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:
https://brainly.com/question/15171681
#SPJ5
What is the key objective of data analysis
Answer: The process of data analysis uses analytical and logical reasoning to gain information from the data. The main purpose of data analysis is to find meaning in data so that the derived knowledge can be used to make informed decisions.
Propane is to be compressed from 0.4 MPa and 360 K to 4 MPa using a two-stage compressor. An interstage cooler returns the temperature of the propane to 360 K before it enters the second compressor. The intermediate pressure is 1.2 MPa. Both adiabatic compressors have a compressor efficiency of 80%.(a) What is the work required in the first compressor per kg of propane?(b) What is the temperature at the exit of the first compressor?(c) What is the cooling requirement in the interstage cooler per kg of propane?
Answer:
a. 81 kj/kg
b. 420.625K
c. 101.24kj/kg
Explanation:
[tex]\frac{t2}{t1} =[\frac{p2}{p1} ]^{\frac{y-1}{y} }[/tex]
t1 = 360
p1 = 0.4mpa
p2 = 1.20
y = 1.13
substitute these values into the equation
[tex]\frac{t2}{360} =[\frac{1.20}{0.4} ]^{\frac{1.13-1}{1.13} }[/tex]
[tex]\frac{t2}{360} =[\frac{1.2}{0.4} ]^{0.1150}\\\frac{t2}{360} =1.1347[/tex]
when we cross multiply
t2 = 360 * 1.1347
= 408.5
a. the work required in the firs compressor
w=c(t2-t1)
c=1.67x10³
t1 = 360
t2 = 408.5
w = 1670(408.5-360)
= 1670*48.5
= 80995 J
= 81KJ/kg
b. [tex]n=\frac{t2-t1}{t'2-t1}[/tex]
n = 80%
t2 = 408.5
t1 = 360
0.80 = 408.5-360 ÷ t'2-360
[tex]0.80 =\frac{48.5}{t'2-360}[/tex]
cross multiply to get the value of t'2
0.80(t'2-360) = 48.5
0.80t'2 - 288 = 48.5
0.8t'2 = 48.5+288
0.8t'2 = 336.5
t'2 = 336.5/0.8
= 420.625
this is the temperature at the exit of the first compressor
c. cooling requirement
w = c(t2-t1)
= 1.67x10³(420.625-360)
= 1670*60.625
= 101243.75
= 101.24kj/kg
Technician A says white smoke in the exhaust of a diesel engine can be the result of a cylinder misfire in a warm engine. Technician B says blue smoke in the exhaust of a diesel engine can be caused by scored cylinder walls. Who is correct?
Answer:
Both
Explanation:
Because of water, fuel does not burn completely. This brings about water fumes that are white in color and looks like white smoke. If engine is cold and water is heating, it leads to steam formation like water vapor. The white times are because of not firing properly in the heated engine. Technician A is right.
Blue fine is caused by this scoring. It is also caused by dirty oil. Technician b is right too
Convert the following indoor air quality standards, established by the U.S. Occupational Safety and Health Administration (OSHA), from ppmv to mg/m3 (at 25°C and 1atm) or vice versa.
a. Carbon dioxide (CO2), 5,000 ppmv
b. Formaldehyde (HCHO), 3.6 mg/m^3
c. Nitric oxide (NO), 25 ppmv
Hi, can anyone draw me an isometric image of this shape?
The floor of a light storage warehouse is made of 6-in.-thick cinder concrete. The floor is a slab having a length of 16 ft and width of 14 ft.
A. Determine the resultant force caused by the dead load.
B. Determine the resultant force caused by the live load.
Explanation:
6/12 = 0.5ft
length = 16
width = 14
The volume of cinder concrete = 0.5 * 16 *14 = 112
the resultant force that is caused by the dead load
density of cinder concrete * volume
density is assumed to be 108
dead load = 108 * 112 = 12096 lb
resultant force caused by the live load
liveload = 125lb/ft2
= 125 * 14 * 16
= 28000 lb
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
A turbine of a fossil fuel burning installation delivers 1,500 hp of mechanical energy to a generator. The generator then converts 80.0% of the mechanical energy into electrical energy. If the terminal potential difference of the generator is 1790 V, what current does it deliver (in A)
Answer:
The generator delivers current of 500.11 A
Explanation:
Given the data in the question;
mechanical energy delivered to the generator = 1500 hp
efficiency η = 80.0 %
terminal potential difference of the generator = 1790 V
we know that;
1 hp = 746 W
so
the mechanical energy delivered to the generator will be
Generator Input = ( 1500 × 746 )W = 1119000 W
So the generator output will be;
Generator Output = Generator Input × η
we substitute
Generator Output = 1119000 W × 80.0 %
Generator Output = 1119000 W × 0.8
Generator Output = 895200 W
So the Current will be;
[tex]I[/tex] = Generator Output / terminal potential difference of the generator
we substitute
[tex]I[/tex] = 895200 W / 1790 V
[tex]I[/tex] = 500.11 A
Therefore, The generator delivers current of 500.11 A
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]
Reynolds Number.Typical values of the Reynolds number for various animalsmoving through air or water are listed below. For which cases is inertia of the fluid important? For which cases do viscous effects dominate? For which cases would the flow be laminar; turbulent? Explain.
