[/tex] and [tex]40^{\circ} \ C[/tex] from the property of superheated [tex]134a[/tex]  

Using the interpolation method,  

Specific volume, [tex]v_1 = 0.0866 -(0.0866 -0.07518) (\frac{0.3-0.28}{0.32-0.28})= 0.08089 \frac{kg}{m^3}[/tex]

Enthalpy, [tex]h_1 = 284.42 - (284.42 - 283.67) (\frac{0.3-0.28}{0.32-0.28}) = 284.05\ \frac{kJ}{kg}\\\\[/tex]

Obtain the following property at pressure [tex]240 \ kPa \ and \ 90^{\circ} \ C[/tex] from the property of superheated [tex]134a[/tex].  

Specific volume,[tex]v_2 = 0.12 \ \frac{kg}{m^3}[/tex]

Enthalpy of superheated [tex]134a, \ \ h_2 = 333 \ \frac{kJ}{kg}[/tex]

For point a:

Calculating the refrigerant weight rate flow:  

[tex]m_1 =\frac{AV_1}{v_1} =\frac{ \frac{ \pi (0.1)^2}{4} \times 25}{0.08089} = \frac{0.196}{0.08089} \\\\ m_1 = 2.42 \frac{kg}{s}[/tex]

Thus, refrigerant weight rate flow is [tex]2.42\ \frac{m^3}{kg}\\\\[/tex]

For point b:

Calculate the exit velocity:

[tex]m_1 = m_2\\\\ \frac{A V_1}{v_1}=\frac{AV_2}{v_2}\\\\ V_2=\frac{V_2}{v_1} V_1\\\\v_2= \frac{0.12038}{ 0.08089} \times 25\\\\ V_2 = 37.20 \frac{m}{s}\\\\[/tex]

Thus, the exit velocity is [tex]37.20\ \frac{m}{s}\\\\[/tex]

For point c:

From the energy rate balance Since, there is no work being done and the pipe is horizontal.  So, the above equation can be written as

[tex]\to Q_{cv} = m [(h_2 - h_1)+ \frac{1}{2}(v_{2}^{2}- v_{1}^{2})] \\\\[/tex]

          [tex]= 2.42 \times [(333.49 - 284.05) +\frac{1}{2}(37.20^2-25^2)]\\\\ = 2.42 \times [49.44 +379.42] \\\\= 119.64\ kW +918.19 \ W |\frac{1 \ KW}{1000\ W}| \\\\= 119.64 \ kW +0.92\ kW\\\\ =120.56\ kW[/tex]

Thus, the heat transfer rate among pipe and its surrounding [tex]120.56\ kW[/tex]

Learn more about Refrigerant:

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Answer:

a) The forward voltage is 0.23 V

b) The current that flows  [tex]I_{d} = (1.45*10^{12}I_{s})A[/tex]

Explanation:

The forward voltage is the minimum voltage that must be applied to a diode before it starts to conduct. The equation is given by:

a) At what forward voltage does a diode conduct a current equal to 10,000 Is ? In terms of Is

[tex]I_{d} = I_{s}(e^{\frac{v_{f} }{0.025} }-1)[/tex]

Where:

Id is the diode current = 10000Is,

Vd is the forward voltage at which the diode begins to conduct,

Is is the saturation current.

[tex]I_{d} = I_{s}(e^{\frac{v_{f} }{0.025} }-1)[/tex]

[tex]10000I_{s} = I_{s}(e^{\frac{v_{f} }{0.025} }-1)[/tex]

Dividing through by Is,

[tex]10000 = (e^{\frac{v_{f} }{0.025} }-1)[/tex]

[tex]10000 +1= e^{\frac{v_{f} }{0.025} }[/tex]

[tex]10001= e^{\frac{v_{f} }{0.025} }[/tex]

Taking the natural logarithm of both sides,

[tex]ln(10001)= {\frac{v_{f} }{0.025} }[/tex]

[tex]9.21= {\frac{v_{f} }{0.025} }[/tex]

multiplying through by 0.025

[tex]{v_{f} }= 0.23[/tex] = 0.23 V

The forward voltage does a diode conduct a current equal to 10,000 Is is 0.23 V

b) what current flows in the same diode when its forward voltage is 0.7 V?

