An AISI 1040 cold-drawn steel tube has an OD 5 50 mm and wall thickness 6 mm. Whatmaximum external pressure can this tube withstand if the largest principal normal stress is notto exceed 80 percent of the minimum yield strength of the material?

Answer:

See the attached java files for code.

Explanation:

See the attached files.

Missing Part of Question

Complete the following tables by entering productivity (in terms of output per worker) for each economy in 2012 and 2042.

Hermes

Year (2002)

Physical Capital: 11 tools per worker

Labour Force: 30 workers

Output: 1,800 Gobs of goo

Productivity:__________

Year (2042)

Physical Capital: 16 tools per worker

Labour Force: 30 workers

Output: 2,160 Gobs of goo

Productivity:__________

Tralfamadore

Year (2002)

Physical Capital: 8 tools per worker

Labour Force: 30 workers

Output: 900 Gobs of goo

Productivity:__________

Year (2042)

Physical Capital: 13 tools per worker

Labour Force: 30 workers

Output: 1,620 Gobs of goo

Productivity:__________

Answer:

The Complete Table is as follows

Hermes

Year (2002)

Physical Capital: 11 tools per worker

Labour Force: 30 workers

Output: 1,800 Gobs of goo

Productivity:60 workers

Year (2042)

Physical Capital: 16 tools per worker

Labour Force: 30 workers

Output: 2,160 Gobs of goo

Productivity:72 workers

Tralfamadore

Year (2002)

Physical Capital: 8 tools per worker

Labour Force: 30 workers

Output: 900 Gobs of goo

Productivity:30 workers

Year (2042)

Physical Capital: 13 tools per worker

Labour Force: 30 workers

Output: 1,620 Gobs of goo

Productivity:54 workers

Explanation:

We calculate the productivity of both Hermes and Tralfamadore by dividing Output by Labor Force. This is given by:

Productivity = Output/Labour Force

For Hermes

In 2002

Productivity = 1800/30 = 60 workers

In 2042,

Productivity = 2160/30 = 72 workers

For Tralfamadore

In 2002

Productivity = 900/30 = 30 workers

In 2042,

Productivity = 1620/30 = 54 workers

Answer:

It is going to take about 22.43 minutes

Explanation:

The Coefficient of Performance of a refrigerator is the ratio of useful cooling to work required to achieve it. The formula would be:

COP = Cooling Effect/Power Input

The COP is given as 2.8 and the power rating is 400 Watts. We can find the cooling effect as follows:

2.8 = Cooling Effect/400

Cooling Effect = 2.8 * 400 = 1120 Watts

Now,

The cooling effect can be done using the formula:

[tex]Q=mc \Delta T[/tex]

Where

Q is the thermal energy (or work)

m is the mass

c is the specific heat capacity

[tex]\Delta T[/tex] is the temperature change

We know

m = 12 kg

c is the specific heat of water, 4187 J/kg*C

[tex]\Delta T[/tex] is 30, from 40C to 10C

Substituting, we get:

[tex]Q=mc \Delta T\\Q=12*4187*30\\Q=1507320[/tex]

This thermal energy is the Work, which is Power * Time required

Thus, we can say:

Time Required = Q/Power

So,

Time Required = 1507320/1120 = 1346 Seconds

To minutes, we can say:

1346/60 = 22.43 minutes

Answer a) roof equilibrium temperature is 400K

Explanation:

The roof surface is sorrounded by ambient air whose temperature is 300K and which has a convective coefficient h of 10W/m2-K.

This means that heat will be conducted to the roof from the air around at an heat Flux of 300K x 10W/m2-K = 3000w/m2

For a clear solar day with solar heat Flux of 500W/m2, total heat Flux on roof will be

Q = 500 + 3500 = 3500W/m2

Q = h(Tr-Ta)

Ts is temperature of roof,

Ta is temperature of air

3500 = 10(Tr - 50)

350 = Tr - 50

Tr = 400k

Answer b): roof equilibrium temperature will be 350K

Explanation:

At night total heat Flux is only due to hot ambient air sorrounding the surface of the roof.

Q = 3000W/m2

Q = h(Tr-Ta)

3000 = 10(Tr-50)

300 = Tr - 50

Tr = 350K

Answer:

34.06 W.

Explanation:

Assumptions : ideal turbine " no loss of work", no pipe or friction losses, all the available energy of water in converted to useful power.

