A manufacturing facility with a wastewater flow of 0.011 m3/sec and a BOD5 of 590 mg/L discharges into the Cattaraugus Creek. The creek has a 10-year, 7-day low flow of 1.7 m3/sec. Upstream of the facility, the BOD5 of the creek is 0.6 mg/L. The BOD rate constants k are 0.115 d-1. Determine: [A] UBOD of wastewater. [B] UBOD of creek. [C] What is the initial ultimate BOD after mixing

Answer:

0.193 m/s

Explanation:

Given :

Density of water = 1000 [tex]kg/m^3[/tex]

Manometric height = 20 mm

Density of SAE(30) = 875.4 [tex]kg/m^3[/tex]

Determining the upstream velocity of the oil, v = [tex]$\sqrt{2gh}$[/tex]    , h = piezometric head.

[tex]$h=\left(\frac{\rho_m}{\rho_l}-1\right)y$[/tex]

[tex]$h=\left(\frac{1000}{875.4}-1\right)\times \frac{20}{1000}$[/tex]

[tex]$h=2.84\times 10^{-3} \ m$[/tex]

Now,

[tex]$v=\sqrt{2 \times 9.81 \times 2.84 \times 10^{-3}}$[/tex]

  = 0.193 m/s  

Answer:

Hence the magnitude of the pure moment m will be [tex]2\times 10^5.[/tex]

Explanation:

Now,

We know that,

[tex]\frac{M}{I} = \frac{E}{R}[/tex]

Thus, [tex]M =\frac{EI}{R}[/tex]

Here

R = 10000 mm  

[tex]I=\frac{1}{12}\times 120\times 10^{3}\\= 10^{4} mm^{4}[/tex]

[tex]E=2\times 10^{5}n/mm^{2}\\\\E=2\times 10^{5}n/mm^{2}\\\\M={(2\times 10^{5}\times 10^{4})/{10000}}\\\\M=2\times 10^{5}[/tex]

Hence, the magnitude of the pure moment m will be [tex]2\times 10^5.[/tex]

Answer:

a) the safety factor according to the maximum normal-stress theory; n = 3

b) the safety factor according to the maximum-shear-stress theory; n = 1.714

c) the safety factor according to the maximum distortion-energy theory is; n = 1.97278

Explanation:

Given the data in the question;

a) What is the safety factor according to the maximum normal-stress theory;

According to the maximum normal-stress theory

n = S[tex]_y[/tex] / σ[tex]_{max[/tex]

since σ₁ = 20 ksi is greater than σ₂ = -15 ksi

σ[tex]_{max[/tex] = 20 ksi and yield strengths in tension and compression S[tex]_y[/tex] = 60 ksi

we substitute

n = 60 ksi / 20 ksi

n = 3

Therefore, the safety factor according to the maximum normal-stress theory; n = 3

b) What is the safety factor according to the maximum-shear-stress theory.

According to maximum-shear-stress theory;

τ[tex]_{max[/tex] = [(σ₁ - σ₂) / 2]

= S[tex]_y[/tex] / 2n

n = S[tex]_y[/tex] / 2[(σ₁ - σ₂) / 2]

n = S[tex]_y[/tex] / (σ₁ - σ₂)

we substitute

n = 60 ksi  / (20 ksi - (-15 ksi))

n = 60 ksi  / (20 ksi +15 ksi)

n = 60 ksi  / 35 ksi

n = 1.714

Therefore, the safety factor according to the maximum-shear-stress theory; n = 1.714

c) the safety factor according to the maximum distortion-energy theory?

By distortion energy theory

σ₁² + σ₂² - σ₁σ₂ = (S[tex]_y[/tex]/n)²

we substitute

(20)² + (-15)² - ( 20 × -15 ) = ( 60 / n )²

400 + 225 + 300 = 3600 / n²

925 =  3600 / n²

n² = 3600 / 925

n = √( 3600 / 925 )

n = 1.97278

Therefore, the safety factor according to the maximum distortion-energy theory is; n = 1.97278

Solution :

Methods for selling pressure of a distillation column :

a). Set, [tex]\text{based on the pressure required to condensed}[/tex] the overhead stream using cooling water.

