An engineer is applying dimensional analysis to study the flow of air through this horizontal sudden contraction for the purpose of characterizing the pressure drop. The flow is being modeled as constant density and steady. What is the functional relationship of the variables that characterize this situation

Explanation:

Light travels in straight lines

Once light has been produced, it will keep travelling in a straight line until it hits something else. Shadows are evidence of light travelling in straight lines. An object blocks light so that it can't reach the surface where we see the shadow.

Answer:

The two factors are Air and Object

Answer:

B. rotor

Explanation:

The correct answer Is rotor because the others are part of a cars drivetrain

The drive train system exists as a critical element of a vehicle and the transmission exists as an integral part of the drive train. B rotors NOT part of a car's drive train.

A drive train exists not really a single part of your car – it's a set of drive train components that interact with the engine to drive the wheels and different regions of the vehicle to thrust it into motion. These components often contain the transmission, differential, driveshaft, axles, CV joints, and wheels.

Therefore, the correct answer is option B rotors.

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Answer:

The answer is "[tex]25\times 10^{-9}[/tex]".

Explanation:

Using formula:

[tex]f_c=\frac{1}{2\pi RC}\\\\w_c= 4 \frac{krad}{sec}\\\\w_c=2\pi fc\\\\R=w\\\\c=\frac{1}{w_c\ R}\\\\[/tex]

  [tex]=\frac{1}{4 \times 10^3 \times 10\times 10^3}\\\\=\frac{1}{40 \times 10^6 }\\\\=0.025 \times 10^{-6 }\\\\=25\times 10^{-9}[/tex]

Answer:

Explanation:

From the given information;

There is no change or any difference in velocity in between the inlet and the outlet.

Therefore by using Bernoulli's equation, we have:

[tex]\dfrac{V_1^2}{2g}+ \dfrac{P_1}{\gamma}+ z_1 + Epump= \dfrac{V_2^2}{2g}+ \dfrac{P_2}{\gamma}+ z_2+ H_L[/tex]

By dividing like terms on both sides, the equation is reduced to:

[tex]z_1 + E_{pump} = z_2+H_L \\ \\ E_{pump} =(z_2-z_1)+H_L[/tex]

where;

[tex]\Delta z = 400[/tex]

[tex]\Delta z = z_2-z_1[/tex]

[tex]\text{total head loss}= 408.5[/tex]

[tex]E_{pump} =(400)+408.5[/tex]

[tex]E_{pump} = 808.5 \ ft[/tex]

The required power input can be determined by using the formula:

[tex]P= \dfrac{\gamma_wQH_{pump}}{\eta}[/tex]

Assuming the missing pump efficiency = 70% and the flow rate Q= 1.34

Then:

[tex]P= \dfrac{62.40\times 1.34 \times 808.5}{0.7}[/tex]

[tex]P = \dfrac{96576.48 \ ft.lb/s}{550\dfrac{ ft*lb/s}{hp}}[/tex]

P = 175.594 hp

Answer:

Rock and orange juice

Explanation:

The mystery matter to be submerged in water must be a solid, therefore we can eliminate the Lemonade and Milk, and Orange juice and Helium, as these pairs do not contain solids. The graduated cylinder is used to measure the volume of a liquid, therefore the only remaining option is Rock and Orange Juice.

Answer:

The International Building Code (IBC) is a model code that provides minimum requirements to safeguard the public health, safety and general welfare of the occupants of new and existing buildings and structures.

Explanation:

Answer:

2

Explanation:

From the formula R=(ro)A/l resistance depends on the length of the wire, the area of the wire(thickness) and the resistivity(ro) which depends on the material and temperature.

Answer:

At the threads due to shear.

Explanation:

Given :

The allowable tensile stress = 207 MPa

The allowable shear stress = 103 MPa

If a tensile force is applied, the maximum shear stress occurs at the threads of  the bolt. The bolt is most likely to fail at the critical section. The critical cross section is the section having the minimum cross sectional area.

The portion of the bolt having threads has the minimum cross sectional area.

So when the bolt is applied with a tensile force, failure is most likely to take place at the threads due to the shearing force.

The per capita GDP of China in the Calendar year 2021 was found to be around 12,359 U.S. dollars.

GDP termed Gross Domestic Product, has been evaluated with the value producing the economy of the region with the values added with the used products formed to be the less of the economy produced. It has been termed as the measure of the income of a region and not the wealth.

The per capita GDP has been the total income earned by a person in a region during a specified period of time. The calculation has been made by dividing the total gross income of the region by the total population.

China has been the world's most populous country in the East Asian region. It has been found that the per capita GDP of China is low because of its large population. In the calendar year 2021, the per capita GDP of China was 12,359 U.S. dollars.