(A) total volume of water per day in cubic meters:

1 cubic meters=1000 L

average flow from each apartment*total apartments/1000= (100/1000)*100

                                                                                                =10 m^3

total mass of water m= density*volume=1000*10=10000 kg

total energy of water at 30 m height= m*g*h= 10000*9.81*30=2943000 J

if all the available energy is converted to power.

power produced per day=total energy of water / time

time in seconds=24*3600=86400 s

power produced in a day=2943000/86400= 34.06 W

Answer:

34.06 W

Explanation:

Assuming ideal conditions which are assuming no fiction or pipe loss is made along the line of extraction of water

Energy of water at ideal condition ( Eₐ ) = 1000 kg/m³

height given = 30 meters

quantity of water = 100 Liters

to calculate the quantity of water in M³/s ( cubic per second )

= [tex]\frac{100*10^{-3} }{24*3600}[/tex] = 1.157 * [tex]10^{-6}[/tex]

power produced by water ( Pw) = energy of water * quantity of water in m^3/s

  = 1.157 * [tex]10^{-6}[/tex]  * [tex]10^{3}[/tex] = 1.157 *[tex]10^{-3}[/tex]

since it is an ideal condition all the power produced by water is converted to power produced by the contraception

Pc = ( Pw * g * h ) * n

H = height = 30

g = 9.81

n = number of apartments = 100

Pc =( 1.157 * [tex]10^{-3}[/tex]  * 9.81 * 30) *100   = 34.06 W

Answer:

the required diameter of the rod is 9.77 mm

Explanation:

Given:

Length = 1.5 m

Tension(P) = 3 kN = 3 × 10³ N

Maximum allowable stress(S) = 40 MPa = 40 × 10⁶ Pa

E = 70 GPa = 70 × 10⁹ Pa

δ = 1 mm = 1 × 10⁻³ m

The required diameter(d)  = ?

a) for stress

The stress equation is given by:

[tex]S = \frac{P}{A}[/tex]

A is the area = πd²/4 = (3.14 × d²)/4

[tex]S = \frac{P}{(\frac{3.14*d^{2} }{4}) }[/tex]

[tex]S = \frac{4P}{{3.14*d^{2} } }[/tex]

[tex]3.14*S*{d^{2}} = {4P}[/tex]

[tex]{d^{2}} =\frac{4P}{3.14*S}[/tex]

[tex]d= \sqrt{\frac{4P}{3.14*S} }[/tex]

Substituting the values, we get

[tex]d= \sqrt{\frac{4*3*10^{3} }{3.14*40*10^{6} } }[/tex]

[tex]d= \sqrt{\frac{12000 }{125600000 } }[/tex]

[tex]d= \sqrt{9.55*10^{-5} }[/tex]

d = (9.77 × 10⁻³) m

d = 9.77 mm

b) for deformation

δ = (P×L) / (A×E)

A = (P×L) / (E×δ) = (3000 × 1.5) / (1 × 10⁻³ × 70 × 10⁹) = 0.000063

d² = (4 × A) / π = (0.000063 × 4) / 3.14

d² = 0.0000819

d = 9.05 × 10⁻³ m = 9.05 mm

We use the larger value of diameter = 9.77 mm

Answer:

the difference in oil levels is 0.850 m

Explanation:

given data

specific gravity ρ = 0.72

pressure across P = 6 kPa = 6000 Pa

solution

we get here difference in oil levels h is

P = ρ × g × h   .................1

here ρ = 0.72 × 1000 = 720 kg/m³

and g is 9.8  

put here value in equation 1  and we get h

6000 = 720  × 9.8 × h

h = [tex]\frac{6000}{720\times 9.8}[/tex]  

h = 0.850 m

so the difference in oil levels is 0.850 m

Using the given specific gravity of oil and the differential pressure, the difference in oil levels in the manometer is calculated as 0.84 meters.

