The 240-ft structure is used to provide various support services to launch vehicles prior to liftoff. In a test, a 12-ton weight is suspended from joints F and G, with its weight equally divided between the two joints. Determine the forces in members GJ and GI. What would be your path of joint analysis for members in the vertical tower, such as AB or KL?

Answer:

Irreversibility = 5.361 kW

Explanation:

From the given information:

By applying ideal gas equation at entry:

PV =  mRT

600 × 0.3 = m × 0.287 × 300      (where R = 0.287 kJ/kg)

180 = m × 86.1

m = 180/86.1

m = 2.0905 kg/min

At the hot end, using the same ideal gas equation:

PV = mRT

100 × V = 1.4905 × 0.287 × 325

V = 139.026/100

V = 1.3903 m³/ min

This implies that: The total entropy change = Entropy of the universe

So,

[tex]m\bigg [ c_p \ In \dfrac{T_2}{T_o}-R \ In \dfrac{P_2}{P_o} \bigg] + m_2 \bigg [ c_p \ In \dfrac{T_2}{T_o}- R \ In\dfrac{P_2}{P_o} \bigg][/tex]

[tex]= 0.6\bigg [ 1.004 \ In \dfrac{245}{300}-0.287 \ In \dfrac{100}{600} \bigg] +1.4905\bigg [1.004 \ In \dfrac{325}{300}- 0.287\ In\dfrac{100}{600} \bigg][/tex]

= 0.6[-0.2033 + 0.5142] + 1.4905 [0.08036 + 0.5142]

= 1.0727 kJ/min.K

= 0.01787 kw/K

Irreversibility = [tex]T_o [ \Delta S][/tex]

Irreversibility = 300 × 0.01787

Irreversibility = 5.361 kW

Answer:

yes it is very hard you should find a reccomended doctor to aid in your situation. But in the meantime how about you give me that lil brainliest thingy :p

Answer:

The mass of oxygen in liquid phase = 14.703 kg

The mass of oxygen in the vapor phase = 20.302 kg

Explanation:

Given that:

The mass of the oxygen [tex]m_{O_2}[/tex] = 35 kg

The mass of the nitrogen [tex]m_{N_2}[/tex] = 40 kg

The cooling temperature of the mixture T = 84 K

The cooling pressure of the mixture P = 0.1 MPa

From the equilibrium diagram for te-phase mixture od oxygen-nitrogen as 0.1 MPa graph. The properties of liquid and vapor percentages are obtained.

i.e.

Liquid percentage of [tex]O_2[/tex] = 70% = 0.70

Vapor percentage of [tex]O_2[/tex] = 34% = 0.34

The molar mass (mm) of oxygen and nitrogen are 32 g/mol and 28 g/mol respectively

Thus, the number of moles of each component is:

number of moles of oxygen = 35/32

number of moles of oxygen =  1.0938 kmol

number of moles of nitrogen = 40/28

number of moles of nitrogen = 1.4286  kmol

Hence, the total no. of moles in the mixture is:

[tex]N_{total} = 1.0938+1.4286[/tex]

[tex]N_{total} = 2.5224 \ kmol[/tex]

So, the total no of moles in the whole system is:

[tex]N_f + N_g = 2.5224 --- (1)[/tex]

The total number of moles for oxygen in the system is

[tex]0.7 \ N_f + 0.34 \ N_g = 1.0938 --- (2)[/tex]

From equation (1), let N_f = 2.5224 - N_g, then replace the value of N_f into equation (2)

∴

0.7(2.5224 - N_g) + 0.34 N_g = 1.0938

1.76568 - 0.7 N_g + 0.34 N_g = 1.0938

1.76568 - 0.36 N_g = 1.0938

1.76568 - 1.0938 = 0.36 N_g

0.67188  = 0.36 N_g

N_g = 0.67188/0.36

N_g = 1.866

From equation (1)

[tex]N_f + N_g = 2.5224[/tex]

N_f + 1.866 = 2.5224

N_f = 2.5224 - 1.866

N_f = 0.6564

Thus, the mass of oxygen in the liquid and vapor phases is:

[tex]m_{fO_2} = 0.7 \times 0.6564 \times 32[/tex]

[tex]m_{fO_2} = 14.703 \ kg[/tex]

The mass of oxygen in liquid phase = 14.703 kg

[tex]m_{g_O_2} = 0.34 \times 1.866 \times 32[/tex]

[tex]m_{g_O_2} = 20.302 \ kg[/tex]

The mass of oxygen in the vapor phase = 20.302 kg

Chemical manufacturers must present Product Identifier, Contact Information for the manufacturer and Hazard pictograms on the product's label.

