Find the velocity and rate of flow of water through a rectangular channel of 6m wide and 3m deep when it's running full. The channel is having a bed slope of 1 in 2000. Take Chezy's coefficient, C=55

Answer:

the factor of safety was used in the design of the cable is 2.6146

Explanation:

Given the data in the question;

Load on the main capable [tex]P_{initial[/tex] = 2600000 lb

number of parallel wires n = 1470

Diameter d = 0.16 in

average ultimate strength [tex]S_{ultimate[/tex] = 230000 psi

First we calculate the Load acting on each cable;

[tex]P_{initial[/tex] = P × n

P = [tex]P_{initial[/tex] / n

we substitute

P = 2600000 lb / 1470

P = 1768.70748 lb

Next we determine the working stress acting in a member;

[tex]S_{working[/tex] = P/A

{ Area A = [tex]\frac{\pi }{4}[/tex]d² }

[tex]S_{working[/tex] = P / [tex]\frac{\pi }{4}[/tex]d²

we substitute

[tex]S_{working[/tex] = 1768.70748 / [tex]\frac{\pi }{4}[/tex](0.16)²

[tex]S_{working[/tex] = 1768.70748 / 0.02010619298

[tex]S_{working[/tex] = 87968.29 psi

Now we calculate the factor of safety F.S

F.S = [tex]S_{ultimate[/tex] / [tex]S_{working[/tex]

we substitute

F.S = 230000 psi / 87968.29 psi

F.S = 2.6145785 ≈ 2.6146

Therefore, the factor of safety was used in the design of the cable is 2.6146

Answer:

the wire that can carry much current is AWG 24

AWG 24 is the  wire can carry a higher current. The diameter of the wire increases with decreasing gauge. The electrical resistance to the signals decreases with increasing wire diameter. Hence, option D is correct.

Any current above 10 mA has the potential to deliver an unpleasant to severe shock, while 200 mA and above is considered lethal. Despite the fact that currents above 200 mA might cause serious burns and unconsciousness, they usually do not cause death if the sufferer receives timely medical attention.

Higher voltage and lower current are frequently more effective combinations. The amount of current that a wire is rated to handle is frequently indicated by its capacity rating. People can use the same cable to drive a load twice as large by doubling the voltage and maintaining the same current.

Thus, option D is correct.

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Answer:

Yes

Explanation:

Answer:

true

Explanation:

hehehe

Answer:

Se obtendrán 116.66 litros de jugo concentrado, y el agua evaporada será por un total de 883.33 litros.

Explanation:

Dado que para conseguir jugo de naranja concentrada, se parte de un extracto con 7% en peso de sólidos el cual se mete a un evaporador al vacío, y en el evaporador se elimina el agua necesaria para que el jugo salga con una concentración del 60% de peso de sólidos, si se introducen al proceso 1000 kg/h de jugo diluido, para calcular la cantidad de agua evaporada y de jugo concentrado saliente se debe realizar el siguiente cálculo;

1000 x 0.07 = 70

60 = 70

100 = X

100 x 70 / 60 = X

7000 / 60 = X

116.66 = X

Por lo tanto, se obtendrán 116.66 litros de jugo concentrado, y el agua evaporada será por un total de 883.33 litros.

here's your answer..

Answer:

Both are correct.

Explanation:

Both the technician A and B are correct. Some relays are equipped with a suppression diode in parallel windings while some relays are equipped with resistors. This is due to different requirements of the electromagnetic objects. Some require resistors to stop the flow of current towards the magnet.

Explanation:

Answer ⬇️

Social and ethical issues in engineering, ethical principles of engineering, professional code of ethics, some specific social problems in engineering practice: privacy and data protection, corruption, user orientation, digital divide, human rights, access to basic services.

➡️dhruv73143(⌐■-■)

Answer:

National Electrical Code

Explanation:

The National Electrical Code (NEC) is the foundational code that establishes the minimum requirements for electrical installation work. The NEC protects individuals and properties from being discriminated against dangers associated with the use of electricity. For electrical installation work, local and state electrical standards and codes should be consulted in addition to the NEC.