[tex]I_{d} = I_{s}(e^{\frac{v_{f} }{0.025} }-1)[/tex]

[tex]I_{d} = I_{s}(e^{\frac{0.7}{0.025} }-1)[/tex]

[tex]I_{d} = I_{s}(1.45*10^{12} -1)[/tex]

[tex]I_{d} = (1.45*10^{12}I_{s})A[/tex]

The question is incomplete! The complete question along with Matlab code and explanation is provided below.

Question

Plot the following trig functions using subplots, choosing an appropriate layout for the number of functions displayed. The subplots should include a title which is the equation displayed. The independent variable (angle) should vary from 0 to 360 degrees and the plots should use a solid red line.

1. cos(u - 45)

2. 3cos(2u) - 2

3. sin(3u)

4. -2cos(u)

Matlab Code with Explanation:

u=[0:0.01:2*pi] % independent variable represents 0 to 360 degrees in steps of 0.01

y1=cos(2*pi*u-45); % function 1

y2=3*cos(2*pi*2*u)-2;  % function 2

y3=sin(2*pi*3*u);  % function 3

y4=-2*cos(2*pi*u);  % function 4

subplot(4,1,1) % 4 rows, 1 column and at position 1

plot(u,y1,'r');  % this function plots y w.r.t u and 'r' is for red color

grid on   % turns on grids

xlabel('u') % label of x-axis

ylabel('y1')  % label of x-axis

title('y1=cos(2*pi*u-45)') % title of the plot

ylim([-3 3]) % limits of y-axis

xlim([0 2*pi]) % limits of x-axis

% repeat the same procedure for the remaining 3 functions

subplot(4,1,2)

plot(u,y2,'r');  

grid on  

xlabel('u')  

ylabel('y2')  

title('y2=3*cos(2*pi*2*u)-2')  

ylim([-6 3])

xlim([0 2*pi])

subplot(4,1,3)  

plot(u,y3,'r');

grid on  

xlabel('u')  

ylabel('y3')  

title('y3=sin(2*pi*3*u)')  

ylim([-3 3])  

xlim([0 2*pi])

subplot(4,1,4)  

plot(u,y4,'r');  

grid on  

xlabel('u')

ylabel('y4')  

title('y4=-2*cos(2*pi*u)')  

ylim([-3 3])

xlim([0 2*pi])

Output Results:

The first plot shows a cosine wave with a phase shift of 45°

The second plot shows that the amplitude of the cosine wave is increased and the wave is shifted below zero level into the negative y-axis because of -2 also there is a increase in frequency since it is multiplied by 2.

The third plot shows that the frequency of the sine wave is increased since it is multiplied by 3.

The fourth plot shows a cosine wave which is multiplied by -2 and starts from the negative y-axis.

Answer:

The viral DNA will be fused into the host's DNA.

Explanation:

When a virus exhibits a lysogenic life cycle, it ensures that its host is not killed rather the viral DNA gets fused into the host's DNA with the viral genes not expressed. The virus releases nucleic acid into the chromosome and becomes part of the host. It undergoes cell division and passes daughter cells while leaving its DNA in the host. The embedded virus goes through the lytic cycle, creating more viruses.

