A single axial load of magnitude P = 15 kips is applied at end C of thesteel rod ABC. Knowing that E = 30 × 106 psi, determine the diameterd of portion BC for which the deflection of point C will be 0.05 in.

Answer:

11.65 minutes.

Explanation:

See the attached picture for detailed explanation.

Answer:

11.65 min

Explanation:

Hydrogen sulfide is very poisonous and as a result, it is essential to reduce the concentration of the gas to the lowest possible value to minimize its effects. The time taken to reduce the amount of hydrogen sulfide in the system to the allowable limit can be estimated as shown below:

t = (V/Q)*ln(Ci/Co) = (160/10)*ln(14/29) = 16*0.728 = 11.65  min

Answer:

m' = 4948.38 kg/s

Explanation:

For a case of 100% efficiency, the power produced must be equal to the rate of potential energy conversion

GIVEN THAT

Power = 100 MW

rate of Potential energy = (m')*g*h

100*10^6 = (m')*9.81*206

m' = 4948.38 kg/s

Answer:

49.484 m³ / s

Explanation:

Volume flow rate = Power in W / (efficiency × density × height × acceleration due to gravity)

Volume flow rate = 100 × 10⁶ / ( 1 × 1000 kg/m³ × 206 m × 9.81 m/s²)

V = 49.484 m³ / s

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:

106600 btu/s

note:

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

Answer:

1.757 kg/s

Explanation:

According to the First Law of Thermodynamics, the physical model for a turbine working at steady state is:

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

The flow rate of steam is:

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

Water enters and exits as superheated steam. After looking for useful data in a property table for superheated steam, specific enthalpies at inlet and outlet are presented below:

[tex]h_{in} = 3451.7 \frac{kJ}{kg} \\h_{out} = 2967.9 \frac{kJ}{kg} \\[/tex]

Finally, the flow rate is calculated:

[tex]\dot m = \frac{100 kW + 750 kW}{3451.7 \frac{kJ}{kg} - 2967.9 \frac{kJ}{kg}}\\\dot m =1.757 \frac{kg}{s}[/tex]

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:

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]

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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]

Answer: Let us use the pickled file - DeckOfCardsList.dat.

Explanation: So that our possible outcome becomes

7♥, A♦, Q♠, 4♣, 8♠, 8♥, K♠, 2♦, 10♦, 9♦, K♥, Q♦, Q♣

HPC (High Point Count) = 16  

To calculate the required cross-sectional area of copper wires needed to connect the DC supply source to the mine lights with less than 5% power loss, calculate the permissible power loss, and then use the relationship between power loss, current, and wire resistance. Considering the resistivity of copper and the round trip length of the wire, the formula A = ρL/(δP/I²) determines the necessary wire thickness.

To determine the required cross-sectional area of copper wires for the lighting equipment, we need to ensure the power loss (δP) is no more than 5% of the lights' power usage (P), which is 5 kW (5000 W). As the power loss in wires is given by δP = I²R, where I is the current and R is the resistance of the wire, we must first calculate the current using P = IV, giving I = P/V = 5000W/120V = 41.67 A. The power loss allowed is thus 5% of 5000 W, equating to 250 W. Given δP, we can find R using δP = I²R, which gives R = δP/I².

The resistance of a copper wire is also given by R = ρL/A, where ρ is the resistivity of copper (1.68 x 10^-8 Ωm), L is the length of the wire (200 meters round trip), and A is the cross-sectional area we need to find. Equating the two expressions for R and solving for A gives A = ρL/(δP/I²). Substituting the given and calculated values yields the required cross-sectional area. Finally, as resistance depends on the entire length of the circuit, remember to double the distance to account for both the outgoing and return paths.

