A 3-ft-diameter duct is used to carry ventilating air ( , ) into a vehicular tunnel at a rate of 11000 ft3/min. Tests show that the pressure drop is 1.2 in. of water per 1500 ft of duct. What is (a) the value of the friction factor for this duct and (b) the approximate size of the equivalent roughness of the surface of the duct

Answer:

true

Explanation:

true An electromagnet is formed when a coil of wire wrapped around an iron core is hooked up to a dry cell battery. The current traveling through the wire sets up a magnetic field around the wire. TRUE or FALSE

This question is not complete, the complete question is;

A thin-walled pressure vessel 6-cm thick originally contained a small semicircular flaw (radius 0.50-cm) located at the inner surface and oriented normal to the hoop stress direction. Repeated pressure cycling enabled the crack to grow larger. If the fracture toughness of the material is [tex]88 Mpam^\frac{1}{2}[/tex] , the yield strength equal to 1250 MPa, and the hoop stress equal to 300 MPa, would the vessel leak before it ruptured

Answer:

length of crack is 5.585 cm

we will observe that, the length of crack (5.585 cm) is less than the vessel thickness (6 cm) Hence, vessel will not leak before it ruptures

Explanation:

Given the data in the question;

vessel thickness = 6 cm

fracture toughness k = [tex]88 Mpam^\frac{1}{2}[/tex]

yield strength = 1250 MPa

hoop stress equal = 300 MPa

we know that, the relation between fracture toughness and crack length is expressed as;

k = (1.1)(2/π)(r√(πa))  

where k is the fracture toughness, r is hoop stress and a is length of crack

so we rearrange to find  length of crack

a = 1/π[( k / 1.1(r)(2/π)]²

a = 1/π[( kπ / 1.1(r)(2)]²

so we substitute  

a = 1/π [( 88π / 1.1(300)(2/π)]²    

a = 1/π[ 0.1754596 ]

a = 0.05585 m

a = 0.05585 × 100 cm

a = 5.585 cm  

so, length of crack is 5.585 cm

we will observe that, the length of crack (5.585 cm) is less than the vessel thickness (6 cm) Hence, vessel will not leak before it ruptures

Answer:

False ( B )

Explanation:

considering that the wind turbine is a horizontal axis turbine

Power generated/extracted by the turbine can be calculated as

P =  n * 1/2 * p *Av^3

where: n = turbine efficiency

           p = air density

           A = πd^2 / 4

           v =  speed

From the above equation it can seen that increasing the Blade radius by 10% will increase the Blade Area which will in turn increase the value of the power extracted by the wind turbine

Answer:

I don't know plead hrdffffdddffff

Answer:

(a) 6.36 mg/L

(b) 5.60 mg/L

Explanation:

(a)

Using the formula below to find the required ultimate BOD of the mixture.

[tex]L_o = \dfrac{Q_wL_w+ Q_rL_r}{Q_W+Q_r}[/tex]

where;

Q_w = volumetric flow rate wastewater

Q_r = volumetric flow rate of the river just upstream of the discharge point

L_w = ultimate BOD of wastewater

Replacing the given values:

[tex]L_o = \dfrac{(1 \ m^3/L ) (40 \ mg/L) + (10 \ m^3/L) (3 \mg/L)}{(1m^3/L) +(10 \ m^3/L)} \\ \\ L_o = 6.36 \ mg/L[/tex]

(b)

The Ultimate BOD is estimated as follows:

Recall that:

[tex]time(t) = \dfrac{distance }{speed}[/tex]

replacing;

distance with 10000 m and speed with [tex]\dfrac{11 \ m^3/s}{55 \ m^2}[/tex]

[tex]time =\dfrac{10000 \ m}{\Bigg(\dfrac{11 \ m^3/s}{55 \ m^2}\Bigg)}\Bigg(\dfrac{1 \ hr}{3600 \ s}\Bigg) \Bigg(\dfrac{1 \ day}{24 hr}\Bigg)[/tex]

time (t) = 0.578 days

Finally; [tex]L_t = L_oe^{-kt}[/tex]

here;

k = rate of coefficient reaction

[tex]L_ t= (6.36) \times e^{-(0.22/day)(0.5758 \ days)}\\ \\ \mathbf{L_t =5.60 \ mg/L}[/tex]

Thus, the ultimate BOD = 5.60 mg/L

Answer:

1040 steel will be a possible candidate for this application since : Yield strength > stress

Explanation:

The 1040 steel would  be a possible candidate for this application because the stress experienced by the load is Lesser than its Yield strength