Section 406.4 of the NEC refers to On the same premises, receptacles linked to circuits with distinct voltages frequencies, or kinds of current (AC or DE) must be designed in such a way that the attachment plugs utilized on these circuits cannot be interchanged.

Answer:

y = 0.834X - 1.58015

Slope = 0.8340 ; Intercept = - 1.5802

y = 40.9539

19.93

0.9765

Explanation:

X: Rainfall volume

6

12

14

16

23

30

40

52

55

67

72

81

96

112

127

Y : Runoff

4

10

13

14

15

25

27

48

38

46

53

72

82

99

100

The scatterplot shows a reasonable linear trend between the Rainfall volume and run off.

The estimated regression equation obtained using a linear regression calculator is :

y = 0.834X - 1.58015

y = Runoff ; x = Rainfall volume

Slope = 0.8340 ; Intercept = - 1.5802

Point estimate for Runoff, when, x = 51

y = 0.834X - 1.58015

y = 0.834(51) - 1.58015

y = 40.95385

y = 40.9539

d.)

Point estimate for standard deviation :

s = 5.145

σ = s * √n

σ = √15 * 5.145

= 19.93

e.)

r² = Coefficient of determination gives the proportion of explained variance in Runoff due to the regression line. From the model output, the r² value = 0.9765. Which means That about 97.65% Runoff is due to Rainfall volume.

Answer:

Resistance of the lamp (r) = 156.52 ohm (Approx.)

Explanation:

Given:

Total power p = 140 W

Resistance of the coffee maker = 300 Ohm

Voltage v = 120 V

Find:

Resistance of the lamp (r)

Computation:

We know that

p = v² / R

SO,

Total power p = [Voltage²/Resistance of the lamp (r)] + [Voltage²/Resistance of the coffee maker]

140 = [120² / r] + [120²/300]

140 = 120²[1/r + 1/300]

140 = 14,400 [1/r + 1/300]

Resistance of the lamp (r) = 156.52 ohm (Approx.)

Answer:

a)  [tex]Tb=1845.05K[/tex]

b)  [tex]Q=1000.25KJ[/tex]

c)  [tex]\mu=0.59[/tex]

Explanation:

From the question we are told that:

Temperature x [tex]Tx=450c=>723K[/tex]

Pressure x [tex]Px=2.5MPa[/tex]

Temperature y [tex]Ty=600c=>873K[/tex]

Pressure y [tex]Py=0.45MPa[/tex]

Let

Air atmospheric temperature be [tex]25c[/tex]

Therefore

Temperature [tex]Ta=25+273=298k[/tex]

Generally the equation for Otto cycle is mathematically given by

 [tex]\frac{Tb}{Tx}=\frac{Ty}{Ta}[/tex]

 [tex]Tb=\frac{873*723}{298}[/tex]

 [tex]Tb=2118.05[/tex]

Therefore the peak cycle temperature (°C)

 [tex]Tb=2118.05k[/tex]

 [tex]Tb=2118.05-273[/tex]

 [tex]Tb=1845.05K[/tex]

Generally the equation for Heat addition is mathematically given by

 [tex]Q=Cv(Tb-Tx)[/tex]

 [tex]Q=Cv(2118.05-723)[/tex]

 [tex]Q=1000.25KJ[/tex]

Generally the equation for Thermal  efficiency is mathematically given by

 [tex]\mu=1-\frac{Ta}{Tx}[/tex]

 [tex]\mu=1-\frac{298}{723}[/tex]

 [tex]\mu=0.59[/tex]

Answer:

An algotherum is a finite set of sequential instructions to accomplish a task where instructions are written in a simple English language

Answer:

I dont kno

Explanation:

Im so sorry

Answer:

Explanation:

here is your answer:

The responsivity will be 0.45 A / W. Then the quantum efficiency will be 80%.

A Si pin photodiode has an active light-receiving area of a diameter of 0.4 mm.

When radiation of wavelength 700 nm (red light and intensity 0.1 mW cm^-2 is incident it generates a photocurrent of 56.6 nA.