Animal Speed Re
(a) large whale 10m/s 300,000,000
(b) tlying duck 20m/s 300,000
(c) large dragonfly 7m/s 30,000
(d) invertebrate larva 1mm/s 0.3
(e) bacterium 0.01mm/s 0.00003
Answer:
i) Cases with Important Inertia
Large whale ( a ) , Flying duck ( b ) , Large dragonfly ( c )
ii) Cases where viscous effects dominate
Invertebrate larva ( d ) , bacterium ( e )
iii) Cases where flow is Laminar ( cases where Re is < 2,100 )
Invertebrate larva ( d ), bacterium ( e )
iv) Cases where flow is turbulent ( case Re is > 2,100 )
Large whale (a) , Flying duck (b), Large dragonfly ( c ),
Explanation:
Reynolds number is the the ratio of Initial forces to viscous forces, hence cases with Large Reynolds number ( > 2100 ) have their inertial forces greater than viscous forces, therefore we can say the Inertia is more important , while cases with smaller Reynolds number ( < 2100 ) have the viscous forces greater than the inertial forces therefore in such case the viscous effect is more important
i) Cases with Important Inertia
Large whale ( a ) , Flying duck ( b ) , Large dragonfly ( c )
ii) Cases where viscous effects dominate
Invertebrate larva ( d ) , bacterium ( e )
iii) Cases where flow is Laminar ( cases where Re is < 2,100 )
Invertebrate larva ( d ), bacterium ( e )
iv) Cases where flow is turbulent ( case Re is > 2,100 )
Large whale (a) , Flying duck (b), Large dragonfly ( c ),
How to Cancel prescription
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
What is An ampere is
Answer:
the SI base unit of electrical current.
Answer:
An ampere is the SI base unit of electrical current
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
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.
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
Answer:
The answer is " [tex]\Delta p = f(V1, p, V2, d, D, L)[/tex]"
Explanation:
Please find the complete question in the attached file.
Its change in temperature in pipes depends on rate heads and loss in pipes owing to pipe flow, contractual loss, etc.
The temperature change thus relies on V1 v2 p d D L.
An intelligence signal is amplified by a 65% efficient amplifier before being combined with a 250W carrier to generate an AM signal. If it is desired to operate at 50% modulation, what must be the dc input power to the final intelligence signal amplifier
Answer:
"192.3 watt" is the right answer.
Explanation:
Given:
Efficient amplifier,
= 65%
or,
= 0.65
Power,
[tex]P_c=250 \ watt[/tex]
As we know,
⇒ [tex]P_t=P_c(1+\frac{\mu^2}{2} )[/tex]
By putting the values, we get
[tex]=P_c(1+\frac{1}{2} )[/tex]
[tex]=1.5 \ P_c[/tex]
Now,
⇒ [tex]P_i=(P_t-P_c)[/tex]
[tex]=1.5 \ P_c-P_c[/tex]
[tex]=\frac{P_c}{2}[/tex]
DC input (0.65) will be equal to "[tex](\frac{P_c}{2} )[/tex]".
hence,
The DC input power will be:
= [tex]\frac{250}{2}\times \frac{1}{0.65}[/tex]
= [tex]\frac{125}{0.65}[/tex]
= [tex]192.3 \ watt[/tex]
. En la facultad de Ingeniería Industrial se realizó una encuesta a 200 personas para saber que lenguaje de programación preferían para aprender al inicio, se obtuvo: 50 prefieren C, 65 prefieren C#, 77 prefieren Python, 100 prefieren C o C#, 105 prefieren C# o Python, 110 prefieren C o Python, 10 personas prefieren C y Python pero no C#.
Answer:
lalalalapumpe
Explanation:
4. An aluminum alloy fin of 12 mm thick, 10 mm width and 50 mm long protrudes from a wall, which is maintained at 120C. The ambient air temperature is 22C. The heat transfer coefficient and conductivity of the fin material are 140 W/m2K and 55 W/mk respectively. Determine a. Temperature at the end of the fin b. Temperature at the middle of the fin. c. Calculate the heat dissipation energy of the fin
Answer:
a) 84.034°C
b) 92.56°C
c) ≈ 88 watts
Explanation:
Thickness of aluminum alloy fin = 12 mm
width = 10 mm
length = 50 mm
Ambient air temperature = 22°C
Temperature of aluminum alloy is maintained at 120°C
a) Determine temperature at end of fin
m = √ hp/Ka
= √( 140*2 ) / ( 12 * 10^-3 * 55 )
= √ 280 / 0.66 = 20.60
Attached below is the remaining answers
Things to be done before isolation
Question
А
Particle of 2kg mass is being pulled across a smooth horizontal
surface by a horizontal force. The force does 24 Joule of work in
increasing
the particle's
velocity from 5m/s
to v m/s. calculate
the value of v and the position of particle
after 15s
Answer:
udhddhdiejebdidjebdhdidh
An air conditioner removes heat steadily from a house at a rate of 750 kJ/min while drawing electric power at a rate of 5.25 kW. Determine (a) the COP of this air conditioner and (b) the rate of heat transfer to the outside air.Answers:(a) 2.38, (b) 1065 kJ/min
Answer:
a) the COP of this air conditioner is 2.38
b) the rate of heat transfer to the outside air is 1065 kJ/min
Explanation:
Given the data in the question;
[ Outdoor ] ← Q[tex]_H[/tex] [ W[tex]_{net, in[/tex] ] Q[tex]_L[/tex] ← [ House ]
Rate of heat removed from the house; Q[tex]_L[/tex] = 750 kJ/min = ( 750 kJ/min × ( 1 kW / 60 kJ/min ) ) = 12.5 kW
Net-work input; W[tex]_{net, in[/tex] = 5.25 kW
a) The coefficient of performance of the air conditioner; COP.