Given that 1040 steel has the following parameter values

Modulus of elasticity ( GPa )  = 205

Yield strength ( Mpa ) =  450

Poisson's ratio = 0.27

limitation of = 1.5 x 10^-2 mm.

stress = Tensile load / area of steel

          = 18,000 N / 4.418 * 10^-5 m^2

          = 407 .424 Mpa

Answer:

3.42 kW

Explanation:

calculate the mass flow rate of the water = density * velocity * area

= 13.3 kg/s

where Area of pipe  =  π/4 * d^2 =  π/4 * 0.075^2 m  =  0.0044 m^2

given that : 1 Kg = 1 N s^2 / m,      1 Watt = 1 N m /s

calculate : power produced by discharge pipe

Discharge Pressure x Volume flow =

Pressure  * Area * velocity

P  * 0.0044 * 3  = 2253 N m /s

calculate: power due the mass of water  

mass of water * 2  = 258 N m/s

where:  mass of water = density * volume

Calculate power produced due the velocity of exiting water

= m * V2/2 = 59.4 N m/s

hence The  power added to fluid = 2253 watts

power added to the fluid by pump = 2253 / 0.75 = 3.42 kW

Answer:

the disadvantages can be people's thoughts about them and also a rival resort nearby which looks more posh pls mark brainliest i need 5 more for next rank thank you

Explanation:

Here are the eight stages of the recruitment process and what Garth might expect to occur or carry out at each stage in the explanation part.

The process of identifying, attracting, and selecting qualified candidates for a job opening in an organisation is known as recruitment.

Here are the eight stages of the recruitment process, as well as what Garth might expect to happen or do at each stage:

Thus, these are the stages of recruitment.

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Technicians A is correct in the given scenario. The correct option is C.

Exhaust gas is a byproduct of combustion that exits the tailpipe of an internal combustion engine.

It consists of a gas mixture that includes carbon dioxide (CO2), carbon monoxide (CO), nitrogen oxides (NOx), hydrocarbons (HC), and particulate matter (PM).

Technician B is mistaken. CO2 levels in the exhaust should be less than 15%, preferably between 13% and 14.5% for petrol engines and 11% to 13% for diesel engines.

High CO2 levels can actually indicate inefficient engine operation, as it means that not all of the fuel in the engine is being burned and is being wasted as exhaust.

Thus, C is the correct answer. A technician is correct.

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Answer:

15625 moles of methane is present in this gas  deposit

Explanation:

As we know,

PV = nRT

P = Pressure = 230 psia = 1585.79 kPA

V = Volume = 980 cuft = 27750.5 Liters

n = number of moles

R = ideal gas constant = 8.315

T = Temperature = 150°F = 338.706 Kelvin

Substituting the given values, we get -

1585.79 kPA * 27750.5 Liters = n * 8.315 * 338.706 Kelvin

n = (1585.79*27750.5)/(8.315 * 338.706) = 15625

Answer:

Wmax =  750 kw  < power developed ( 1000kw ) for a reversible    the cycle is Impossible

Explanation:

Hot reservoir Temperature = 1000 K

Cold reservoir Temperature = 500 K

Heat transfer ( energy received by Hot reservoir ) ( Q ) = 1500 kW

Heat transfer ( energy received by Cold reservoir via Hot reservoir ) = 1000 Kw

Calculate the rate of entropy production

The higher the entropy production the less efficient the system

Δs = Cp In ( T2 / T1 )

power developed = 1000 kW

considering that the cycle is reversible and the constant volume or constant pressure of the substance in the thermodynamic cycle is not given we will use the efficiency to determine if the cycle is possible or not

Л = efficiency

∴Л = 1 - T2 / T1 = 1 - ( 500 / 1000 ) = 0.5

note as well that;  Л = work output / work input = Wmax / Q

                                  = 0.5 = Wmax / Q

∴ rate of entropy production = Q ( 0.5 ) = 1500 * 0.5 = 750 kw

Given that Wmax =  750 kw  < power developed ( 1000kw ) for a reversible    the cycle is Impossible

Answer:

  (√6)/3 ≈ 0.8165

Explanation:

The RMS value is the square root of the mean of the square of the waveform over one period. It will be ...