The difference in oil levels in a manometer when the differential pressure is 6kPa, and the specific gravity of the oil is 0.72, can be calculated using the formula: Δh = ΔP/(ρ*g), where Δh is the height difference, ΔP is the differential pressure, ρ is the density of the oil, and g is the acceleration due to gravity. This calculation assumes standard gravity (g = 9.81 m/s^2). The density of oil can be obtained from its specific gravity and the density of water (1000 kg/m^3). Thus:

ρ = 0.72 * 1000 kg/m^3 = 720 kg/m^3

Substituting the values into the formula, we get:

Δh = 6000 Pascals / (720 kg/m^3*9.81 m/s^2) = 0.84 meters

Hence, the difference in the oil levels in the manometer is approximately 0.84 meters.

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Answer:

the change in length of the elastic rod is 0.147 mm and the vertical displacement of the point b is 0.1911 mm.

Explanation:

As the complete question is not visible , the complete question along with the diagram is found online and is attached herewith.

Part a

Take moments at point A to calculate the tensile force on the steel

rod at point C

[tex]\sum M_A=0\\T_c*2.5-P(2.5+0.75)-q*2.5*2.5/2=0\\T_c*2.5-10(3.25)-5*3.125=0\\T_c=19.25 kN\\[/tex]

Calculate the change in length of the elastic rod

[tex]\Delta l=\dfrac{T_c*L}{A*E}\\\Delta l=\dfrac{19.25\times 10^3*0.75}{\pi/4\times 0.025^2*200\times 10^9}\\\Delta l=0.000147 m \approx 0.147 mm[/tex]

So the change in length of the elastic rod is 0.147 mm.

As by the relation of the similar angles the vertical displacement of the point B is given as

[tex]\Delta B=\Delta l\dfrac{AC+BC}{AC}[/tex]

Here AC=2.5 m

BC=0.75 m

Δl=0.147 mm so the value is as

[tex]\Delta B=\Delta l\dfrac{AC+BC}{AC}\\\Delta B=0.147 mm\dfrac{2.5+0.75}{2.5}\\\Delta B=0.147 mm\dfrac{3.25}{2.5}\\\Delta B=0.1911 mm[/tex]

So the vertical displacement of the point b is 0.1911 mm.

Answer:

I=0.3636

Explanation:

See the attached picture for explanation.

Answer:

D. Charge is moved along an equipotential line

Explanation:

This means that the potential will be the same sown each equipotential line, which implies that the charge does not require any work before it moves anywhere along the line. Although work is required for a charge to move from an equipotential line to another, in all situations equipotential lines are vertical to electric field lines.

Answer:

Complete question for your reference is below.

A milling operation was used to remove a portion of a solid bar of square cross section. Forces of magnitude P = 19 kN are applied at the centers of the ends of the bar. Knowing that a = 28 mm and sigma_all = 130 MPa, determine the smallest allowable depth d of the milled portion of the bar

Explanation:

Please find attached file for complete answer solution and explanation.

To find the smallest allowable depth 'd', we use the stress formula and the data provided in the question. By substituting these values into the formula, we can solve for 'd'.

The question asks for the smallest allowable depth 'd' of the milled portion in an Engineering situation involving a solid square bar, forces, and a given stress limit. First, we need to realize this is a problem of Stress Analysis. According to the formula for stress in a bar under axial load, Stress (σ) = Force (P) / Area. The cross-sectional area of the reduced portion of the bar is now a*d (width*depth), hence σ = P / (a*d). Here, 'a' is 28 mm and 'P' is 19 kN. As per the problem, the maximum allowable stress (σ_all) is 130 MPa. By substitifying known values into formula, we get 130 = 19,000 / (28 * d), and solving this equation will give the minimum value for 'd'.

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Answer:

 Average power dissipated by the 700-Ω resistor=47W

Explanation:

average power dissipated by the 700-Ω resistor

[tex]\frac{1}{t} \int\limits^t_0 {P(t)} \, dx \\=\frac{1}{30} (\int\limits^18_0 {55} \, dx + \int\limits^30_18 {35} \, dx )\\\\=\frac{1}{30} (55*(18-0) + 35(30-18))\\\\=\frac{1410}{30}\\[/tex]

=47W

Answer:

[tex]P=47[/tex] [tex]W[/tex]

Explanation:

The average power dissipated in a resistor is given by

[tex]P=\frac{1}{T} \int\limits^T_0 {} \, p(t)dt[/tex]

Where T is the time taken by the sine wave to complete one cycle.