Diameter (d) = 0.4 mm = 0.04 cm

Then the area will be

Area (A) = π / 4 x d²

Area (A) = π / 4 x 0.04²

Area (A) = 1.26 x 10⁻³ square cm

Then the incident power will be

P₀ = intensity of light x area

P₀ = 1.26 x 10⁻³ x 0.1 x 10⁻³

P₀ = 0.126 x 10⁻⁶ μW

Then the responsivity will be

R = photocurrent / power

R = 56.6 x 10⁻⁹ / 0.126 x 10⁻⁶

R = 0.4492

R ≅ 0.45 A / W

Then the quantum efficiency will be

η = Rhc / qλ

h = plank constant

c = speed of light

q = charge of an electron

Then we have

η = 0.45 x 6.62 x 10⁻³⁴ x 3 x 10⁸ / 1.6 x 10⁻¹⁹ x 700 x 10⁻⁹

η = 0.7979

η = 0.8

η = 80%

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a.No it will be not an easy matter for Peter to prove that the Er staff caused peter to lose his promotion

Here's why?

1)The boss may think he might not able to stretch himself in emergency situations

2)He can think he can't handle normal pressure

b.What precautions can be taken to avoid giving confidential information to medical personnel who have no need to see it?

1)Peter should not let the medical personnel about his personal life

2)He might have not said extra information about his job promotion

3)Will it be easy matter for Peter to prove that the ER staff caused Peter to lose his promotion? Explain your answer

No, it will not be easy, Peter has to prove in one or in another situation that he can face the difficulties and be able to handle anxiety attacks in the future.

Hence above answers are suitable for the given situation

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Answer:

The answer is "shear".

Explanation:

Every transformation could be shown by some kind of conventional matrix.

Its standard matrices of shape k here could be any real value enabling shear transformation to parallel to the y axis.

[tex]\left[\begin{array}{cc}1&0\\k&1\\\end{array}\right][/tex]

This coefficient matrix A is the standard matrix during transformation. With the use of a the transformation could be entered into T(x)=Ax

And standard transfer function is standard.

[tex]\left[\begin{array}{cc}1&0\\6&1\\\end{array}\right][/tex]

The matrix in the form of the shear matrix.

Answer:

wa 4tefaw gfawe f

Explanation:

so you do e Fse gaewr gware gfawe agfawe gwar g

Answer:

i don't know

Explanation:

sorry

Answer:

The units on the rate of heat transfer are Joule/second, also known as a Watt.

Explanation:

Heat flow is calculated using the rock thermal conductivity multiplied by the temperature gradient. The standard units are mW/m2 = milli Watts per meter squared. Thus, think of a flat plane 1 meter by 1 meter and how much energy is transferred through that plane is the amount of heat flow.

hope it helps .

The rate of heat transfer is measured in Joules per second, also known as Watts.

Heat transfer is a thermal engineering discipline that deals with the generation, use, conversion, and exchange of thermal energy between physical systems.

Heat transfer mechanisms include thermal conduction, thermal convection, thermal radiation, and energy transfer via phase changes.

The rate of heat transfer through a unit thickness of material per unit area per unit temperature difference is defined as thermal conductivity. Thermal conductivity varies with temperature and is measured experimentally.

Heat is typically transferred in a combination of these three types and occurs at random. Heat transfer rate is measured in Joules per second, also known as Watts.

Thus, Joules per second or watts is the unit of rate of heat transfer.

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Answer:

1+1×1 multiplay then you get the answer

Answer:

the rate of entropy change of the air is -0.1342 kW/K

the assumptions made in solving this problem

- Air is an ideal gas.

- the process is isothermal ( internally reversible process ). the change in internal energy is 0.

- It is a steady flow process

- Potential and Kinetic energy changes are negligible.

Explanation:

Given the data in the question;

From the first law of thermodynamics;

dQ = dU + dW ------ let this be equation 1

where dQ is the heat transfer, dU is internal energy and dW is the work done.

from the question, the process is isothermal ( internally reversible process )

Thus, the change in internal energy is 0

dU = 0

given that; Air is compressed by a 40-kW compressor from P1 to P2

since it is compressed, dW = -40 kW

we substitute into equation 1

dQ = 0 + ( -40 kW )

dQ = -40 kW

Now, change in entropy of air is;

ΔS[tex]_{air[/tex] = dQ / T

given that T = 25 °C = ( 25 + 273.15 ) K = 298.15 K

so we substitute

ΔS[tex]_{air[/tex] =  -40 kW / 298.15 K

ΔS[tex]_{air[/tex] =  -0.13416 ≈ -0.1342 kW/K

Therefore, the rate of entropy change of the air is -0.1342 kW/K

the assumptions made in solving this problem

- Air is an ideal gas.