Answer:

[tex]P_{atm} = 87.5\,kPa[/tex]

Explanation:

The heat required to boil the water in the pan is:

[tex]Q = \eta_{e}\cdot \dot W_{e} \cdot \Delta t[/tex]

[tex]Q = 0.75 \cdot (2\,kW)\cdot (30\, min)\cdot (\frac{60\,sec}{1\, min} )[/tex]

[tex]Q = 2700\,kJ[/tex]

Since the pan is accompained with a poorly fitting lid, the heating process is isobaric and change on specific enthalpy is obtained by following expression:

[tex]\Delta h = \frac{Q}{\Delta m}[/tex]

[tex]\Delta h = \frac{2700\,kJ}{1.19\,kg}[/tex]

[tex]\Delta h = 2268.908\,\frac{kJ}{kg}[/tex]

Then, the local atmospheric pressure can be estimated by looking for the saturation pressure related to the change on specific enthalpy at property tables for saturated water:

[tex]P_{atm} = 87.5\,kPa[/tex]

Answer:

106600 btu/s

note:

solution is attached due to error in mathematical equation. please find the attachment

The output voltage from a Wheatstone bridge with a change in temperature of 20℃ in one of its platinum resistance temperature sensor arms is calculated to be approximately 0.233 V, taking into account the specific temperature coefficient of resistance for platinum.

The student's question involves calculating the output voltage from a Wheatstone bridge when a platinum resistance temperature sensor, which forms one arm of the bridge, changes its resistance due to a temperature change. With a temperature coefficient of resistance for platinum of 0.0039/K and an initial balance condition at 0℃ with each arm having a resistance of 120 Ω, the temperature change of 20℃ will lead to a change in resistance in the platinum arm, affecting the bridge's balance and generating an output voltage.

To calculate the change in resistance (ΔR) for the platinum sensor due to the temperature change: ΔR = Ro·α·ΔT, where Ro is the initial resistance (120 Ω), α is the temperature coefficient of resistance (0.0039/K), and ΔT is the temperature change (20℃). Therefore, ΔR = 120·0.0039·20 = 9.36 Ω. The new resistance of the platinum sensor at 20℃ is 120 Ω + 9.36 Ω = 129.36 Ω.

Given the supply voltage (Vs) is 6.0 V, and considering the bridge was initially balanced, the output voltage (Vo) from the bridge can be calculated using the formula derived from the Wheatstone bridge principles: Vo = Vs · (ΔR / (2Ro + ΔR)). Substituting the values gives Vo = 6.0 · (9.36 / (240 + 9.36)) = 0.233 V. Thus, the output voltage from the bridge for a change in temperature of 20℃ is approximately 0.233 V.

Answer:

Part (a)

K = 0.00406 m / s

T = 0.14 m2 / s

Part (b)

R = Radius of influence of Pumping well = 237.94 m

Part (c)

S = Drawdown at well = 3.68 m

Explanation:

General formula for wells at confined aquifer is

Where ,

    Q = Discharge

    K = Hydraulic conductivity

    B = Thickness of aquifer

    S1 = Draw-down at point 1

    S2 = Draw-down at point 2

    R = Radius of influence of well

    r =  Radius of well

Part (a)

We will use

Where,

    r 1 = Distance of first observation well from main well

    r2 = Distance of 2nd observation well from main well

Given data:  Q = 0.5 m3/s       B = 34 m            r 1 = 50 m  

                     r2 = 100 m          S1 = 0.9 m          S2 = 0.4 m

Put all these values in equation 1

  0.5 = 2*3.1416*K*34*(0.9 - 0.5) / {2.303 log(100 / 50)}

Write Equation in terms of K = hydraulic conductivity

 K = {0.5*2.303 log (100 / 50)} / {2*3.1416*34*(0.9 - 0.5)}

 K = 0.00406 m / s

Now Transmissivity

  T = K*B

  T = 0.14 m2 / s

Part (b)

To calculate radius of influence, use equation (A)  

  Q = [tex]{2*3.1416*K*B*(S1-S2)} / {2.303 log (R/r)}[/tex]

  At distance R from main well drawdown (S2) = 0

  use r = r1 = 50 m and S1 = 0.9

  use ln ( R / r) = 2.303 log (R / r)

  0.5 = 2* 3.1416 * 0.00406*34* (0.9 - 0) / ln(R / 50)

  Write equation in terms of R

  ln( R / 50) =  2* 3.1416 * 0.00406*34* (0.9 - 0) / 0.5

  ln( R / 50) = 1.56

  Take anti log (e) of above equation

  R / 50 = 4.76

  R = Radius of influence of Pumping well = 237.94 m

Part (c)

 Use equation A

 Q = [tex]{2*3.1416*K*B*(S1-S2)} / {2.303 log (R/r)}[/tex]

 S1 = ?            