  (minimum of approximate 45°C condenser temperature)

b). Set, [tex]\text{based on highest temperature}[/tex] of bottom product that avoids decomposition or reaction.

c). Set, [tex]\text{based on available highest }[/tex] not utility for reboiler.

Running the distillation column above the ambient pressure because :

The components to be distilled have very high vapor pressures and the temperature at which they can be condensed at or below the ambient pressure.

Run the reactor at an evaluated temperature because :

a). The rate of reaction is taster. This results in a small reactor or high phase conversion.

b). The reaction is endothermic and equilibrium limited increasing the temperature shifts the equilibrium to the right.

Run the reaction at an evaluated pressure because :

The reaction is gas phase and the concentration and hence the rate is increased as the pressure is increased. This results in a smaller reactor and /or higher reactor conversion.

The reaction is equilibrium limited and there are few products moles than react moles. As increase in pressure shifts the equilibrium to the right.

Answer:

450.000 USD

Explanation:

Đây,

Cấu trúc hữu cơ 3/1 có nghĩa là ¾ một phần vốn được chuyển vào chi phí cố định và ¼ một phần đi vào chi phí biến đổi.

Tiền lương của người lao động là chi phí cố định và do đó, mức lương

= ¾ * 600.000 USD

= 450.000 USD

Answer:

the answer how you analyzs the problwm

Answer:

To access a missions, simply log on to your CoderZ account and navigate to one of the courses you are assigned, choose a pack, and then choose one of the missions!

Answer:

i) 3750 veh/hr/ln

ii) 100 veh/mi/In

iii) 37.5 mph

Explanation:

number of lanes = 3

sf for both directions = 75 mph  ( free mean speed )

Dj for both directions = 200 veh/mi/In

Calculate the value of  S0, D0 (veh/mi/ln) and maximum Vm (veh/hr)

For either direction we will consider the total volume = 3 lanes

value of Dj = 3 lanes * 200 = 600 veh/mi/

i) value of SO

=  ( Dj * sf ) / 4 = ( 600 * 75 ) / 4 = 11250 veh/hr  = 3750 veh/hr/lane

ii) Value of DO

DO = Dj / 2 = 200 /2 = 100 veh/mi/In

iii) Value of Vm

= sf /2 = 75 / 2 = 37.5 mph

Answer:

Following are the response to the given question:

Explanation:

In the given question, A curve one would be to "reject" them. Within this way, people are directly non-confrontationally off from their amorous interests or advances. The curve looks much like a current through the unidirectional load, and it has only a good maximum value. It seems like the total rating of the complete wave rectifier that is illustrated below, the present vas temporal curve please find it.

Answer:

0.320

Explanation:

Productivity is measured as the total output divided by total input.

From the given information, the output is taken into consideration when we finish subtracting the defective materials.

The labor hours usually regarded as the input = 7 hours per day × 5 = 35

For the prior productivity, we have:

[tex]\mathbf{Prior \ Productivity = \dfrac{60 - 10}{35}}[/tex]

[tex]\mathbf{Prior \ Productivity = \dfrac{50}{35}}[/tex]

[tex]\mathbf{Prior \ Productivity = 1.43}[/tex]

For the current productivity:

[tex]\mathbf{current \ productivity= \dfrac{70-4}{35}}[/tex]

[tex]\mathbf{current \ productivity= \dfrac{66}{35}}[/tex]

[tex]\mathbf{current \ productivity= 1.89}[/tex]

Thus, the growth rate i.e. the increased change in  productivity is:

[tex]= \dfrac{1.89 - 1.43}{1.43}[/tex]

= 0.320

Answer:

No puedo entenderte solo en español

Answer:

Almost done

Explanation:

I am just finishing up my work

Answer:

[tex]I=3A[/tex]

Explanation:

From the question we are told that:

Number of lamps [tex]N=4[/tex]

Potential difference [tex]V=60v[/tex]

Total Resistance of the lamp is [tex]R= 20ohms[/tex]

Generally the equation for Current I is mathematically given by

 [tex]I=\frac{V}{R}[/tex]

 [tex]I=\frac{60}{20}[/tex]

 [tex]I=3A[/tex]

Answer:

a) 53 MPa,  14.87 degree

b) 60.5 MPa  

Average shear = -7.5 MPa

Explanation:

Given

A = 45

B = -60

C = 30

a) stress P1 = (A+B)/2 + Sqrt ({(A-B)/2}^2 + C)

Substituting the given values, we get -

P1 = (45-60)/2 + Sqrt ({(45-(-60))/2}^2 + 30)

P1 = 53 MPa

Likewise P2 = (A+B)/2 - Sqrt ({(A-B)/2}^2 + C)

Substituting the given values, we get -

P1 = (45-60)/2 - Sqrt ({(45-(-60))/2}^2 + 30)

P1 = -68 MPa

Tan 2a = C/{(A-B)/2}

Tan 2a = 30/(45+60)/2

a = 14.87 degree

Principal stress

p1 = (45+60)/2 + (45-60)/2 cos 2a + 30 sin2a = 53 MPa

b) Shear stress in plane

Sqrt ({(45-(-60))/2}^2 + 30) = 60.5 MPa

Average = (45-(-60))/2 = -7.5 MPa

Answer:

Compute the first four central moments for the following data:

i xi

1 45

2 22

3 53

4 84Explanation:

Answer:

ok i will on day

Explanation:

11 [2-bit]

011 [3-bit]

0011 [4-bit]

________

1 + 1 + 1 = 3

________

3 = 2 + 1

2¹ 2⁰

3 = (.. × 0) + (2¹ × 1) + (2⁰ × 1)

3 = ..011

Since 2³, 2⁴, 2⁵, .. are not used, they are represented as 0.

[ 2⁷ 2⁶ 2⁵ 2⁴ 2³ 2² 2¹ 2⁰ ]

[ 128 64 32 16 8 4 2 1 ]

Answer:

The right solution is "2625 kN".

Explanation:

According to the question,

The average pressure will be:

= [tex]density\times g\times \frac{h}{2}[/tex]

By putting values, we get

= [tex]1000\times 9.8\times \frac{12.2}{2}[/tex]

= [tex]1000\times 9.8\times 6.1[/tex]

= [tex]59780[/tex]

hence,

The average force will be:

= [tex]Pressure\times Area[/tex]

= [tex]59780\times 3.6\times 12.2[/tex]

= [tex]2625537 \ N[/tex]

Or,

= [tex]2625 \ kN[/tex]

Answer: Hello your question is incomplete attached below is the complete question

answer :

Attached below

Explanation:

Given data:

The inputs of two registers are controlled by a 2-to-1 multiplexer.

The multiplexer select line and the register load enable inputs are controlled by inputs Co, C1, and C2.

Using the required transfers in the question to complete the detailed logic diagrams ( attached below )

This is a test. Please ignore

Explanation:

Test

Complete Question

Complete Question is attached below

Answer:

Option A

Explanation:

From the question we are told that:

inner Diameter of pipe [tex]d_i=100^c[/tex]

Thickness [tex]t=50mm[/tex]

Outer diameter of pipe [tex]d_o=1.1m[/tex]

Length [tex]l=5m[/tex]

Temperature [tex]T_i=130^oC[/tex]

Generally the equation for Heat Balance is mathematically given by

[tex]q*\pi d_oL=mC_p(T_i-T_o)[/tex]

Therefore

[tex]T_o=T_i+\frac{q*\pi d_oL}{mC_p}[/tex]

[tex]T_o=130+\frac{100*3.142 *1.1*5}{0.5*4000}[/tex]

[tex]T_o=129.136^oC[/tex]

Therefore the exit temperature of the water.is [tex]T_o=129.136^oC[/tex]

Option A

Answer:

The plot for percentage error as a function of fractional displacement ( [tex]\frac{R_{L} }{R_{C} }[/tex]) for the values of 0.1,1.0,10.0 is shown in image attached below.