- the process is isothermal ( internally reversible process ). the change in internal energy is 0.

- It is a steady flow process

- Potential and Kinetic energy changes are negligible.

Answer:

the theoretical fracture strength of the brittle material is 5.02 × 10⁶  psi

Explanation:

Given the data in the question;

Length of surface crack α = 0.25 mm

tip radius ρ[tex]_t[/tex] = 1.2 × 10⁻³ mm

applied stress σ₀ = 1200 MPa

the theoretical fracture strength of a brittle material = ?

To determine the the theoretical fracture strength or maximum stress at crack tip, we use the following formula;

σ[tex]_m[/tex] = 2σ₀[tex]([/tex] α / ρ[tex]_t[/tex] [tex])^{\frac{1}{2}[/tex]

where α is the Length of surface crack,

ρ[tex]_t[/tex] is the tip radius,

and σ₀ is the applied stress.

so we substitute

σ[tex]_m[/tex] = (2 × 1200 MPa)[tex]([/tex] 0.25 mm / ( 1.2 × 10⁻³ mm ) [tex])^{\frac{1}{2}[/tex]

σ[tex]_m[/tex] = 2400 MPa × [tex]([/tex] 208.3333 [tex])^{\frac{1}{2}[/tex]

σ[tex]_m[/tex] = 2400 MPa × 14.43375

σ[tex]_m[/tex] = 34641 MPa

σ[tex]_m[/tex] = ( 34641 × 145 )psi

σ[tex]_m[/tex] = 5.02 × 10⁶  psi

Therefore, the theoretical fracture strength of the brittle material is 5.02 × 10⁶  psi

Answer:

Explanation:

The missing diagram attached to the question is shown in the attached file below:

The very first thing we need to do in other to solve this question is to determine the mass of both the tractor and the mass of the gravel

For tractor, the mass is:

[tex]m_1 = \dfrac{2400 \ lb }{32.2 \ ft/s^2}[/tex]

[tex]m_1 = 74.53 \ lb.s^2/ft[/tex]

For gravel, the mass is:

[tex]m_2 = \dfrac{900 \ lb}{32.2 \ft/s^2}[/tex]

[tex]m_2 = 27.95 \ lb.s^2/ft[/tex]

From the diagram, let's consider the force along the horizontal components and vertical components;

So,

[tex]\sum F_x = ma_x \\ \\ 2F = (m_1+m_2) a \\ \\ F = \dfrac{1}{2}(74.53 4 + 27.950)lb.s^2/ft(2 \ ft/s^2) \\ \\ F = 102.484 \ lb[/tex]

[tex]\sum F_y = 0 \\ \\ 2N_A+2N_B - 2400 -900 = 0 \\ \\ N_A +N_B = 1650 \ lb[/tex]

Consider the algebraic sum of moments in the plane of A, with counter-clockwise moments being positive.

[tex]\sum M_A = I_o \alpha + \sum ma (d) \\ \\ = -2400 (20) + 2N_B (60) -900(110) = 0 - (74.534)(2)(20) - (27.950)(2)(40)[/tex]

[tex]=-48000 + 2N_B (60) -99000 = -2981.36-2236 \\ \\ = + 2N_B (60) = -2981.36-2236+48000+99000 \\ \\ = + 2N_B (60) = 141782.64 \\ \\ N_B = \dfrac{141782.64}{120} \\ \\ N_B = 1181.522 \ lb[/tex]

Replacing the value of 1181.522 lb for [tex]N_B[/tex] in equation (1)

[tex]N_A[/tex] + 1181.522 lb = 1650 lb

[tex]N_A[/tex] = (1650 - 1181.522)lb

[tex]N_A[/tex] = 468.478 lb

The net reaction on each of the rear wheels now is:)

