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)
For a pipe system with a pump (pumping uphill), the change in elevation is 400 feet and the total head loss is 408.5 feet. Assuming gage pressure at the entrance and exit and no difference in velocity between the entrance and exit, determine the total energy transferred to the water. Estimate the required power input if the pump efficie
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
The purpose of the international residential code is to
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:
Engineers design products or processes to meet desired needs. Your desired need or goal (hopefully) is to graduate with your Bachelor of Science degree in engineering. But what is the process you need to apply to be successful in achieving this goal?
A cylindrical buoy is 2m in diameter and 2.5m long and weight 22kN . The specific weight of sea water is 10.25kN/m^3 . (I) Show that buoy does not float with its axis vertical. (II). What minimum pull should be applied to a chain attached to the center of the base to keep the buoy vertical?
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
Why does the ceramic made from Thorium and Oxygen have the chemical ratio of 2 oxygen atoms to every thorium atom (ThO2)
Water flows through a converging pipe at a mass flow rate of 25 kg/s. If the inside diameter of the pipes sections are 7.0 cm and 5.0 cm, find the volume flow rate and the average velocity in each pipe section.
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]
If you deposit $ 1000 per month into an investment account that pays interest at a rate of 9% per year compounded quarterly.how much will be in your account at the end of 5 years ?assume no interpèriod compounding
Answer:
5,465.4165939453
Explanation:
formula
A=P(1+r/n)^n(t)
p=1000
r=0.09
n=4
t=5
Resistance depends on which three properties of a wire?
Color and texture are not directly related to a wire’s resistance.
1. color, thickness, texture
2. thickness, length, temperature
3. length, texture, temperature
4. temperature, color, texture
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.
Vince is trying to figure out the volume of two mystery matters. The volume of one of the substances needs to be measured by submerging it in water and the other needs to be measured using a graduated cylinder. Based on the properties of two mystery matters, what are they? (2 points)
Group of answer choices
A rock and orange juice
Helium and a golf ball
Lemonade and milk
Orange juice and helium
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.
Things to be done before isolation
A 2.0-in-thick slab is 10.0 in wide and 12.0 ft long. Thickness is to be reduced in three steps in a hot rolling operation. Each step will reduce the slab to 75% of its previous thickness. It is expected that for this metal and reduction, the slab will widen by 3% in each step. If the entry speed of the slab in the first step is 40 ft/min, and roll speed is the same for the three steps, determine: (a) length and (b) exit velocity of the slab after the final reduction
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
A square plate of titanium is 12cm along the top, 12cm on the right side, and 5mm thick. A normal tensile force of 15kN is applied to the top side of the plate. A normal tensile force of 20kN is applied to the right side of the plate. The elastic modulus, E, is 115 GPa for titanium.If the left and bottom edges of the plate are fixed, calculate the normal strain and elongation of both the TOP and RIGHT side of the plate.
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]
Hi, can anyone draw me an isometric image of this shape?
The number of pulses per second from IGBTs is referred to as
Water enters and leaves a pump in pipelines of the same diameter and approximately the same elevation. If the pressure on the inlet side of the pump is 30 kPa and a pressure of 500 kPa is desired for the water leaving the pump, what is the head that must be added by the pump
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.
An aggregate blend consists of 65% of aggregate A and 35% of aggregate B. The bulk specific gravities of aggregate A and B are 2.45 and 3.25, respectively. What is the bulk specific gravity of the blend?
a) 2.45
b) 2.68
c) 2.73
d) 2.92
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
State two factors that shows that light travels in a straight line
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
can help me with this circuit question?
the last word is (cutoff)
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]
Can some people answer these questions so i can get to know the age group i an making my target market for DT-GSCE thankyou if you do my deadline is tomorrow :D
Answer:
I think you might have forgotten to post the problems
explain all the characteristics of computer
Determine the horsepower required to compress 1 lbm/min of ethylene oxide from 70 oF and 1 atm to 250 psia. The compressor has an efficiency of 75%. The molar heat capacity of ethylene oxide is given by Cp
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]
A cylindrical block of wood 1 m in diameter and 1 m long has a specific weight of 7500 N/m^3. Will it float in water with its axis vertical?
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.
A brittle material is subjected to a tensile stress of 1.65 MPa. If the specific surface energy and modulus of elasticity for this material are 0.60 J/m2 and 2.0 GPa, respectively. What is the maximum length of a surface flaw that is possible without fracture
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]
What is the per capita GDP of China? Be sure to indicate the calendar year that this information represents.
The per capita GDP of China in the Calendar year 2021 was found to be around 12,359 U.S. dollars.
What is GDP?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.
Learn more about the GDP, here:
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Which of the following is NOT part of a car's drive train?
A axle
B rotor
C differential
D transmission
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.
Which is not part of the 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.
To learn more about the part of a car's drive train
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The statement that is correct about the relation between the velocity boundary layer and heat transfer for flow over a flat plate that is uniform in temperature is
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.
A particle which moves in two-dimensional curvilinear motion has coordinates in millimeters which vary with time t in seconds according to X=2t^2 +3t–1 and y = 5t - 2. Determine the coordinates of the center of curvature C at time t = 1s.
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]
The allowable tensile stress for a 6.25 mm diameter bolt with a thread length of 5.5 mm is 207 MPa. The allowable shear stress of the material is 103 MPa. Where and how will such a bolt be most likely to fail if placed in tension
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.
Okay bro let’s go man yes yes
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
What possible scenarios may happen if you do the task without using PPE?