A product label means a display of written, printed or graphic material that is printed on or affixed to a product or its immediate container.

Let's consider which of the following information must be presented on the product's label by chemical manufacturers.

Chemical manufacturers must present Product Identifier, Contact Information for the manufacturer and Hazard pictograms on the product's label.

Learn more about hazard pictograms here: https://brainly.com/question/22130371

Answer:

a) 6 coulombs

b) The electrons carrying the charge will enter at point b with respect to element and this is because electrons follow in opposite direction of current

c) = -72 joules

Energy is taken from element

Explanation:

Given data:

V ab = -12 v

I ab = 3A

period ( t ) = 2 seconds

a) determine how much charge moves through the element

q = I * t

  = 3 * 2 = 6 coulombs

b) The electrons carrying the charge will enter at point b with respect to element and this is because electrons follow in opposite direction of current

c) determine how much energy is transferred

= Vab * Iab * t

= -12 * 3 * 2

= -72 joules

Energy is taken from element

Answer:

a( contractors need to know the amount of markup)

because the contractor should have a long term vision for a proper satisfaction to the people.

Answer:

The required heat flux = 12682.268 W/m²

Explanation:

From the given information:

The initial = 25°C

The final = 75°C

The volume of the fluid = 0.2 m/s

The diameter of the steel tube = 12.7 mm = 0.0127 m

The fluid properties for density [tex]\rho[/tex] = 1000 kg/m³

The mass flow rate of the fluid can be calculated as:

[tex]m = pAV[/tex]

[tex]m = \rho \dfrac{\pi}{4}D^2V[/tex]

[tex]m = 1000 \times \dfrac{\pi}{4} \times ( 0.0127)^2 \times 0.2[/tex]

[tex]m = 0.0253 \ kg/s[/tex]

To estimate the amount of the heat by using the expression:

[tex]q = mc_p(T_{final}-T_{initial})[/tex]

q = 0.0253 × 4000(75-25)

q = 101.2 (50)

q = 5060 W

Finally, the required heat of the flux is determined by using the formula:

[tex]q" = \dfrac{q}{A_s}[/tex]

[tex]q" = \dfrac{q}{\pi D L}[/tex]

[tex]q" = \dfrac{5060}{\pi \times 0.0127 \times 10}[/tex]

q" =  12682.268 W/m²

The required heat flux = 12682.268 W/m²

Answer:

Follows are the solution to this question:

Explanation:

Calculating the area under the curve:  

A = as

   [tex]=\frac{1}{2}(3 +6 \frac{m}{s^2})(100 \ m)+ \frac{1}{2}(6+4 \frac{m}{s^2})(100 m) \\\\=\frac{1}{2}(9 \frac{m}{s^2})(100 \ m)+ \frac{1}{2}(10\frac{m}{s^2})(100 m) \\\\=\frac{1}{2}(900 \frac{m^2}{s^2})+ \frac{1}{2}(1,000\frac{m^2}{s^2}) \\\\=(450 \frac{m^2}{s^2})+ (500\frac{m^2}{s^2}) \\\\= 950 \ \frac{m^2}{s^2}[/tex]

Calculating the kinematics equation:

[tex]\to v^2 = v^2_{o} + 2as\\\\[/tex]

        [tex]=0+ \sqrt{2as}\\\\ = \sqrt{2(A)}\\\\= \sqrt{2(950 \frac{m^2}{s^2})}\\\\= 43.59 \frac{m}{s}[/tex]

Calculating the value of acceleration:  

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

[tex]=\frac{dv}{ds}(\frac{ds}{dt}) \\\\=v\frac{dv}{ds}\\\\\to \frac{dv}{ds}=\frac{a}{v}[/tex]

[tex]\to \frac{dv}{ds} =\frac{4 \frac{m}{s^2}}{43.59 \frac{m}{s}} \\\\[/tex]

         [tex]=\frac{0.092}{s}[/tex]

Answer:

Instantaneous velocity = 8m/s

Explanation:

Given the following data;

Distance = 2t² - 4t

Time, t = 3 secs.