  [tex]\displaystyle\sqrt{\frac{1}{T}\left(\int_{\frac{T}{4}}^{\frac{3T}{4}}{1^2}\,dt+\int_{\frac{3t}{4}}^{\frac{5t}{4}}{(\frac{-4}{T}}(t-T))^2\,dt\right)}=\sqrt{\frac{1}{T}\left(\frac{T}{2}+\left.\frac{16}{T^2}\cdot\frac{1}{3}(t-T)^3\right|_{\frac{3t}{4}}^{\frac{5t}{4}}\right)}\\\\=\sqrt{\frac{1}{T}\left(\frac{T}{2}+\frac{T}{6}\right)}=\sqrt{\frac{2}{3}}=\boxed{\frac{\sqrt{6}}{3}\approx0.8165}[/tex]

Answer:

C. throttle

Explanation:

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Answer:

prove that | S | = | E | ; every element of S there is an Image on E , while not every element on E has an image on S

Explanation:

Given that S = { p q |p, q are prime numbers greater than 0}

                    E = {0, −2, 2, −4, 4, −6, 6, · · · }

To prove  by constructing a bijection from S to E

detailed  solution attached below

After the bijection :

prove that | S | = | E | :  every element of S there is an Image on E , while not every element on E has an image on S

∴ we can say sets E and S are infinite sets

Answer:

hello your question has some missing values below are the missing values

Mirror Radius (mm) Bending Failure Stress (MPa)

.603                                         225

.203                                         368

.162                                          442

answer : 191 mPa

Explanation:

Determine the stress present at the time of fracture for the original plate

Bending stress ∝ 1 / ( mirror radius )^n ------ ( 1 )

at 0.603  bending stress = 225

at 0.203  bending stress = 368

at 0.162  bending stress = 442

applying equation 1   determine the value of n for several combinations

 ( 225 / 368 ) = ( 0.203 / 0.603 )^n

hence : n = 0.452

also

 ( 368/442 ) = ( 0.162 / 0.203 ) ^n

hence : n = 0.821

also

( 225 / 442 ) = ( 0.162 / 0.603 ) ^n

hence : n = 0.514

Next determine the average value of n

n ( mean value ) =  ( 0.452 + 0.821 + 0.514 ) / 3 = 0.596

Calculate estimated stress present at the time of fracture for the original plate

= bending stress at x =  0.796 / bending stress at x = 0.603

= x / 225 = ( 0.603 / 0.796 ) ^ 0.596

therefore X ( stress present at the time of fracture of original plate )

     = 225 * 0.84747

     =  191 mPa

Solution :

Given :

Velocity, [tex]$V_1 =100$[/tex] cm/sec

Pressure,  [tex]$P_1 = 120 $[/tex] mm Hg

Then, [tex]$P_1 = \rho_1 g h$[/tex]

[tex]$P_1 = 0.120 \times 13.6 \times 1000 \times 9.81$[/tex]

    = 16.0092 kPa

[tex]$P_2 = 110 $[/tex] mm Hg

[tex]$P_2 = \rho_2 g h$[/tex]

    [tex]$= 0.110 \times 13.6 \times 1000 \times 9.81$[/tex]

    = 14.675 kPa

Then blood is incompressible,

[tex]$A_1v_1=A_2v_2$[/tex]

[tex]$\frac{\pi}{4}(25)^2\times 100=\frac{\pi}{4}(21)^2\times v_2$[/tex]

[tex]$v_2=141.72 \ cm/s$[/tex]

Then the linear momentum conservation fluid :

(Blood ) in y - direction

[tex]$P_1A_1+ P_2A_2-F_g = m_2v_2-m_1v_1$[/tex]

[tex]$m_1=m_2=P_1A_1v_1$[/tex]

              [tex]$=1.50 \times \frac{\pi}{4}\times (0.025)^2 \times 1.00$[/tex]

              = 0.515 kg/ sec

Then the linear conservation of momentum of blood in y direction.

[tex]$P_1A_1+ P_2A_2-F_g = m_2v_2-m_1v_1$[/tex]

[tex]$16.0092 \times 1000 \times \frac{\pi}{4} \times (0.025)^2+14.675 \times 1000\times \frac{\pi}{4}\times (0.021)^2$[/tex]

[tex]$-F_y=0.515(-1.4172-1)$[/tex]

7.858+5.0828- Fy = 0.515(-2.4172)

Fy = 14.1856 N

Answer:

Option A is the correct answer. The bus traveled at 50 mph for 20 minutes

Explanation:

The complete question is

Which of the following choices best describes velocity of a bus traveling along its route? A. The bus traveled at 50 mph for 20 minutes. B. The airplane traveled southwest at 280 mph. C. The car went from 35 mph to 45 mph. D. The train made several stops, with an average rate of 57 mph.

Solution

In option A the bus is travelling at a speed of 50 miles per hour. This describes the velocity of bus along its route.