We have instantaneous power P(t) = 55 W for 0 < t < 18s and P(t) = 35 W for 18 < t < 30s

So the time period is T = 30 seconds

Now we will integrate both of the instantaneous powers

[tex]P=\frac{1}{30} (\int\limits^b_a {} \, 55dt + \int\limits^b_a {35} \, dt )[/tex]

[tex]P=\frac{1}{30} (55t +{35} t )[/tex]

[tex]P=\frac{1}{30} (55(18-0) +{35(35-18)} )[/tex]

[tex]P=\frac{1}{30} (55(18) +{35(12}))[/tex]

[tex]P=\frac{1}{30} (990 +420)[/tex]

[tex]P=\frac{1410}{30}[/tex]

[tex]P=47[/tex] [tex]W[/tex]

Answer:

50 μsec

Explanation:

See the attached pictures for detailed answer.

Answer: 67.392s

Explanation: detailed calculation is shown below

Answer:

Time(t) = 2592.91 seconds

Explanation:

From the question, we have;

ρ = 7800 kg/m

c =480 J/kg⋅K

k = 50 W/m⋅K

Ti = 300°C

Ts = 25°C

x = 25 mm or in meters; = 0.025 m

From formula, we know that;

(T(x,t) - T(s))/(T(i) - T(s)) = erf(x/(2√(αt))

Where erf is error function

And α = k/(ρc)

So, solving we have;

(50 - 25)/(300 - 25) = erf(x/(2√(αt))

erf(x/(2√(αt)) = 0.0909

From the error function table which i attached, 0.0909 will give us approximately 0.0806 upon interpolation.

Thus, (x/(2√(αt)) = 0.0806

Let's make "t" the subject of the formula;

x² = 0.0806²(2²)(αt)

t = x²/0.026α

Let's find α

From earlier, we saw that;

α = k/(ρc)

Thus; α = 50/(7800 x 480) = 1.335 x 10^(-5) m²/s

Also, from the question, x = 30mm or 0.03m

So, t = 0.03²/(0.026 x 1.335 x 10^(-5)) = 2592.91 seconds

Answer:

a) s = 42 m

b) V = 26 m/s

c) a = 11 m/s²

Explanation:

The velocity(V) = 2 - 4t + 5t^3/2 where t is in seconds and V is in m/s

a) To get the position, we have to integrate the velocity with respect to time.

We get the position(s) from the equation:

[tex]V=\frac{ds}{dt}[/tex]

[tex]ds=Vdt[/tex]

Integrating both sides,

[tex]s=\int\ {V} \, dt[/tex]

[tex]s=\int\ {(2-4t+5t^{\frac{3}{2} }) \, dt[/tex]

Integrating, we get

[tex]s=(2t-2t^{2} +2t^{\frac{5}{2} }+c)m[/tex]

But at t=0, s(0) = 2 m

Therefore,

[tex]s(0)=2(0)-2(0)^{2} +2(0)^{\frac{5}{2} }+c[/tex]  

[tex]2=0+0+0+c[/tex]

c = 2

Therefore

[tex]s=(2t-2t^{2} +\frac{10}{5}t^{\frac{5}{2} }+2)m[/tex]

When t = 4,

[tex]s=2(4)-2(4)^{2} +2(4)^{\frac{5}{2} }+2[/tex]

[tex]s=8-32+64+2[/tex]

s = 42 m

At t= 4s, the position s = 42m

b) The velocity equation is given by:

[tex]V=(2-4t+5t^{\frac{3}{2} })m/s[/tex]

At t = 4,

[tex]V=2-4(4)+5(4)^{\frac{3}{2} }[/tex]

V = 2 - 16 + 40 = 26 m/s

The velocity(V)at t = 4 s is 26 m/s

c) The acceleration(a) is given by:

[tex]a=\frac{dv}{dt}[/tex]

[tex]a=\frac{d}{dt}(2-4t+5t^{ \frac{3}{2}})[/tex]

Differentiating with respect to t,

[tex]a=-4+\frac{15}{2}t^{\frac{1}{2} }[/tex]

[tex]a=(-4+\frac{15}{2}t^{\frac{1}{2} })m/s^{2}[/tex]

At t = 4 s,

[tex]a=-4+\frac{15}{2}(4)^{\frac{1}{2} }[/tex]

[tex]a=-4+15[/tex]

a = 11 m/s²

The position s, velocity v, and acceleration a when t = 4s are 40m, 26m/s and 11m/s² respectively.