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Answer:

[tex]GM<0[/tex]

So the bouy does not float with its axis vertical

Explanation:

From the question we are told that:

Diameter [tex]d=2m[/tex]

Length [tex]l=2.5m[/tex]

Weight [tex]W=22kN[/tex]

Specific weight of sea water [tex]\mu= 10.25kN/m^3[/tex]

Generally the equation for weight of cylinder is mathematically given by

Weight of cylinder = buoyancy Force

[tex]W=(pwg)Vd[/tex]

Where

[tex]V_d=\pi/4(d)^2y[/tex]

Therefore

[tex]22*10^3=10.25*10^3 *\pi/4(2)^2y\\\\\22*10^3=32201.3247y\\\\\y=1.5m[/tex]

Therefore

Center of Bouyance B

[tex]B=\frac{y}{2}=0.26m\\\\B=0.75[/tex]

Center of Gravity

[tex]G=\frac{I.B}{2}=2.6m[/tex]

Generally the equation for\BM is mathematically given by

[tex]BM=\frac{I}{vd}\\\\BM=\frac{3.142/64*2^4}{3.142/4*2^2*0.5215}\\\\BM=0.479m\\\\[/tex]

Therefore

[tex]BG=2.6-0.476\\\\BG=0.64m[/tex]

Therefore

[tex]GM=BM-BG\\\\GM=0.479m-0.64m\\\\GM=-0.161m\\\\[/tex]

Therefore

[tex]GM<0[/tex]

So the bouy does not float with its axis vertical

Complete Question:

Problem 8 Determine the horsepower required to compress 1 lbm/min of ethylene oxide from 70 °F and 1 atm to 250 psia. The compressor has an efficiency of 75%. The molar heat capacity of ethylene oxide is given by

C_p=10.03+0.0184T  C_p[=]Btu/lbmole- "F ; T[=] °F C,

Answer:

[tex]P'=0.377hp[/tex]

Explanation:

From the question we are told that:

Initial Temperature T_1=70 F

Final Temperature [tex]T_2=250pisa =114.94F[/tex]

Efficiency [tex]E=75\%=0.75[/tex]

Generally the equation for Work-done  is mathematically given by

 [tex]W=\int C_pT[/tex]

 [tex]W=10.03(114.94-70 )+0.0184((114.94)^2-70^2 )[/tex]

 [tex]W=527.21btu/ibmole[/tex]

 [tex]W=11.982btu/ibm[/tex]

Generally the equation for Efficiency  is mathematically given by

 [tex]E=\frac{isotropic Power}{Actual P'}[/tex]

 [tex]E=\frac{P}{P'}[/tex]

Since

Isotropic Power

  [tex]P=0.0167*11.982btu/ibm[/tex]

 [tex]P=0.2btu/s[/tex]

Therefore

 [tex]P'=\frac{0.2}{0.75}[/tex]

 [tex]P'=0266btu/s[/tex]

Since

 [tex]1btu/s=1.4148hp[/tex]

Therefore

 [tex]P'=0.377hp[/tex]

Answer:

The block will float with its axis vertical.

Explanation:

For it to float, the upward force on the cylindrical block must be equal to the weight of an equal volume of water. Also, this upward force must be greater than or equal to the weight of the cylindrical block for it to float.

So, weight of cylindrical block, W = specific weight × volume

specific weight = 7500 N/m³

volume = πd²h/4 where d = diameter of block and h = height of block

volume = π(1 m)² × 1 m/4 = π/4 m³ = 0.7854 m³

W = 7500 N/m³ × 0.7854 m³ = 5890.5 N

Since the density of water = 1000 kg/m³, its specific weight W' = 1000 kg/m³ × 9.8 m/s² = 9800 N/m³

Since the volume of the cylinder = volume of water displaced, the weight of water displaced W' = upward force = specific weight of water × volume of water displaced =  9800 N/m³ × 0.7854 m³ = 7696.92 N

Since W' = 7696.92 N > W = 5890.5 N, the block will float with its axis vertical since the upward force is greater than the weight of the cylindrical block.

Answer:

The answer is "22.501,-22.899"

Explanation:

Just as in the previous problems find the angle the velocity makes with the  x-axis and radius of curvature.

[tex]x= 2t^2 + 3t — 1\\\\y=5t-2\\\\x=4t+3\\\\y=5\\\\\tan \alpha (t = 1) =\frac{y}{x}=\frac{5}{4+3}=\frac{5}{7} \to alpha=35.54^{\circ}\\\\[/tex]

For the radius of curvature, we can use the expression from the last two  problems, but first express the position and derivatives as y(x).