The other options are about the velocity of airplane, car and train

Answer:

Wind energy is converted to Mechanical energy  which is then converted in to  electrical energy

Explanation:

In a wind mill the following energy conversions take place

a) Wind energy is converted into Mechanical energy (rotation of rotor blades)

b) Mechanical energy is converted into electrical energy (by using electric motor)

This electrical energy is then used for transmission through electric lines.

Answer:

a)  927 MPa

b)The specimen will experience fracture

Explanation:

a) Calculate critical stress for brittle fracture

σ = fracture toughness / (y √ π * surface crack)

  = 45  /  ( 1  [tex]\sqrt{\pi * ( 0.75*10^-3)}[/tex]  )

  = 927 MPa

b) since critical stress( 927 MPa)  < 1000 MPa

hence : The fracture will occur

Answer:

a. 41

b. i. 15 mA ii. 625 mA

c. 192 Ω

Explanation:

Here is the complete question

A high-voltage discharge tube is often used to study atomic spectra. The tubes require a large voltage across their terminals to operate. To get the large voltage, a step-up transformer is connected to a line voltage (120 V rms) and is designed to provide 5000 V (rms) to the discharge tube and to dissipate 75.0 W. (a) What is the ratio of the number of turns in the secondary to the number of turns in the primary? (b) What are the rms currents in the primary and secondary coils of the transformer? (c) What is the effective resistance that the 120-V source is subjected to?

Solution

(a) What is the ratio of the number of turns in the secondary to the number of turns in the primary?

For a transformer N₂/N₁ = V₂/V₁

where N₁ = number of turns of primary coil, N₂ =number of coil of secondary, V₁ = voltage of primary coil = 120 V and V₂ = voltage of secondary coil = 5000 V

So,  N₂/N₁ = V₂/V₁

N₂/N₁ = 5000 V/120 V = 41.6 ≅ 41 (rounded down because we cannot have a decimal number of turns)

(b) What are the rms currents in the primary and secondary coils of the transformer?

i. The rms current in the secondary

We need to find the current in the secondary from

P = IV where P = power dissipated in secondary coil = 75.0 W, I =rms current in secondary coil and V = rms voltage in secondary coil = 5000 V

P = IV

I = P/V = 75.0 W/5000 V = 15 × 10⁻³ A = 15 mA

ii. The rms current in the primary

Since N₂/N₁ = V₂/V₁ = I₁/I₂

where N₁ = number of turns of primary coil, N₂ =number of coil of secondary, V₁ = voltage of primary coil = 120 V, V₂ = voltage of secondary coil = 5000 V, I₁ = current in primary coil and I₂ = current in secondary coil = 15 mA

So, V₂/V₁ = I₁/I₂

V₂I₂/V₁ = I₁

I₁ = V₂I₂/V₁

= P/V₁

= 75.0 W/120 V

= 0.625 A

= 625 mA

(c) What is the effective resistance that the 120-V source is subjected to?

Using V = IR where V =  voltage = 120 V, I = current in primary = 0.625 A and R = resistance of primary coil

R = V/I

= 120 V/0.625 A

= 192 V/A

= 192 Ω

Answer:

12ma²

Explanation:

The moment of inertia I = ∑mr² where m = mass and r = distance from shaft

Since we have three masses,

So, I = m₁r₁² + m₂r₂² + m₃r₃² where m₁ = first mass = m, m₂ = second mass = 3m and m₃ = third mass = 2m. Also, r₁ = distance of first mass from shaft = a, r₂ = distance of second mass from shaft = a and r₃ = distance of third mass from shaft = 2a.

I = m₁r₁² + m₂r₂² + m₃r₃²

I = ma² + 3ma² + 2m(2a)²

I = ma² + 3ma² + 2m(4a²)

I = ma² + 3ma² + 8ma²

I = 12ma²

Answer:

[tex]\mathbf{C_{10} = 137.611 \ kN}[/tex]

Explanation:

From the information given:

Life requirement = 40 kh = 40 [tex]40 \times 10^{3} \ h[/tex]

Speed (N) = 520 rev/min

Reliability goal [tex](R_D)[/tex] = 0.9

Radial load [tex](F_D)[/tex] = 2600 lbf

To find C10 value by using the formula:

[tex]C_{10}=F_D\times \pmatrix \dfrac{x_D}{x_o +(\theta-x_o) \bigg(In(\dfrac{1}{R_o}) \bigg)^{\dfrac{1}{b}}} \end {pmatrix} ^{^{^{\dfrac{1}{a}}[/tex]

where;