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:

note:

please find the attached code

Answer:

note:

solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment

The voltage of the source can be determined by dividing the total energy transferred (sum of the paddle-wheel work and heat required to evaporate water) by the product of current and time. The process will be seen as a horizontal line on a piston-cylinder diagram

The heat transferred to the water can be calculated from the energy supplied by the electrical source and the work done by the paddle wheel. Since power is the energy transferred per unit time, we can use the equation Power = Voltage x Current. We know that the current is 8 A and the energy transferred is the sum of the work done by the paddle wheel and the heat required to evaporate half of the water, which is 45 minutes (converted to seconds) times the power. Solving for voltage we get:

Voltage = (Energy transferred) / (Current x Time)

In terms of a piston-cylinder diagram, the process will appear as a horizontal line since the pressure is constant. The line will move upwards as heat is added and some of the water is turned into steam.

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Answer:

The answer to this question is attached fully with the explanation.

Answer:

Please find attached file for complete answer solution and explanation of same question.

Explanation:

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:

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

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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:

[tex]W= 3.22 \mu m[/tex]

Explanation:

the transistor In saturation drain current region is given by:

[tex]i_D}=K_a(V_{GS}-V_{IN})^2[/tex]

Making [tex]K_a[/tex] the subject of the formula; we have:

[tex]K_a=\frac {i_D} {(V_{GS} - V_{IN})^2}[/tex]

where;

[tex]i_D = 1.2m[/tex]

[tex]V_{GS}= 3.0V[/tex]

[tex]V_{TN} = 0.6 V[/tex]

[tex]K_a=\frac {1.2m} {(3.0 - 0.6)^2}[/tex]

[tex]K_a = 208.3 \mu A/V^2[/tex]

Also;

[tex]k'_n}=\frac{\mu n (\frac{cm^2}{V-s} ) \epsilon _{ox}(\frac{F}{cm} ) }{t_{ox}(cm)}[/tex]

where:

[tex]\mu n (\frac{cm^2}{V-s} ) = 600[/tex]

[tex]\epsilon _{ox}=3.9*8.85*10^{-14}[/tex]

[tex]{t_{ox}(cm)=200*10^{-8}[/tex]

substituting our values; we have:

[tex]k'_n}=\frac{(600)(3.988.85*10^{-14})}{(200*10^{-8})}[/tex]

[tex]k'_n}=103.545 \mu A/V^2[/tex]

Finally, the width can be calculated by using the formula:

[tex]W= \frac{2LK_n}{k'n}[/tex]

where;

L = [tex]0.8 \mu m[/tex]

[tex]W= \frac{2*0.8 \mu m *208.3 \mu}{103.545 \mu}[/tex]

[tex]W= 3.22 \mu m[/tex]

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

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:

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.

Answer:

The purpose of the algorithm is to print the least digit among a total of 15 digita

Explanation:

input somenum

Repeat the following steps for 14 times

input variable1

if variable1 < somenum then

somenum = variable1

print somenum

On line 1, the algorithm takes an input through variable1

An iteration is started on line 2 and ends on line 6

Line 3,4,5 re performed repeatedly;

On line 3, the algorithm accepts another input through somenum and it keep accepting it till the end of the iteration.

On line 4, the algorithm tests if variable1 is lesser than somenum.

If yes, line 5 is executed and the value of variable1 is assigned to somenum

Else, line 5 is skipped; the iteration moves to line 3 as long as the condition is still valid.

At the end of the iteration, the least value stored in somenum is printed through

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.

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.

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:

The solution is given in the attachments

Answer:

Please find attached file for complete answer.

Explanation:

Answer:

59.78 m

Explanation:

Data:

PI station = 250 will allow a speed of 65 mph

The maximum superelevation will be  = 0.08 ft/ft

Central angle of the curve                    =  35 degrees

A figure will be used to present the information. This gives the height of 59.78 m as the maximum elevation for the safe speed of the vehicle.

The length of Bridge B is expressed as m - 5555 feet, where m represents the length of Bridge A. For example, if Bridge A is 10,000 feet long, then Bridge B would be 4445 feet long.

Bridge B is 5555 feet shorter than Bridge A, so we can represent the length of Bridge B as m - 5555 feet. This expression represents the length of Bridge B in terms of m, where m represents the length of Bridge A. For example, if Bridge A is 10,000 feet long, then Bridge B would be 10,000 - 5555 = 4445 feet long.

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