The velocity, v, of the particle at any time, t, is given by;

v = 2 - 4t + 5[tex]t^{\frac{3}{2} }[/tex]          ----------(i)

Analysis 1: To get the position, s, of the particle at any time, t, we integrate equation (i) with respect to t as follows;

s = ∫ v dt

Substitute the value of v into the above as follows;

s = ∫ (2 - 4t + 5[tex]t^{\frac{3}{2} }[/tex]) dt          

s = 2t - [tex]\frac{4t^2}{2}[/tex] + [tex]\frac{5t^{5/2}}{5/2}[/tex]

s = 2t -2t² + 2[tex]t^{\frac{5}{2} }[/tex] + c       [c is the constant of integration]            ------------(ii)

According to the question;

when t = 0, s = 2m

Substitute these values into equation (ii) as follows;

2 = 2(0) -2(0)² + 2[tex](0)^{\frac{5}{2} }[/tex] + c  

2 = 0 - 0 - 0 + c

c = 2

Substitute the value of c = 2 back into equation (ii) as follows;

s = 2t -2t² + 2[tex]t^{\frac{5}{2} }[/tex] + 2      --------------------------------(iii)

Analysis 2: To get the acceleration, a, of the particle at any time, t, we differentiate equation (i) with respect to t as follows;

a = [tex]\frac{dv}{dt}[/tex]

Substitute the value of v into the above as follows;

a = [tex]\frac{d(2 - 4t + 5t^{3/2})}{dt}[/tex]

a = -4 + [tex]\frac{15}{2}[/tex][tex]t^{1/2}[/tex]                 ----------------------------------(iv)

Now;

(a) When t = 4s, the position s, of the particle is calculated by substituting t=4 into equation (iii) as follows;

s = 2(4) -2(4)² + 2([tex]4^{\frac{5}{2} }[/tex]) + 2

s = 8 - 32 + 2(4)²°⁵

s = 8 - 32 + 2(32)

s = 8 - 32 + 64

s = 40m

(b) When t = 4s, the velocity v, of the particle is calculated by substituting t=4 into equation (i) as follows;

v = 2 - 4(4) + 5([tex]4^{\frac{3}{2} }[/tex])

v = 2 - 16 + 5(4)¹°⁵

v = 2 - 16 + 5(8)

v = 2 - 16 + 40

v = 26m/s

(c) When t = 4s, the acceleration a, of the particle is calculated by substituting t = 4 into equation (iv) as follows;

a = -4 + [tex]\frac{15}{2}[/tex]([tex]4^{1/2}[/tex])

a = -4 + [tex]\frac{15}{2}[/tex](2)

a = -4 + 15

a = 11m/s²

Therefore, the position s, velocity v, and acceleration a when t=4s are 40m, 26m/s and 11m/s² respectively.

Answer:

The maximum power dissipation of the zener diode 112mV.

Explanation:

The minimum zener current should be:

5 * Iza= 5 * 1=  5 mA.

Since the load current can be at maximum 15 mA, we should select R so that, IL= 15 mA.

A zener current of 5 mA is available, Thus the current should be 20 mA, which leads to,

R = [tex]\frac{15 - 5.6}{20 mA}[/tex] = 470 Ω.

Maximum power dissipated in the diode occours when, IL=0 is

Pmax = 20 * [tex]10^{3}[/tex] * 5.6 = 112mV.

The net work done during the Stirling cycle is 2.7 kJ.

In order to determine the net work done during the Stirling cycle, we first need to calculate the initial and final volumes of the air in the piston-cylinder assembly.

Given:

The compression ratio is given by:

r = V1/V2

where V2 is the final volume. Solving for V2, we get:

V2 = V1/r = 0.03/7.5 = 0.004 m3

Next, we can use the ideal gas law to relate the initial and final pressures and volumes:

P1V1/T1 = P2V2/T2

Since the process is isothermal, T1 = T2 = TH = 1200 K.