Explanation:

Electrical loading non linearity error (percentage) is shown below.

[tex]E=\frac{(\frac{v_{o} }{v_{r} }-\frac{Q}{Q_{max} } )}{\frac{Q}{Q_{max} } }[/tex]×[tex]100[/tex]

where Q= displacement of the slider arm

[tex]Q_{max}=[/tex] maximum displacement of a stroke

[tex]\frac{v_{o} }{ v_{r} } =[/tex][tex]\frac{(\frac{Q}{Q_{max} }(\frac{R_{L} }{R_{C} } ) )}{(\frac{R_{L} }{R_{C} } ) +(\frac{Q}{Q_{max} })-(\frac{Q}{Q_{max} })^{2} }[/tex]

here [tex]R_{L}=load resistance[/tex]

[tex]R_{C}=[/tex]total resistance of potentiometer.

Now the nonlinearity error in percentage is

[tex]E=\frac{(\frac{(\frac{Q}{Q_{max} }(\frac{R_{L} }{R_{C} } ) )}{(\frac{R_{L} }{R_{C} } ) +(\frac{Q}{Q_{max} })-(\frac{Q}{Q_{max} })^{2} }-\frac{Q}{Q_{max} } )}{\frac{Q}{Q_{max} } }[/tex]×[tex]100[/tex]

The following attached file shows nonlinear error in percentage as a function of [tex]\frac{R_{L} }{R_{C} }[/tex]  displacement with given values 0.1, 1.0, 10.0. The plot is drawn using MATLAB.

The MATLAB code is given below.

clear all ;

clc ;

ratio=0.1 ;

i=0 ;

for zratio=0:0.01:1 ;

i=i+1 ;

tratioa (1,i)=zratio ;

E1(1,i)=((((zratio*ratio)/(ratio+zratio-zratio^2))-zratio)/zrtio)*100 ;

end

ratio=1.0 :

i=0 ;

for zratio=0:0.01:1 ;

i=i+1 ;

tratiob (1,i)=zratio ;

E2(1,i)=((((zratio*ratio)/(ratio+zratio-zratio^2))-zratio)/zratio)*100 ;

end

ratio=10.0 :

i=0 ;

for zratio=0:0.01:1 ;

i=i+1 ;

tratioc (1,i)=zratio ;

E3(1,i)=((((zratio*ratio)/(ratio+zratio-zratio^2))-zratio)/zrtio)*100 ;

end

k=plot(tratioa,E1,tratiob,E2,tratioc,E3)

grid

title({non linear error in % as a function of R_L/R_C})

k(1). line width = 2;

k(1).marker='*'

k(1).color='red'

k(2).linewidth=1;

k(2).marker='d';

k(2).color='m';

k(3).linewidth=0.5;

k(3).marker='h';

k(3).color='b'

legend ('location', 'south east')

legend('R_L/R_C=0.1','R_L/R_C=1.0','R_L/R_C=10.0')

Answer:

R = 20Ω

L = 0.1 H

C = 1 × 10⁻⁵ F

Explanation:

Given the data in the question;

Vs = 10∠30°V   { peak value }

V"s[tex]_{rms[/tex] = 10/√2 ∠30° V

resonance freq w₀ = 10³ rad/s

Average Power at resonance Power[tex]_{avg[/tex]  = 2.5 W

Q = 5

values of R, L, and C = ?