[tex]y(x)=2(\frac{y+2}{5})^2+3(\frac{y+2}{5})-1=\frac{1}{25}(2y^2+23y+13)\\\\y'(x)=\frac{1}{25}(4y+23)\\\\y''(x)=\frac{4}{25}\\\\\rho(t=1)=\frac{[1+(\frac{dy}{dx})^2]^{\frac{3}{2}}}{\frac{d^2y}{dx^2}}=\frac{(1+(\frac{35}{25})^2)^{\frac{3}{2}}}{4}25=31.828[/tex]

The position for the center of the radius of curvature [tex]\vec{r}[/tex], (finding this expression is easy and is left as an exercise for the reader.)

[tex]\to \vec{r} = \hat{x}(x + \rho \sin \alpha) + \hat{y}(y- \rho \cos \alpha)\\\\= (4 + 18.501, 3-25.899)\\\\=(22.501, -22.899)[/tex]

Answer:

Following are the application to the given question:

Explanation:

To,

          The Chief Operating Officer,

           Maidan Hospitals, USA

02.10.2020

Subject: HIV/AIDS disease spreading control

Respected Mr Dolman,

The media as well as the individuals living in the area have indeed been notified again that the spread of Hiv / Aids must be regulated and managed immediately. A proper environment must be created artificially to heal this sickness and help stop it from spreading.

Its fight for the sickness ought to be a forward end to bring it through to the high levels instead of being fearful of disease. The disease also is seen in the hospital as it has grown via infusion so that the nursing staff must keep a check mostly on the issue frequently.

This position taken through distributing petitions in the community proposes that a small number of clinics must be established, as this could help to stop the disease from becoming spread. They have also seen that hospital nurses eat, sleep, washing dishes, etc. in such a single place.

To stop this same nursing staff that lives in the health facilities shortly, I suggest that perhaps it should be broken into combinations of 4 individuals and relocated to newly built hospitals, in which they can stay individually and retain hygiene, preserve the atmosphere, help educate, etc. Those who should be accountable. Each clinic and physicians visit every day to examine their clients.

Many patients might well be healed without great faith using these specific HIV/AIDS measures because all norms and laws are observed and mainly social distance is maintained of each patient. Its clinics, as well as the environment, must be sanitized but kept from outside so that the community members recover the confidence and trust of the health departments beforehand. More importantly, a regulatory body should be established that can monitor the operations properly, handle cash, etc. to run smoothly.

Here are a few of my suggestions and promise that it could be successful if it is kept & rigorously implemented.

Thank you and regards,

Joy Roy.

(PR. Director)

Answer: the heat flux increases as the velocity boundary layer transitions to laminar to turbulent.

Explanation:

The correct statement about the relation between the velocity boundary layer and heat transfer for flow over a flat plate that is uniform in temperature is that the heat flux increases as the velocity boundary layer transitions to laminar to turbulent.

It should be noted that the heat goes in a streamline direction in a laminar flow, thereby the molecules less collide with each other. On the other hand, the direction is zig zag in a turbulent heat flux and this will bring about more collision of the molecules which leads to a rise in the heat flux.

Answer:

5,465.4165939453

Explanation:

formula

A=P(1+r/n)^n(t)

p=1000

r=0.09

n=4

t=5

Answer:

2.68

Explanation:

Percentage by Mass of each Aggregate :

Pa = 65% ; Pb = 35%;

Bulk Specific gravity of each aggregate :

Ga = 2.45 ; Gb = 3.25

Gsb = (Pa + Pb) / (Pa/Ga + Pb/Gb)

Gsb = (65 + 35) / (65/2.45 + 35/3.25)

Gsb = (65 + 35) / 37.299843

Gsb = 100 / 37.299843

Gsb = 2.68

Answer:

volumetric flow rate = [tex]0.0251 m^3/s[/tex]

Velocity in pipe section 1 = [tex]6.513m/s[/tex]

velocity in pipe section 2 = 12.79 m/s

Explanation:

We can obtain the volume flow rate from the mass flow rate by utilizing the fact that the fluid has the same density when measuring the mass flow rate and the volumetric flow rates.

The density of water is = 997 kg/m³

density = mass/ volume

since we are given the mass, therefore, the  volume will be mass/density

25/997 = [tex]0.0251 m^3/s[/tex]

volumetric flow rate = [tex]0.0251 m^3/s[/tex]

Average velocity calculations:

Pipe section A:

cross-sectional area =

[tex]\pi \times d^2\\=\pi \times 0.07^2 = 3.85\times10^{-3}m^2[/tex]

mass flow rate = density X cross-sectional area X velocity

velocity = mass flow rate /(density X cross-sectional area)

[tex]velocity = 25/(997 \times 3.85\times10^{-3}) = 6.513m/s[/tex]