Solving for P2, we get:

P2 = (P1V1T2)/V2T1 = (100,000 x 0.03 x 1200)/(0.004 x 1200) = 100,000 Pa

Finally, we can calculate the net work done using the formula:

Wnet = (P1V1) - (P2V2)

Substituting the values, we get:

Wnet = (100,000 x 0.03) - (100,000 x 0.004) = 2700 J = 2.7 kJ

Answer:

A) The quality of the refrigerant at the evaporator inlet = 0.48

B) The refrigeration load = 5.39 kW

C) COP = 2.14

Explanation:

A) From the refrigerant R-144 table I attached,

At P=60kpa and interpolating at - 34°C,we obtain enthalpy;

h1 = 230.03 Kj/kg

Also at P= 1.2MPa which is 1200kpa and interpolating at 65°C,we obtain enthalpy ;

h2 = 295.16 Kj/Kg

Also at P= 1.2MPa which is 1200kpa and interpolating at 42°C,we obtain enthalpy ;

h3 = 111.23 Kj/Kg

h4 is equal to h3 and thus h4 = 111.23 Kj/kg

We want to find the refrigerant quality at the evaporation inlet which is state 4 and P= 60 Kpa.

Thus, from the table attached, we see that hf = 3.84 at that pressure and hg = 227.8

Now, to find the quality of the refrigerant, we'll use the formula,

x4 = (h4 - hg) /(hf - hg)

Where x4 is the quality of the refrigerant. Thus;

x4 = (111.23 - 3.84)/(227.8 - 3.84) = 0.48

B) The mass flow rate of the refrigerant can be determined by applying a 1st law energy balance across the

condenser. Thus, the water properties can be obtained by using a saturated liquid at the given temperatures;

So using the first table in the image i attached; interpolating at 18°C; hw1 = hf = 75.54 kJ/kg

Also interpolating at 26°C; hw2 = hf = 109.01 kJ/kg

Now;

(mr) (h2 − h3)= (mw) (hw2 − hw1)

mr is mass flow rate

Making mr the subject, we get;

mr = [(mw) (hw2 − hw1)] /(h2 − h3)

mr = [(0.25 kg/s)(109.01 − 75.54) kJ/kg

] /(295.13 − 111.37) kJ/kg

mr = 8.3675/183.76

mr = 0.0455 kg/s

Formula for refrigeration load is;

QL = mr(h1 − h4)

Thus,

QL = (0.0455 kg/s)(230.03 − 111.37) kJ/kg = 5.39 kW

C) The formula for specific work into the compressor is;

W(in) = [(h2 − h1)] − (Q(in

)/mr)

= (295.13 − 230.03) kJ/kg − (0.450kJ/s

/0.0455 kg/s)

= 65.10 Kj/kg - 9.89 Kj/kg

55.21 kJ/kg

Formula for COP is;

COP = qL

/W(in)

Thus; COP = (h1 − h4)/W(in

)

= [(230.03 − 111.37) kJ/kg

] /55.24 kJ/kg

= 2.14

Answer:

#include <iostream>

#include <cmath>

#include <iomanip>

using namespace std;

double piValue() {

       double p = 1;

       int i = 1;

       while(true) {

               double prev = p;

               p += pow(-1, i) * (1.0 / (1 + 2*i));

               if(abs(prev - p) < 0.00005) {

                       break;

               }

               i++;

       }

       return 4*p;

}

int main() {

       cout << piValue() << endl;

}

**************************************************

Answer and Explanation:

The answer is attached below

Answer:

80 mi/h

Explanation:

To get the adequate design speed the assumed design speed ( Fs ) for the Horizontal curve has to produce the necessary Middle ordinate distance that is equal to the provided Middle ordinate

The provided middle ordinate is gotten form the formula

Ms = Available clearance - lane width - lane width / 2  ( equation 1 )

available clearance = 52 ft

lane width = 12 ft

hence equation 1 becomes

Ms = 52 - 12 - 12/2

      = 40 - 6 = 34 ft

To check if the assumed design speed will produce the necessary middle ordinate distance that is equal to the provided middle ordinate the formula below is used

Ms₍necessary₎  = Rv ( 1 - cos[tex]\frac{90SSD}{\pi Rv }[/tex] ) (equation 2 )