We know that;

Power[tex]_{avg[/tex] = |V"s[tex]_{rms[/tex]|² / R

{ resonance circuit is purely resistive }

we substitute

2.5 = (10/√2)² × 1/R

2.5 = 50 × 1/R

R = 50 / 2.5

R = 20Ω

We also know that;

Q = w₀L / R

we substitute

5 = ( 10³ × L ) / 20

5 × 20 = 10³ × L

100 = 10³ × L

L = 100 / 10³

L = 0.1 H

Also;

w₀ = 1 / √LC

square both side

w₀² = 1 / LC

w₀²LC = 1

C = 1 / w₀²L

we substitute

C = 1 / [ (10³)² × 0.1 ]

C = 1 / [ 1000000 × 0.1 ]

C = 1 / [ 100000 ]

C = 0.00001 ≈ 1 × 10⁻⁵ F

Therefore;

R = 20Ω

L = 0.1 H

C = 1 × 10⁻⁵ F

Answer:

Explanation:

do you have any other questions besides "tyuuyiopopiouyttrrtrffrlkl,k;;';'l.l"

Answer:

the total charge which crosses into the collector is 60 nC

Explanation:

Given the data in the question;

current flowing into the collector lead of the bipolar junction transistor (BJT); i = 1 nA = 10⁻⁹ A

no charge was transferred in or out of the collector lead prior to t = 0

the current flow time t = 1 min = 60 sec

Now we write the relation between current, charge, and time;

i = dq / dt

where i is current, q is charge and t is time. { d refers to change }

Now,

[tex]q=\int\limits^t_{t=0} {i(t)} \, dt[/tex]

[tex]q=\int\limits^{t=60}_{t=0} { (10^{-9}) } \, dt[/tex]

[tex]q = ( 10^{-9}) (t)_0^{60[/tex]

[tex]q = ( 10^{-9}) ( 60 - 0 )[/tex]

q = 60 × 10⁻⁹ C

q = 60 nC

Therefore, the total charge which crosses into the collector is 60 nC

Answer:

Hi, there your answer is hydraulics

Explanation:

Answer:

The mass of fuel added, which is 10,166.2 kg is less than 22,300 kg which is the mass of fuel required to travel from Toronto to Edmonton, the plane therefore crashed.

Explanation:

Since density ρ = m/v where m = mass of fuel and v = volume of fuel, we need to find the mass of each volume of fuel.

So, m = ρv now ρ = specific gravity × density of water = 0.803 × 1000 kg/m³ = 803 kg/m³.

To find the mass of the 7,682 L of fuel, its volume is 7,682 dm³ = 7,682 dm³ × 1 m³/1000 dm³ = 7.682 m³.

It's mass, m = 803 kg/m³ × 7.682 m³ = 6168.646 kg

To find the mass of the extra 4,916 L of fuel added, we have

m' = ρv' where v' = 4,916 L = 4,916 dm³ = 4916 dm³ × 1 m³/1000 dm³ = 4.916 m³

m' =  803 kg/m³ × 4.916 m³ = 3947.548 kg

So, the total mass of the fuel is m" = m + m' = 6168.646 kg + 3947.548 kg = 10116.194 kg ≅ 10,166.2 kg

Since this mass of fuel added, which is 10,166.2 kg is less than 22,300 kg which is the mass of fuel required to travel from Toronto to Edmonton, the plane therefore crashed.

Answer:

64640.92 psi

Explanation:

True stress ( psi )       True strain

49300                          0.11

61300                           0.21

Determine the true stress necessary to produce a true plastic strain of 0.25

бT1 = 49300

бT2 = 61300

бT3 = ?

∈T1 = 0.11

∈T2 = 0.21

∈T3  = 0.25

note : бTi = k ∈Ti^h

∴ 49300 = k ( 0.11 )^h ----- ( 1 )

   61300 = k ( 0.21)^h ------ ( 2 )

solving equations 1 and 2 simultaneously

49300/61300 = ( 0.11 / 0.21 )^h

0.804 = (0.52 )^h

next step : apply logarithm

log  ( 0.804 ) = log(0.52)^h

h = log 0.804 /  log (0.52)

  =  0.33

back to equation 1

49300 = k ( 0.11 )^0.33

k = 49300 / (0.11)^0.33

 = 102138

therefore бT3 = K (0.25)^h

                        = 102138 ( 0.25 )^ 0.33  

                         = 64640.92