The 1000-lb elevator is hoisted by the pulley system and motor M. The motor exerts a constant force of 500 lb on the cable. The motor has an efficiency of ε = 0.65. Determine the power that must be supplied to the motor at the instant the load has been hoisted s = 27 ft starting from rest.

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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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:

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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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:

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

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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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  

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:

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

Kinetic energy lost in collision is 1097.95 J

Explanation:

Given,

Mass,  [tex]m_{1}[/tex]= 60 g = 0.006 kg

Speed, [tex]v_{1}[/tex] = 605 m/s

[tex]m_{2}[/tex] = 54 kg

[tex]v_{2}[/tex]= 0

Kinetic energy lost, K×E = ?

During collision, momentum is conserved.

So,

[tex]m1v1 + m2v2 = (m1 + m2)v\\\\0.006 X 605 + 54 X 0 = (0.006 + 54) v\\\\v = \frac{3.63}{54.006}\\ \\v = 0.067m/s[/tex]

Before collision, the kinetic energy is

[tex]\frac{1}{2}* m1 * (v1)^2 + \frac{1}{2} * m2 * (v2)^2[/tex]

[tex]=\frac{1}{2} X 0.006 X (605)^2 + 0\\\\= 1098.075J[/tex]

Therefore, kinetic energy before collision is 1098 J

Kinetic energy after collision:

[tex]\frac{1}{2}* (m1+m2) * (v)^2 + KE(lost)[/tex]

By plugging in the values, we get

[tex]\frac{1}{2} * (0.006 + 54) * (0.067)^2 + KE(lost)[/tex]

[tex]0.1212J + KE(lost)[/tex]

Since,

initial Kinetic energy = Final kinetic energy

1098.075 J = 0.1212 J + K×E(lost)

K×E(lost) = 1098.075 J - 0.121 J

K×E(lost) = 1097.95 J

Therefore, kinetic energy lost in collision is 1097.95 J

Answer:

(a) maximum positive reaction at A = 64.0 k

(b) maximum positive shear at A = 32.0 k

(c) maximum negative moment at C = -540 k·ft

Explanation:

Given;

dead load  Gk = 400 lb/ft

live load Qk = 2 k/ft

concentrated live load Pk =8 k

(a) from the influence line for vertical reaction at A, the maximum positive reaction is

[tex]A_{ymax}[/tex] = 2*(8) +(1/2(20 - 0)* (2))*(2 + 0.4) = 64 k

See attachment for the calculations of (b) & (c) including the influence line

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.

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

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:

note:

please find the attached code

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]

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 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:

The solution is given in the attachments

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.

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:

brainly.com/question/25114255

Answer:

1. Alkenes Can Be Nucleophiles! But How Do We Draw The Curved Electron-Pushing Arrows?

2. The Conventional Approach For Drawing Electron-Pushing Curved.

3. Arrows In Alkene Addition Reactions Is Slightly Ambiguous Modified Electron-Pushing Arrow Convention #1: “Bouncy” Arrows.

4. Modified Curved Arrow Convention #2: “Pre-bonds”.

Explanation:

1. Alkenes Can Be Nucleophiles! But How Do We Draw The Curved Electron-Pushing Arrows?

Alkenes are a lot more exciting than they’re often given credit for. That means that given a sufficiently frisky electrophile, they can donate their pair of π electrons to form a new sigma bond.

Like this!

However, there’s one little problem here. See that curved arrow? What does it really mean? If you weren’t given the product, would you be able to draw it, given that curved arrow?

See the problem here: Which atom of the alkene is actually forming the bond to hydrogen? When we were dealing with lone pairs, it was easy: atoms clearly “own” their lone pairs, and we can tell exactly which atom is forming a bond to which. With alkenes, it’s different: since they “share” that pair of electrons, we’re going to have to somehow show which atom gets the new atom and which is left behind as a carbocation.

2. The Conventional Approach For Drawing Electron-Pushing Curved Arrows In Alkene Addition Reactions Is Slightly Ambiguous

Here’s the conventional way it’s done. If we want to show the bottom carbon forming the bond, the usual way to do this is to draw this loop like this, to show the “path” of the electrons coming in an arc from this direction. The carbon on the alkene “closest” to the hydrogen is the one that ends up bonded to it.

Similarly, if we wanted to show the left carbon forming the bond, we’d “arc” the bond like this:

One problem with this: it’s kind of a kludge. The curved arrow notation is limited in that all we can really do is decide where the tail should go (at the π bond, obviously) and where the head should go (to form the new bond). But the question of which carbon forms the bond is still ambiguous.

And if there’s one thing organic chemists hate, it’s ambiguity.

Give me clear definitions or give me death!

To try and deal with this issue, organic chemists have come up with two potential solutions. They’re worth looking at if you’re finding this issue confusing.

3. Modified Electron-Pushing Arrow Convention #1: “Bouncy” Arrows.

Instead of showing the curved arrow as a big sweeping arc, one solution is to put an extra bounce into the arrow. The idea here is that we’re showing the pair of electrons travelling to the carbon in question, and from there moving on to form the new sigma bond. No more ambiguity here. [Literature reference]

This solves the ambiguity problem at the expense of putting in an extra hump in the arrow. Although it doesn’t seem like a big deal, the extra bounce has likely been the reason why this convention hasn’t taken off. However well intentioned, the trouble with a convention like this is humanity’s natural tendency towards laziness: taking the time to consistently draw an extra hump into the arrow – even if it takes only 5 seconds – represents extra work that is skipped unless absolutely necessary. Behavioral change is very difficult.

4. Modified Curved Arrow Convention #2: “Pre-bonds”.

Another way of dealing with this is to insert the equivalent of “training wheels” into our curved arrows. Since the curved arrow is itself ambiguous, to clarify things we put in a dashed line that precisely delineates where the new bond is forming. Then, we draw the arrow with the tail coming from the electron source (the π bond) and the head going to the new bond. We can put the arrow right on the dashed line itself. This has the advantage of not modifying the curved arrow convention itself, just adding in an optional “guide” that makes its application more clear. [For an application of this technique I recommend checking out Dr. Peter Wepplo’s blog, where I first found this convention used]

dotted line convention for alkene addition resolves ambiguity

If you find yourself confused following the movement of electrons in the reactions of alkenes with electrophiles, these supplementary conventions might be of use to you.

Personally, even though conventional curved arrows suffer from a bit of ambiguity, that’s generally not enough to make me stop using them. YMMV.

In the next post we’ll resume our regularly scheduled program on alkenes and carbocations.

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 and Explanation:

The answer is attached below

Answer:

106600 btu/s

note:

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

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)

Read more about Lagrange Multipliers at; https://brainly.com/question/4609414

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:

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.