Assuming a design speed ( Fs ) of 80 mi/h and referring to the engineering table and also considering the super elevation of 0.06 ( 6% ) the values of

Rv = 3050 ft

SSD = 910 ft

substituting these values into (equation 2) Ms ( necessary ) becomes

= 3050 ( 1 - cos [tex]\frac{90*910}{\pi * 3050 }[/tex] )

= 3050 ( 1 - cos [tex]\frac{81900}{\pi * 3050 }[/tex] )

= 33.876 ft

33.876 ft is approximately equal to the provided middle ordinate of 34 ft hence the design speed used would be 80mi/h

Answer:

A). Check the results obtained with a leaner setting of the mixture.

Explanation:

Answer:

Under the explained circumstances, the most logical initial action would be:

A. Check the results obtained with a leaner setting of the mixture.

Explanation:

Check the results obtained with a leaner setting of the mixture, as it has been enriched by the carburetor-heated air causing engine roughness at that high altitude.

Answer b is wrong because the pilot should first try the runup with a leaner mixture and answer c is not correct because detonation is the result of a mixture that is to lean.

Answer:

The ratio of the difference of the pressure at the top and bottom of the cylinder to dynamic pressure is given as

[tex]\dfrac{4a (U\omega- g)}{U^2}[/tex]

Explanation:

As the value of the diameter is given as d=2a

The velocity is given as v=U

The rotational velocity is given as ω rad/s

Point A is at the top of the cylinder and point B is at the bottom of the cylinder

Such that the point A is at the highest point on the circumference and point B is at the bottom of the cylinder

Now the velocity at point A is given as

[tex]v_A=U-\dfrac{d}{2}\omega\\v_A=U-\dfrac{2a}{2}\omega\\v_A=U-a\omega\\[/tex]

Now the velocity at point B is given as

[tex]v_B=U+\dfrac{d}{2}\omega\\v_B=U+\dfrac{2a}{2}\omega\\v_B=U+a\omega\\[/tex]

Considering point B as datum and applying the Bernoulli's equation between the point A and B gives

[tex]\dfrac{P_A}{\rho g}+\dfrac{v_A^2}{2 g}+z_A=\dfrac{P_B}{\rho g}+\dfrac{v_B^2}{2 g}+z_B[/tex]

Here P_A and P_B are the local pressures at the point A and point B.

v_A and v_B are the velocities at the point A and B

z_A and z_B is the height of point A which is 2a and that of point B is 0

Now rearranging the equation of Bernoulli gives

[tex]\dfrac{P_A-P_B}{\rho g}=\dfrac{v_B^2-v_A^2}{2 g}+z_B-z_A[/tex]

Putting the values

[tex]\dfrac{P_A-P_B}{\rho g}=\dfrac{v_B^2-v_A^2}{2 g}+z_B-z_A\\\dfrac{P_A-P_B}{\rho g}=\dfrac{(U+a\omega)^2-(U-a\omega)^2}{2 g}+0-2a\\\dfrac{P_A-P_B}{\rho g}=\dfrac{(U^2+a^2\omega^2+2Ua\omega)-(U^2+a^2\omega^2-2Ua\omega)}{2g}-2a\\\dfrac{P_A-P_B}{\rho g}=\dfrac{U^2+a^2\omega^2+2Ua\omega-U^2-a^2\omega^2+2Ua\omega)}{2g}-2a\\\dfrac{P_A-P_B}{\rho g}=\dfrac{4Ua\omega}{2g}-2a\\\dfrac{P_A-P_B}{\rho g}=\dfrac{2Ua\omega}{g}-2a\\P_A-P_B=\dfrac{2Ua\omega}{g}*\rho g-2a*\rho g\\[/tex]

[tex]P_A-P_B=2Ua\omega\rho-2a\rho g[/tex]

Now the dynamic pressure is given as

[tex]P_D=\dfrac{1}{2}\rho U^2[/tex]

[tex]\dfrac{P_A-P_B}{P_D}=\dfrac{2Ua\omega\rho-2a\rho g}{1/2 \rho U^2}\\\dfrac{P_A-P_B}{P_D}=\dfrac{2a\rho (U\omega- g)}{1/2 \rho U^2}\\\dfrac{P_A-P_B}{P_D}=\dfrac{4a\rho (U\omega- g)}{\rho U^2}\\\dfrac{P_A-P_B}{P_D}=\dfrac{4a (U\omega- g)}{U^2}[/tex]