Put S2 = 0         R = 237.94           r = 0.4

0.5 = 2* 3.1416* 0.00406*34*(S1 - 0) / 2.303 log(237.94 / 0.4)

Write equation in terms of S1

S1 = 0.5* 2.303 log(237.94 / 0.4) / 2*3.1416*0.00406*34

S1 = Drawdown at well = 3.68 m

The hydraulic conductivity and transmissivity of the aquifer are respectively; 0.0032 m/s and 0.11 m³/s

A) We are given;

Pump rate; Q = 0.5 m³/s

thickness of quifer; b = 34 m

depth 1; r₁ = 50 m

depth 2; r₂ = 100 m

distance 1; S₁ = 0.9 m

distance 2; S₂ = 0.4 m

Formula for the pump rate is;

Q = 2π × b × k × (S₁ - S₂)/(In r₂/r₁)

making k the subject gives;

k = Q(In r₂/r₁)/(2π × b × (S₁ - S₂))

k = 0.5(In 100/50)/(2π × 34 × (0.9 - 0.4))

Solving for K gives;

Hydraulic conductivity is; k = 0.0032 m/s

Transmissivity is;

T = K * b

T = 0.0032 * 34

T = 0.11 m³/s

B) Formula for radius of incfluence is;

S_w = S₁ - [(Q/2π × b × k) In (r_w/r₁)]

Plugging in the relevant values gives;

S_w = 4.338 m

C) Formula for expected drawdown is;

R = r₁ e^(2πbk(S_w - S₁)/Q)

R = 100 * e^(2π*34*0.0032(-78.9)/0.5)

R = 147.7 m

Read more about Hydraulic conductivity at; https://brainly.com/question/26411935

Answer:

note:

please find the attached code

Answer: $15.34

Explanation: see image below

To find the yearly energy cost of the washing machine, half the estimated energy usage is taken due to halved weekly loads, resulting in 175 kW-hrs per year. Multiplying this by the local electricity cost gives an annual operating cost close to $15.34.

To calculate the annual energy usage of the compact washing machine, you should first adjust the estimated yearly electricity use based on the difference in the number of loads washed per week. Since you are washing half the number of loads (4 instead of 8), you should cut the energy usage in half:

Annual Energy Usage = 0.5 × 350 kW-hrs = 175 kW-hrs per year.

To calculate the annual energy cost of operating the machine, you multiply the adjusted energy usage by your local electricity cost:

Annual Energy Cost = 175 kW-hrs × $0.086 per kW-hr = $15.05

Thus, the answer closest to the calculated annual energy cost is (c) $15.34.

Answer:

Explanation:

The method or principle applied is the Routh- hurtwitz criterion for solving characteristics equation.

The steps by step analysis and appropriate substitution is carefully shown in the attached file.

Complete Question

The complete question is shown on the second uploaded image

Answer:

a

The optimal x coordinate is 50.76 and the optimal y coordinate is 46.34

b

The Scatter plot is shown on the second uploaded image

Explanation:

In this question we are given the annual demand and the annual cost per mile  per ton. Now to obtain the annual cost per mile for the whole inventory  we multiply the demand  with the cost per mile per ton.