Pipe section B:

cross-sectional area =

[tex]\pi \times d^2\\=\pi \times 0.05^2= 1.96\times10^{-3}m^2[/tex]

mass flow rate = density X cross-sectional area X velocity

velocity = mass flow rate /(density X cross-sectional area)

[tex]velocity = 25/(997 \times 1.96\times10^{-3}) = 12.79m/s[/tex]

Answer:

The maximum length of a surface flaw that is possible without fracture is

[tex]2.806 \times 10^{-4} m[/tex]

Explanation:

The given values are,

σ=1.65 MPa

γs=0.60 J/m2

E= 2.0 GPa

The maximum possible length is calculated as:

[tex]\begin{gathered}a=\frac{2 E \gamma_{s}}{\pi \sigma^{2}}=\frac{(2)\left(2 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}\right)(0.60 \mathrm{~N} / \mathrm{m})}{\pi\left(1.65\times 10^{6} \mathrm{~N} / \mathrm{m}^{2}\right)^{2}} \\=2.806 \times 10^{-4} \mathrm{~m}\end{gathered}[/tex]

The maximum length of a surface flaw that is possible without fracture is

[tex]2.806 \times 10^{-4} m[/tex]

Answer:

26.02 ft

86.7690 ft/min

Explanation:

After 3 steps

0.75³(2.0 thickness)

T = 0.84375

W = (1+0.03)³10

= 10.92727 inches

A To get length

2.0 x 10 x 12 x12 = 0.84375 x 10.92727x lf

= 2880 = 9.21988Lf

Lf = 2880/9.21988

= 312.368 inches

Convert to feet

322.368 x 0.0833

= 26.02 ft

B.

= 2 x 10 x 40 = 0.84375 x 10.92727 x vf

800 = 9.21988vf

Vf = 800/9.21988

Vf = 86.7690 ft/min

Answer:

[tex]X_t=2.17391304*10^{-4}[/tex]

[tex]X_r=2.89855072*10^{-4}[/tex]

[tex]e_t=0.0026[/tex]

[tex]e_r=0.0035[/tex]

Explanation:

From the question we are told that:

Dimension [tex]12*12[/tex]

Thickness [tex]l_t=5mm=5*10^-3[/tex]

Normal tensile force on top side [tex]F_t= 15kN[/tex]

Normal tensile force on right side  [tex]F_r= 20kN[/tex]

Elastic modulus, [tex]E=115Gpap=>115*10^9[/tex]

Generally the equation for Normal Strain X is mathematically given by

 [tex]X=\frac{Force}{Area*E}[/tex]

Therefore

For Top

 [tex]X_t=\frac{Force_t}{Area*E}[/tex]

Where

 [tex]Area=L*B*T[/tex]

 [tex]Area=12*10^{-2}*5*10^{-3}[/tex]

 [tex]Area=6*10^{-4}[/tex]  

 [tex]X_t=\frac{15*10^3}{6*10^{-4}*115*10^9}[/tex]

 [tex]X_t=2.17391304*10^{-4}[/tex]

For Right side[tex]X_r=\frac{Force_r}{Area*E}[/tex]

Where

Area=L*B*T

 [tex]Area=12*10^{-2}*5*10^{-3}[/tex]

 [tex]Area=6*10^{-4}[/tex]  

 [tex]X_r=2.89855072*10^{-4}[/tex]

 [tex]X_r=2.89855072*10^{-4}[/tex]

Generally the equation for elongation is mathematically given by

 [tex]e=strain *12[/tex]

For top

 [tex]e_t=2.17391304*10^{-4}*12[/tex]

 [tex]e_t=0.0026[/tex]

For Right

 [tex]e_r=2.89855072*10^{-4} *12[/tex]

 [tex]e_r=0.0035[/tex]

Answer:do me ti

Why not me

Why not me

do me ti

Why not me

Why not me

Do me ti

Why not me

Why not me

Explanation:

mitski

Answer:

The head added by the pump is approximately 51.8 meters

Explanation:

The given parameters of the water are;

The initial diameter of the pump ≈ The final diameter of the pump

The inlet side pressure, p₁ = 30 kPa

The intended outlet side pressure, p₂ = 500 kPa

The power of the pump, P = ρ·g·Q·H

Where, the pressure added by the pump, Δp = p₂ - p₁ = ρ·g·H

ρ = The density of the water ≈ 997 kg/m³

g = The acceleration due to gravity ≈ 9.81 m/s²

H = The head added by the pump

Therefore, we have;

500 kPa - 30 kPa = 997 kg/m³ × 9.81 m/s² × H

H = (500 kPa - 30 kPa)/(997 kg/m³ × 9.1 m/s²) ≈ 51.8 m

The head added by the pump, H ≈ 51.8 meters.