So the ratio of the difference of the pressure at the top and bottom of the cylinder to dynamic pressure is given as

[tex]\dfrac{4a (U\omega- g)}{U^2}[/tex]

Answer:

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SELECT VendorID FROM Invoices

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Explanation:

Answer:

attached below

Explanation:

Explanation:

Tension in the rope

[tex]\begin{aligned}T_{1} &=0.4 \times x \\T_{2} &=100+0.4 \times 10 \\&=104\end{aligned}[/tex]

[tex]M s=0.3[/tex]  ∅  [tex]=2 \cdot 5(2 \pi)=5 \pi[/tex]

[tex]\begin{aligned}\frac{T_{1}}{T_{2}}=e^{\mu \theta} & \Rightarrow \frac{104}{0.4 x}=e^{0.3(5 \pi)} \\& \Rightarrow x=2.34 \mathrm{H}\end{aligned}[/tex]

NOTE : Refer the image

Answer:

[tex]\dot Q = 524.957 W[/tex], [tex]T_{out} = 317.048 K[/tex].

Explanation:

a) The rate of heat loss is determined by following expression:

[tex]\dot Q = \frac{T_{ins, in} - T_{air}}{R_{th}}[/tex]

Where [tex]R_{th}[/tex] is the thermal resistance throughout the system:

[tex]R_{th} = R_{cond} + R_{conv || rad}[/tex]

Since convection and radiation phenomena are occuring simultaneously, the equivalent thermal resistance should determined:

[tex]\frac{1}{R_{conv||rad}} = \frac{1}{R_{conv}} + \frac{1}{R_{rad}}[/tex]

[tex]R_{conv||rad} = \frac{R_{conv}\cdot R_{rad}}{R_{conv} + R_{rad}}[/tex]

Where:

[tex]R_{conv} = \frac{1}{h_{conv} \cdot A} \\R_{rad} = \frac{1}{h_{rad} \cdot A}[/tex]

Outer surface area is given by:

[tex]A = 2 \cdot \pi \cdot r_{out} \cdot L[/tex]

Heat resistance associated to conduction through a hollow cylinder is:

[tex]R_{cond} = \frac{\ln \frac{r_{out}}{r_{in}} }{2 \cdot \pi L \cdot k}[/tex]

According to an engineering database, calcium silicate has a thermal conductivity k = [tex]0.085 \frac{W}{m \cdot K}[/tex]. Then, needed variables are calculated (L = 1 m):

[tex]r_{out} = 0.08 m, r_{in} = 0.06 m[/tex]

[tex]A \approx 0.503 m^2[/tex]

[tex]h_{conv} = 25 \frac{W}{m^{2}\cdot K}\\h_{rad} = 30 \frac{W}{m^2 \cdot K}[/tex]

[tex]R_{conv} = 0.080 \frac{K}{W}\\R_{rad} = 0.066 \frac{K}{W}[/tex]

[tex]R_{conv||rad} = 0.036 \frac{K}{W}[/tex]

[tex]R_{cond} = 0.539 \frac{K}{W}[/tex]

[tex]R_{th} = 0.575 \frac{K}{W}[/tex]

[tex]\dot Q = 524.957 W[/tex]

The temperature on the outside surface of the calcium silicate can be determined from the following expression:

[tex]\dot Q = \frac{T_{in}-T_{out}}{R_{cond}}[/tex]

Then,

[tex]T_{out}=T_{in}-\dot Q \cdot R_{cond}\\T_{out} = 317.048 K[/tex]

Answer:

Qcd=0.01507rad

QT= 0.10509rad

Explanation:

The full details of the procedure and answer is attached.

Answer:

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

Explanation:

Answer:

Please find attached solution

Explanation:

Answer: Tech B

Explanation:

Tech A is wrong by saying that a clutch must engage completely, all at once. Engaging the clutch completely will affect the clutch and may get it spoilt.

Tech B says that the clutch must slip slightly during engagement. That is the right way to engage the clutch.