[tex]x_D = \text{bearing life in million revolution} \\ \\ x_D = \dfrac{60 \times L_h \times N}{10^6} \\ \\ x_D = \dfrac{60 \times 40 \times 10^3 \times 520}{10^6}\\ \\ x_D = 1248 \text{ million revolutions}[/tex]

[tex]\text{The cyclindrical roller bearing (a)}= \dfrac{10}{3}[/tex]

The Weibull parameters include:

[tex]x_o = 0.02[/tex]

[tex](\theta - x_o) = 4.439[/tex]

[tex]b= 1.483[/tex]

∴

Using the above formula:

[tex]C_{10}=1.4\times 2600 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{1}{\dfrac{10}{3}}}[/tex]

[tex]C_{10}=3640 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{3}{10}}[/tex]

[tex]C_{10} = 3640 \times \bigg[\dfrac{1248}{0.9933481582}\bigg]^{\dfrac{3}{10}}[/tex]

[tex]C_{10} = 30962.449 \ lbf[/tex]

Recall that:

1 kN = 225 lbf

∴

[tex]C_{10} = \dfrac{30962.449}{225}[/tex]

[tex]\mathbf{C_{10} = 137.611 \ kN}[/tex]

Answer:

a) the tubular resistance of the blood vessel is 6.88 Gpa.s/m³.

b) the flow rate is 5.8 ml/min

Explanation:

Given the data in the question;

Length of blood vessel L = 1 m

radius r = 1 mm = 0.001 m

blood viscosity μ = 2.7 × 10⁻³ pa.s = 2.7 × 10⁻³ × 10⁻⁹ Gpa.s = 2.7 × 10⁻¹² Gpa.s

Now, we know that Resistance = 8μL / πr⁴

so we substitute

Resistance = [8 × (2.7 × 10⁻¹²) × 1] / [π(0.001)⁴]

Resistance = [2.16 × 10⁻¹¹] / [3.14159 × 10⁻¹²]

Resistance = 6.8755 ≈ 6.88 Gpa.s/m³

Therefore, the tubular resistance of the blood vessel is 6.88 Gpa.s/m³.

b)

blood pressure at the inlet of the vessel = 43 mm Hg

blood pressure at the outlet of the vessel = 38 mm Hg

flow rate = ?

we know that;

flow rate Q = ΔP / R

where ΔP is change in pressure and R is resistance.

ΔP = Inlet pressure - Outlet pressure = 43 - 38 = 5 mm Hg =  665 pa

R = 6.8755 Gpa.s/m³ = { 6.8755 × 10⁹ / 60 × 10⁶ } = 114.5916 pa.min.ml⁻¹

so we substitute

Q =  665 pa / 114.5916 pa.m.ml⁻¹

Q = 5.8 ml/min

Therefore, the flow rate is 5.8 ml/min

Answer:

A) 1040 steel is not a possible candidate for this application

B) 35.94%

Explanation:

Initial length = 100 mm =  0.1 m

Initial diameter ( d ) = 7.5 mm = 0.0075 m

Tensile load ( p ) = 18,000 N

Condition : The 1040 steel must not experience plastic deformation or a diameter reduction of more than 1.5 * 10^-5 m

A) would the 1040 steel be a possible candidate for this application

Yield strength of 1040 steel < stress  ( in order to be a possible candidate )

stress = p / A0 = ( 18000 ) / ( [tex]\frac{\pi }{4}[/tex] ) * 0.0075^2

                      = 18,000 / (4.418 * 10^-5 )   =  407.424 MPa

Yield strength of 1040 steel = 450 MPa

stress = 407.424 MPa

∴ Yield strength ( 450 MPa ) > stress ( 407.424 MPa )  

Therefore 1040 steel is not a possible candidate for this application

B) Determine How much cold work would be required to reduce the diameter of the steel to 6.0 mm

Area1 = ( [tex]\frac{\pi }{4}[/tex] ) ( 0.006 )^2 = 2.83 * 10^-5 m^2

therefore % of cold work done = ( A0 - A1 ) / A0  * 100 = 35.94%

Answer:

surface temp of fuel rod = 678.85 K

Explanation:

Given data :

D1 = 25 mm

D2 = 50 mm

T2 = 335 k

T∞ = 300 k

hconv = 0.15 w/m^2.k

ε2 = 0.05

ε1 = 1

Determine energy at Q23

Q23 = Qconv + Qrad

attached below is the detailed solution

Insert given values into equation 1 attached below to obtain the surface temperature of the fuel rod ( T1 )