COP = Q[tex]_L[/tex] / W[tex]_{net, in[/tex]
we substitute
COP = 12.5 kW / 5.25 kW
COP = 2.38
Therefore, the COP of this air conditioner is 2.38
b) the rate of heat transfer to the outside air.
Q[tex]_H[/tex] = Q[tex]_L[/tex] + W[tex]_{net[/tex]
we substitute
Q[tex]_H[/tex] = 12.5 kW + 5.25 kW
Q[tex]_H[/tex] = 17.75 kW
Q[tex]_H[/tex] = ( 17.75 × 60 ) kJ/min
Q[tex]_H[/tex] = 1065 kJ/min
Therefore, the rate of heat transfer to the outside air is 1065 kJ/min
A cylindrical rod of copper originally 16.0 mm in diameter is to be cold worked by drawing; the circular cross section will be maintained during deformation. A cold-worked yield strength in excess of 250 MPa and a ductility of at least 12% EL are desired. Furthermore, the final diameter must be 11.3 mm. Explain how this may be accomplished:
(1) show why it cannotbe realized by single drawing and (2) suggest proper procedures to fulfill the requirements.You will need to use following figuresto solve the problem.
Solution :
Given data:
Diameter of the copper cylindrical rod = 16 mm
Yield strength = 250 MPa
Calculating the percent cold work
[tex]$\text{Percentage Cold Work} = \frac{\pi\left(\frac{d_0}{2}\right)^2-\pi\left(\frac{d_d}{2}\right)^2}{\pi\left(\frac{d_0}{2}}\right)^2} \times 100$[/tex]
[tex]$ = \frac{\pi\left(\frac{16}{2}\right)^2-\pi\left(\frac{11.3}{2}\right)^2}{\pi\left(\frac{16}{2}}\right)^2} \times 100$[/tex]
= 50% CW
Therefore, at [tex]50\% \ CW[/tex], the yield strength of copper will be of the order of 330 MPa.
The ductility will be 4% elongation (EL).
Rather than performing drawing in single operation, we draw some of the fraction of total deformation, and then the anneal them to recrystallize and also finally w do cold work on the material for the second time to achieve its final diameter, ductility and yield strength.
[tex]21\% \ CW[/tex]is required for a yield strength of [tex]250 \ MPa[/tex]. Similarly, a maximum of [tex]23\% \ CW[/tex] is required for[tex]12\% \ EL[/tex].
The average of the two values is [tex]22\% \ CW[/tex]. To achieve both the [tex]\text{specified yield strength and ductility,}[/tex] the copper should be deformed to[tex]22\% \ CW[/tex]. The[tex]\text{ final diameter}[/tex] after the first drawing and the initial diameter for the second drawing is [tex]$d_0'$[/tex] , then
[tex]$22\% \ CW = \frac{\pi\left(\frac{d_0'}{2}\right)^2-\pi\left(\frac{11.3}{2}\right)^2}{\pi\left(\frac{d_0'}{2}}\right)^2} \times 100$[/tex]
[tex]$d_0'=\frac{11.3}{\sqrt{1-\frac{22\% \ CW}{100}}}$[/tex]
[tex]$d_0'=12.8 \ mm$[/tex]
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.
An AM signal having a carrier frequency of 460 kHz is to be mixed with a local oscillator signal at a frequency of 1135 kHz. What does the output of the IF amplifier consist of
Answer:
the output of the IF amplifier consist of 675 kHz
Explanation:
Given the data in the question;
AM signal carrier frequency [tex]_{RF[/tex] = 460 kHz
Local oscillator frequency[tex]_{lo[/tex] = 1135 kHz
Now, The output of the IF amplifier consists of difference of local oscillator frequency & AM carrier signal frequency;
FREQUECY[tex]_{IF[/tex] = FREQUECY[tex]_{lo[/tex] - FREQUECY[tex]_{RF[/tex]
so we substitute in our given values
FREQUECY[tex]_{IF[/tex] = 1135 kHz - 460 kHz
FREQUECY[tex]_{IF[/tex] = 675 kHz
Therefore, the output of the IF amplifier consist of 675 kHz
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
Even though the content of many alcohol blends doesn’t affect engine drive ability using gasoline with alcohol in warm weather may cause
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