[tex]F_R = \sqrt{N_A^2 +F^2}[/tex]

[tex]F_R = \sqrt{(468.478)^2 + (102.484)^2}[/tex]

[tex]\mathbf{F_R =479.6 \ lb}[/tex]

Now, we can determine the angle at the end of the rear wheels at which the resultant reaction force is being made in line with the horizontal

[tex]\theta = tan ^{-1}( \dfrac{468.478 }{102.484})[/tex]

[tex]\theta = 77.7^0[/tex]

Finally, the net reaction on each of the front wheels is:

[tex]F_B = N_B[/tex]

[tex]F_B =[/tex] 1182 lb  

Answer:

[tex]\mu_{max}=200Mpa[/tex]

Explanation:

From the question we are told that:

Internally pressurized [tex]P_i=100MPa[/tex]

Tangential Stress [tex]P_t=400mpa[/tex]

Axial stress [tex]P_a=200mpa[/tex]

Generally the equation for maximum normal and shear stresses are experienced by the outer surface is mathematically given by

 [tex]\mu_{max}=|\frac{P_t-P_a}{2}|,|\frac{P_t}{2}|,|\frac{P_t}{2}|[/tex]

Therefore

 [tex]\mu_{max}=|\frac{400-200}{2}|,|\frac{400}{2}|,|\frac{200}{2}|[/tex]

 [tex]\mu_{max}=200Mpa[/tex]

Answer: hello there is a Missing information below is the missing information

surface dimensions of  10m * 8m

answer :

a) 640 m^3/hr

b) 1334.66 m^3

Explanation:

a) Determine the flow rate ( m^3/s ) of the filter handle

Va = Q /A[tex]_{f}[/tex]

where ; Va = filtration rate ( 8.00 m/h ) ,  Q = flow rate of filter handle , A[tex]_{f}[/tex] = surface area ( 10 m * 8 m  )

Q = 8 * ( 10m * 8m ) = 640 m^3 / hr

b) Determine the volume of water needed for backwashing plus rinsing the filter in each filter cycle

Лf = 0.96  ( production efficiency )

Vb + Vr = 0.04 Vf

∴ Vf  =  ( Vb + Vr ) / 0.04  ------ ( 1 )

next step ; determine the volume of filtered water making use of effective filtration rate

Ref = ( Vf - Vb - Vr ) / A[tex]_{f}[/tex]T[tex]_{c}[/tex]

therefore : Vb + Vr = Vf - ( 80 * 52 * 7.7 ) ----  ( 2 )

Input equation 1 into 2

Vb + Vr = ( ( Vb + Vr ) / 0.04  )  - 32032  ---- ( 3 )

Resolve equation 3

hence the Volume of water needed for Backwashing  and rinsing the filter in each filter cycle

= 1334.66 m^3

Answer:

35.92 kpsi

Explanation:

Given data:

diameter of the steel bar d = 0.875 in

Area A = πd^2/4 = π(0.875)^2/4

length L = 15.0 ft

Load P = 21.6 kip

Modulus of elesticity E = 29×10^6 Psi

Assume we are asked to determine axial stress in the bar which is given as

[tex]\sigma = Load, P/ Area, A[/tex]

[tex]\sigma = 4P/\pi d^2[/tex]

substitute the value

[tex]\sigma = \frac{4\times 21.6}{\pi \times (0.875)^2} \\=35.92\ kpsi[/tex]

Answer:

1700 W

Explanation:

The heat transfer rate P = kA(T - T')/d where k = thermal conductivity of wall = 1.7 W/m-K, A = area of wall = 0.5 m × 1.2 m = 0.6 m², T = temperature of inner surface = 1400 K, T = temperature of outer surface = 1150 K and d = thickness of wall = 0.15 m

So, P = kA(T - T')/d

substituting the values of the variables into the equation, we have

P = 1.7 W/m-K × 0.6 m²(1400 K - 1150 K)/0.15 m

P = 1.7 W/m-K × 0.6 m² × 250 K/0.15 m

P = 255 Wm/0.15 m

P = 1700 W

So, the heat transfer rate through the wall is 1700 W