To find the instantaneous velocity;

[tex] Velocity = \frac {distance}{time} [/tex]

[tex] V(t) = \frac {dd}{dt} [/tex]

We would differentiate the equation for the distance with respect to time, t.

[tex] \frac {dd}{dt} = \frac {d(2t^{2} - 4t)}{dt}[/tex]

[tex] \frac {dd}{dt} = 4t - 4[/tex]

Substituting the value of "t" into the above equation, we have;

[tex] V(3) = 4(3) - 4[/tex]

[tex] V(3) = 12 - 4[/tex]

Instantaneous velocity = 8m/s

Answer:

1.16 k/ft

Explanation:

From the given information;

Using table 1.3 for Minimum design dead loads;

For 12-in clay brick,

the obtained min. design dead load = 115 psf

For Fiberboard 1/2 in. ceilings, the minimum design dead load is 0.75 psf

To start with the load that is being exerted on the floor as a result of the clay brick wall ([tex]L_1[/tex] ), we have:

[tex]L_1 = Load \times h[/tex]

[tex]L_1 = 115 \times 10[/tex]

[tex]L_1 = 1150 \ lb/ft[/tex]

To calculate the load exerted on the floor as a result of the 1/2 fireboard, we have:

[tex]L_2 = Load \times h[/tex]

[tex]L_2 = 0.75 \times 10[/tex]

[tex]L_2 = 7.5 \ lb/ft[/tex]

The total load exerted on the floor = [tex]L_1 + L_2[/tex]

The total load exerted on the floor = 1150 + 7.50

The total load exerted on the floor = 1157.50 lb/ft

To (k/ft), we get:

[tex]= 1157.50 \ lb/ft \times \dfrac{1 \ k}{1000 \ lb}[/tex]

= 1.157 k/ft

≅ 1.16 k/ft

Answer:

-0.1006Kw/K

Explanation:

The rate of entropy change in the air can be reduced from the heat transfer and the air temperature. Hence,

ΔS = Q/T

Where T is the constant absolute temperature of the system and Q is the heat transfer for the internally reversible process.

S(air) = - Q/T(air) .......1

Where S.air =

Q = 30-kW

T.air = 298k

Substitute the values into equation 1

S(air) = - 30/298

= -0.1006Kw/K

Answer: 98.5% of pearlite was formed during the equilibrium cooling

Explanation:

First we calculate the fraction of pro-eutectoid phase which forms for equilibrium cooling of the 1085 steel from 1000°C at room temperature;

we know that in 1085 steel, last two digits denotes the carbon percentage

so 1085 steel contains 0.85% carbon.

Now from the diagram, carbon percentage is greater than the eutectoid com[psition

i.e 0.85 > 0.76

it is a hyper eutectoid steel

so

fraction of pro eutectoid phase W_Fe₃C = (0.85 - 0.76) / ( 6.7 - 0.76)

= 0.09 / 5.94 = 0.015 = 1.5%

Now, the amount of pearlite formed during the equilibrium cooling of the 1055 steel from 1000°C to room temperature will be;

pearlite (C') = (1 - W_Fe₃C)

= 1 - 0.015

= 0.985 = 98.5%

Therefore 98.5% of pearlite was formed during the equilibrium cooling

Answer: its A

Explanation:

Answer:

A)

- Q ( kw ) for vapor =  -1258.05 kw

- Q ( kw ) for liquid = -1146.3 kw

B )

- Q ( kj ) for vapor  = -1258.05 kJ

- Q ( KJ ) for liquid = - 1146.3 KJ

Explanation:

Given data :

45.00 % mole of methane

55.00 % of ethane

attached below is a detailed solution

A) calculate - Q(kw)

- Q ( kw ) for vapor =  -1258.05 kw

- Q ( kw ) for liquid = -1146.3 kw

B ) calculate  - Q ( KJ )

- Q ( kj ) for vapor  = -1258.05 kJ

- Q ( KJ ) for liquid = - 1146.3 KJ

since combustion takes place in a constant-volume batch reactor

Answer:

0

Explanation:

Given that:

The peak of the voltage [tex]V_{peak}[/tex] = 12 V

The frequency f = 50 Hz

At the time t  = 10 ms

[tex]V = V_p \ sin \ \omega t[/tex]

where;

[tex]\omega = 2 \pi f[/tex]

[tex]V = 1 2\ V \times sin \ ( 2 \pi \times 50 \times 10 \ )[/tex]

V = 12V sin (180)

V = 12 V × 0

V = 0

Answer: hydraulic retention time,τ=28.67 hours

Explanation:

The hydraulic retention time  τ (tau),  is given as  The volume of the settling tank(V) divided by the influent flowrate(Q)

τ =V/Q

But Volume is not known  and is given as

Volume =  surface area  x depth of the tank

= 8000 ft² X 12 ft

= 96,000 ft³

Also, the influent flow rate is in mgd ( million gallons per day), we change  it to  ft³/sec so as to be in same unit with the volume in ft³

1 million gallons/day = 1.5472286365101 cubic feet/second

0.6mgd =  1.5472286365101 cubic feet/second  x 0.6

=0.93cubic feet/second

τ =V/Q

96,000 ft³/0.93 ft³/sec

τ=103,225.8 secs

changing to hours

103,225.8 /3600 =28.67 hours

The hydraulic retention time =28.67 hours

According to the scenario, the pounds per square inch required by a bubbler system to produce bubbles at a depth of 4 ft 7 in water is found to be approximately 2.0 Psig.

The bubbler system may be defined as a type of system that significantly measures water level based on the amount of pressure it takes to push an air bubble out of an orifice line and into the water body. This pressure often referred to as the “line pressure”, requires changes in the elevation of the water.

According to the context of this question,

The depth of bubbles produced by the bubbler system = 4 ft 7 inches.

The pounds per square inch = 2.31 Psig.

∴ The pounds per square inch is required by a bubbler system to produce bubbles at a depth of 4 ft 7 in water = 4 ft 7 inches/2.31 = 2.03 Psig ≅ 2Psig.

Therefore, according to the scenario, the pounds per square inch required by a bubbler system to produce bubbles at a depth of 4 ft 7 in water is found to be approximately 2.0 Psig.

To learn more about the Bubbler system, refer to the link:

https://brainly.com/question/24317138

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Answer:

bcde!!

Explanation:

They also tend to be traditional, which means that they enjoy working in structured environments and are typically organized and detail-oriented. Thus option B,C,D, E is correct.

Carpenters continue to have a bright future in their profession. According to Job Outlook, the carpentry industry is expanding quickly. The advantages of carpentry lead to employment security and a long-term career with this kind of industrial growth.

Carpentry is a physically demanding line of work that calls for endurance. You frequently spend the most of your shift standing, moving slowly, and crouching.

Therefore, As well as using hand tools to shape and cut wood, lifting big objects, moving heavy beams, furniture, or equipment are all possible.

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Answer:Decay rate constant,k  = 0.00376/hr

Explanation:

IsT Order  Rate of reaction is given as

In At/ Ao = -Kt

where [A]t is the final concentration at time  t  and  [A]o  is the inital concentration at time 0, and  k  is the first-order rate constant.

Initial concentration = 80 mg/L

Final concentration = 50 mg/L

Velocity = 40 m/hr

Distance= 5000 m

Time taken = Distance / Time

              5000m / 40m/hr = 125 hr

In At/ Ao = -Kt

In 50/80 = -Kt

-0.47 = -kt

- K= -0.47 / 125

k = 0.00376

Decay rate constant,k  = 0.00376/hr

Answer:

The energy yield for one gallon of ethanol is 2.473 %.

Explanation:

The net energy yield ([tex]\% e[/tex]), expressed in percentage for one gallon of ethanol is the percentage of the ratio of the difference of the provided energy ([tex]E_{g}[/tex]), measured in Btu, and the energy needed to produce the ethanol ([tex]E_{p}[/tex]), measured in Btu, divided by the energy needed to produce the ethanol. That is:

[tex]\% e =\frac{E_{g}-E_{p}}{E_{p}} \times 100\,\%[/tex] (1)

If we know that [tex]E_{g} = 76330\,Btu[/tex] and [tex]E_{p} = 74488\,Btu[/tex], then the net energy yield of 1 gallon of ethanol:

[tex]\%e = \frac{76330\,Btu-74488\,Btu}{74488\,Btu}\times 100\,\%[/tex]

[tex]\%e = 2.473\,\%[/tex]

The energy yield for one gallon of ethanol is 2.473 %.