This shown on the third uploaded image

To obtain the optimal new coordinate let consider the location of the existing coordinates and the annual cost of the existing facilities

   Mathematically the optimal x coordinate = (Summation of the old x coordinates multiplied by the annual cost per mile) /(Summation of the annual cost per mile  )

i.e optimal new x coordinate = [tex](40* 6250 +50*4400+70*9500 +25*4350)/(6250+4400+9500+4350)[/tex]

[tex]=(250000+220000+665000+108750)/(24500) = 1243750/24500 = 50.76[/tex]  

 For y

the optimal y coordinate = (Summation of the old y coordinates multiplied by the annual cost per mile) /(Summation of the annual cost per mile  )

      [tex]=(20*6250 +25*4400+65*9500+65*4350)/(6250+4400+9500+4350)[/tex]

[tex]=1135250/24500 = 46.35[/tex]

On the Scatter plot the existing location are in green while the optimal location is in white

The table that shows the given and obtained data is shown on the fourth uploaded image  

Answer:

Wavelength = 5,000,000 m

Explanation:

Power line has extremely low frequency and produce and transmit magnetic and electric over long distance. The radiation from power line are electromagnetic radiation since it can be classified as Radiowave. Hence using the formula:

Velocity of propagation=frequency

× wavelength

We use Velocity= c = 3 × 10 ^8 m/s

f = 60Hz

wavelength = 3 ×10^8/60

Wavelength = 5,000,000 m

We can ignore the transmission line effect when the line length is less than (1/50)wavelength or (1/20)wavelength.

Seeing the transmission line length given is 1000m, we ignore the effect of transmission line as negligible

The magnitude of the impulse imparted to the bullet is 2.932 lb s. The magnitude of the average force acting on the bullet during the given time interval is 2666 lb.

Calculate the Impulse

Impulse is defined as the change in momentum of an object. It is given by the equation:

[tex]\[ \text{Impulse} = \Delta p = m \Delta v \][/tex]

Conversion factor for pound to slug (since force in pounds and velocity in ft/s, mass should be in slugs, where 1 slug = 32.174 lb):

[tex]\[ m \text{ (in slugs)} = \frac{0.02857 \text{ lb}}{32.174 \text{ lb/slug}}\\ = 0.000888 \text{ slugs} \][/tex]

[tex]\[ \text{Impulse} = m \Delta v = 0.000888 \text{ slugs} \times 3300 \text{ ft/s} \\= 2.9324 \text{ slug ft/s} \][/tex]

The impulse imparted to the bullet is:

[tex]\[ \text{Impulse} = 2.9324 \text{ lb s} \][/tex]

Calculate the Average Force

The average force can be calculated using the formula:

[tex]\[ F_{avg} = \frac{\Delta p}{\Delta t} \][/tex]

Given:

Time interval, [tex]\( \Delta t = 0.0011 \text{ s} \).[/tex]

[tex]\[ F_{avg} = \frac{2.9324 \text{ lb s}}{0.0011 \text{ s}} \\= 2665.82 \text{ lb} \][/tex]

Answer:

Explanation:

Given f(x, y) = 5x + y + 2 and g(x, y) = xy = 1

The step by step calculation and appropriate substitution is clearly shown in the attached file.

The local extreme values for the given function are;

minimum value is 2 - (2√5) while the maximum value is 2 + (2√5)

We are given the functions;

f(x, y) = 5x + y + 2 and g(x, y) = xy = 1

The general formula for lagrange multiplier is;

L(x, λ) = f(x) - λg(x)

From lagrange multipliers, we know that;

∇f = λ∇g  ----(1)

Since g(x, y) = xy = 1, then;

f_x = λg_x   -----(2)

f_y = λg_y   -----(3)

From eq(2), we have;

λ = 5/y    ------(4)

From eq 3, we have;

λ = 1/x    -----(5)

Combining eq 4 and 5 gives us;

5x = y

Put 5x for y into xy = 1 to get;

5x² = 1 and so;

x = ±1/√5

Put ±1/√5 for x in xy = 1 to get;

y = ±√5

Thus, f has extreme values at;

(1/√5, √5), (-1/√5, -√5), (1/√5, -√5), (-1/√5, √5)

At (1/√5, √5), f(x, y) becomes 2 + (2√5)

At (-1/√5, -√5), f(x, y) becomes 2 - (2√5)

At (1/√5, -√5), f(x, y) becomes 2

At (-1/√5, √5), f(x, y) becomes 2

Thus, in conclusion we can say that;

The minimum value is 2 - (2√5) at the point (-1/√5, -√5) while the maximum value is 2 + (2√5) at (1/√5, √5)

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Answer and Explanation:

• 1 thread awaits the incoming request

• 1 thread responds to the request

• 1 thread reads the hard disk

A multithreaded file server is better than a single-threaded server and a finite-state machine server because it provides better response compared to the rest and can make use of the shared Web data.

Yes, there are circumstances in which a single-threaded server might be better. If it is designed such that:

- the server is completely CPU bound, such that multiple threads isn't needed. But it would account for some complexity that aren't needed.

An example is, the assistance number of a telephone directory (e.g 7771414) for an community of say, one million people. Consider that each name and telephone number record is sixty-four characters, the whole database takes 64 MB, and can be easily stored in the server's memory in order to provide quick lookup.

NOTE:

Multiple threads lead to operation slow down and no support for Kernel threads.

Explanation:

Δ[tex]L_{BC}[/tex] = Δ[tex]L_{AB}[/tex]

[tex]\frac{(Q - 4000)(0.5)}{3.14* 0.03 *0.03 *70*10^{9} }[/tex]     (1)

= [tex]\frac{4000*0.4}{3.14*0.01*0.01*70*10^{9} }[/tex]

Q = 32,800 N

now put this value in equation 1.

Deflection of B = [tex]\frac{(32800-4000)(0.5)}{3.14*0.03*0.03*70*10^{9} }[/tex]

                       = 0.0728 mm

Answer: 0.98kJ/kg

Explanation: for a kilogram mass of this system, raising the system at a height h gives it a potential energy of the magnitude gh,

Where g is gravitational acceleration,

and h is the height difference.

This system will have an energy change of

PE = 100*9.8 = 980 J/kg

This becomes 980/1000 kJ/kg

= 0.98kJ/kg

The rubber-tired Herrywampus machine would need to make 288 trips to backfill the space with the given geometrical volume, considering both volume and weight limitations.

The Breakdown

we need to consider the volume and weight limitations of the rubber-tired Herrywampus machine.

Geometrical volume of the space to be backfilled: 5400 cubic yards

Maximum capacity of the machine: 30 cubic yards (heaped) or 40 tons

Compaction shrinkage: 25% (relative to bank measure)

Swell factor: 20% (relative to bank measure)

- Material weight: 3,000 lb/cu yd (bank)

Calculate the actual volume of material required to backfill the space.

Actual volume = Geometrical volume / (1 - Compaction shrinkage)

Actual volume = 5400 cubic yards / (1 - 0.25)

Actual volume = 7200 cubic yards

Calculate the volume of material to be loaded into the machine, considering the swell factor.

Swelled volume = Actual volume × (1 + Swell factor)

Swelled volume = 7200 cubic yards × (1 + 0.20)

Swelled volume = 8640 cubic yards

Calculate the number of trips required based on the machine's volume capacity.

Number of trips = Swelled volume / Machine capacity

Number of trips = 8640 cubic yards / 30 cubic yards

Number of trips = 288 trips

Check the weight limitation.

Weight of material per trip = Machine capacity × Material weight

Weight of material per trip = 30 cubic yards × 3,000 lb/cu yd

Weight of material per trip = 90,000 lb

Total weight of material = Swelled volume × Material weight

Total weight of material = 8640 cubic yards × 3,000 lb/cu yd

Total weight of material = 25,920,000 lb

Number of trips based on weight limitation = Total weight of material / Weight of material per trip

Number of trips based on weight limitation = 25,920,000 lb / 90,000 lb

Number of trips based on weight limitation = 288 trips

Therefore, the rubber-tired Herrywampus machine would need to make 288 trips to backfill the space with the given geometrical volume, considering both volume and weight limitations.

Answer and Explanation:

The answer is attached below

Answer:

The solution is given in the attachments

Force P is 11304 N and normal stress is 400 N/mm²

Explanation:

Given-

Length, l = 9 m = 9000 mm

Diameter, d = 6 mm

Radius, r = 3 mm

Stretched length, Δl= 18 mm

Modulus of elasticity, E = 200 GPa = 200 X 10³MPa

Force, P = ?

According to Hooke's law,

Stress is directly proportional to strain.

So,

σ ∝ ε

σ = E ε

Where, E is the modulus of elasticity

We know,

ε = Δl / l

So,

σ = E X Δl/l

σ =

[tex]200 X 10^3 * \frac{18}{9000} \\\\ = 400N/mm^2[/tex]

We know,

σ = P/A

And A = π (r)²

σ = P / π (r)²

[tex]400 N/mm^2 = \frac{P}{3.14 X (3)^2} \\\\400 = \frac{P}{28.26} \\\\P = 11304N[/tex]

Therefore, Force P is 11304 N and normal stress is 400 N/mm²

The magnitude of the force is 11.3KN and the normal stress is 400 MPa

Given that length (L) = 9 m, diameter (d) = 6 mm = 6 * 10⁻³ m, extension (δ) = 18 mm = 18 * 10⁻³ m, E = 200 GPa = 200 * 10⁹ Pa

The area of the wire (A) is:

[tex]A=\pi*\frac{diameter^2}{4}=\pi*\frac{(6*10^{-4})^2}{4} =28*10^{-6}\ m^2[/tex]

[tex]\delta=\frac{PL}{AE} \\\\P=\frac{AE\delta}{L}=\frac{28*10^{-6}*200*10^9*18*10^{-3}}{9}=11300N\\\\\\Normal\ stress(\sigma)=\frac{P}{A} =\frac{11300}{28*10^{-6}} =400*10^6\ Pa[/tex]

The magnitude of the force is 11.3KN and the normal stress is 400 MPa

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Answer:

Explanation:

Find attached the figure to solve this problem (taken from a problem, in the internet, with the same statement, but different mass for the blok).

The block will stop when all its kinetic energy is absorbed by the friction and the spring.

1. Initial kinetic energy of the blockm [tex]KE_i[/tex]

   [tex]KE_i=\dfrac{1}{2}mass\times (speed)^2\\\\\\KE_i=\dfrac{1}{2}(5.6kg)\times (5m/s)^2=70J[/tex]

2. Work of friction

The friction force is the product of the normal force by the coefficient of kinetic friction ,  [tex]\mu_k=0.25[/tex] .

Since, the only vertical force is the force of gravity, the normal force, [tex]F_N[/tex] , is the weight of the block:

       [tex]F_N=5.6kg\times 9.8m/s^2=54.88N[/tex]

Then, the friction force,   [tex]F_f[/tex]  , is:

      [tex]F_f=0.25\times 54.88N=13.72N[/tex]

The distance run by the block before stopping is the 2 meters distance plus the amount the spring compresses. Calling x the distance the spring compresses, the friction work is:

      [tex]W_f=13.72N\times (2+x)[/tex]

3. Energy absorbed by the spring

The energy absorbed by the spring is the elastic potential energy, PE, which is given by the formula:

      [tex]PE=\dfrac{1}{2}kx^2[/tex]

Where k is the elasticity constant of the spring (200B/m, according to the figure), and x is the distance the spring compresses.

Substituting:

          [tex]PE=\dfrac{1}{2}\times 200N/m\times x^2\\\\\\PE=100N/m\cdot x^2[/tex]

4. Final equation

Now you can write your equation to find the compression of the spring, x:

        [tex]70=13.72(2+x)+100x^2[/tex]

Solving:

           [tex]70=27.44+13.72x+100x^2\\\\100x^2+13.72x-42.56[/tex]

Use the quadratic formula:

        [tex]x=\dfrac{-13.72\pm \sqrt{(13.72)^2-4(100)(-42.56)}}{2(100)}[/tex]

There is one negative solution, which you discard, and the positive solution is 0.59.

The initial kinetic energy of a moving block gets converted into potential energy along with overcoming friction. By setting kinetic energy equal to the work done against friction plus potential energy in the compressed spring, we can rearrange the equation to solve for the compression in the spring when the block momentarily stops.

To find the compression in the spring when the block momentarily stops, we use the principles of energy conservation. The initial kinetic energy of the block is converted into potential energy in the spring while overcoming the frictional forces. We can write this relationship as follows:

(1/2)m(v^2) = μkmgd + (1/2)kd^2

where m is the mass of the block, v is the initial speed of the block, μk is the coefficient of kinetic friction, g is the gravitational acceleration, d is compression in the spring, and k is the spring constant. Since we need to find d (the compression of the spring when the block stops), you will need to isolate d in the equation.

Remember that frictional force does work against the motion of the block and potential energy stored in the spring is the block’s initial kinetic energy minus the work done by friction.

Assuming we know the spring constant k, we can solve the equation to find d, the compression in the spring when the block momentarily stops.

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Vector addition is used to calculate the speeds of the swimmer and the water in the river, understanding relative velocity is key in solving the problem.

To determine the speed of the water in the river, we first calculate the resultant velocity of the swimmer using vector addition. The swimmer's speed with respect to a friend at rest on the ground is found by considering the swimmer's velocity and the water current's velocity. By understanding the concepts of relative velocity and vector addition, we can accurately calculate the required speeds.

Answer:

Radius = 1565ft  ;  PC = 245 + 11.13  ; PT = 255 + 48.88

Explanation:

1. Accordingly to the law of mechanics;

Centrifugal factor =S+F = V²/15R

Where; S = Super-elevation slope = 0.07ft/ft

F= Slide friction factor

V= Design speed=65mi/h

R= Radius

From the graph (see attached), at design speed of 65mi/h, coefficient of slide friction factor, F =0.11

Applying the figures in the equation above;

0.07+0.11 = 65²/15R

R=281.67/0.18

R=1564.8

Approximately Radius = 1565ft

2. Stationing PC = Stationing PI - T

where T = Tangent distance

T= Rtan(Δ/2) where Δ = central angle = 38° & Stationing PI = 250 + 50

T=1565tan19°

T=538.87ft

Stationing PC = 250 + 50 - (5 + 38.87)

PC = 245 + 11.13

3. Stationing PT = Stationing PC + L

Where L = Length of the circular curve

L = π/180*(RΔ)

L=0.01745*1565*38

L=1037.75

Therefore;

Stationing PT= 245 + 11.13 + (10 +37.75)

PT = 255 + 48.88

In real-time applications where tasks can be completed faster than necessary, energy savings can be achieved either by running at full speed and then turning the system off, or more effectively, by reducing the voltage and frequency by half, which reduces power consumption and enhances energy efficiency without affecting task completion.

When designing a system for a real-time application where deadlines must be met, and finding that your system can execute the necessary code twice as fast as necessary, the energy savings can be approached in two ways. First, executing at the current speed and then turning off the system can save energy because energy consumed is the product of power and time. Second, reducing the voltage and frequency to half can also save energy. Lowering the clock frequency reduces power consumption because dynamic power consumption in digital electronic circuits is directly proportional to the product of the capacitance being switched per cycle, the square of the voltage, and the frequency of operation. Therefore, halving the voltage and frequency not only reduces power consumption but, since the task can be completed within the deadline even at reduced speed, energy efficiency can be significantly enhanced without compromising performance.

Overall, the goal is to minimize energy consumption by either completing tasks faster and shutting down or, more effectively, by operating the system more efficiently at lower power levels. These strategies align with broader energy conservation principles, highlighting the importance of designing systems that require minimal energy to meet their performance requirements.