This question is incomplete, the complete question is;

Air flow is measured in a Venturi meter that has a large diameter of 1.5 m and a small diameter of 0.9 m. A 1:12 scale model with water as the fluid is used to calibrate the meter. For the model, it is determined that when the volume flowrate is 0.07 m3 /s, the pressure drop from the large diameter portion (Section 1) to the small diameter portion (Section 2) is 172 kPa.

Calculate The corresponding flowrate in the prototype.

Assume a water temperature of 15°C and standard properties of air

Answer:  The corresponding flowrate in the prototype is 10.21 m³/s

Explanation:

Given that;

Lm = Lp/12 and lp = 12Lm,   Qm = 0.07 m³/s,   ΔPm = 172 Kpa

properties of water at 15°C ---- Vm = 1.2015 × 10⁻⁶ m²/s,   Sm = 1001.2 kg/m₂

Also for Air---- Vp = 1.46041 × 10⁻⁵,  Sp = 1.225 kg/m³

Now by Using Reynold's model law; (Vm × Lm)/Vm = (Vp × Lp)/Vp

(Vm × Lm) / 1.2015 × 10⁻⁶ = (Vp ×12 × Lm) / 1.46041 × 10⁻⁵

Vm/Vp = 0.9872

we know that

Discharge = Area × Velocity

Qm/Qp = Lm²/Lpl × Vm/Vp

= (1/12)² × 0.9872

= 6.856 × 10⁻³

so Qp = (0.07 / 6.856 × 10⁻³) = 10.21 m³/s

Therefore The corresponding flowrate in the prototype is 10.21 m³/s

Answer:

attached below is the detailed solution

A) 8288.77 cu.ft

B) 4.96 hours

C) Vss = 131.21 IbVss/day

   Tss = 164 IbTss/day

D) attached below

E ) 0.2

F) 287.23 Ib/day

Explanation:

A) Determine the aeration basin volume

Given

∅c = 20 days

Y = 0.8Ib VSS/Ib BOD

Q = 0.3 mgd

So = 120 mg/l

Se = 2 mg/l

X = 5000 mg/l

Kd = 0.04 per day

attached below is the detailed solution

B) Determine the detention time using this relation

t = ( V / Q )* 24

  = ( 0.062 / 0.3 ) * 24  = 4.96 hours

C ) Determine Vss and Tss

we first calculate the excess biomass Px then assuming Vss ratio to be 80% for sludge age of 20 days

Vss = 131.21 IbVss/day

Tss = 164 IbTss/day

D ) determine BOD loading

  Q = 0.3 mgd , BOD = 220mg/l , V = 8288.77 cu.ft

solution attached below

e) food to microorganism ratio

F/M  = 0.2

solution attached below

f) determine the waste solid

waste solid = Q * SS * % removal of suspended solids

where : Q = 0.3 , SS = 220mgl , % = 50 %

waste solids = 0.3 * 230 * 0.5 * 8.34  = 287.23 Ib/day

Answer:

Total voltage = 24 V

Explanation:

Given:

Volts per rpm = 0.01

Total rpm = 2400

Find:

Total voltage

Computation:

Total voltage = Volts per rpm x Total rpm

Total voltage = 0.01 x 2400

Total voltage = 24 V

Answer:

6 Ω

Explanation:

given data :

Voltage ( v ) = 12cos ( 60t + 45° )

L = 0.1 H

calculate the inductor's impedance Z  

Z = jXL

  = jx = 60

  = L = 0.1

hence Z = 60 * 0.1 =  6 Ω

Answer:

The answer is "0.187 lbm and 1 lbf".

Explanation:

The mass = [tex]0.031080997078386\ slug[/tex]

Calculating mass on Mars:

[tex]\to m=m_g\frac{g}{g_e}[/tex]

        [tex]=0.031080997078386 \times \frac{32.2}{5.35}\\\\=0.187 \ lbm[/tex]

[tex]\to W=mg_e[/tex]

        [tex]=0.187 \times 5.35\\\\=1 \ lbf[/tex]

Answer:

hmmm.........

Explanation: