The pressure gage on a 2.5-m3 oxygen tank reads 500 kPa. Determine the amount of oxygen in the tank (mass in kg) if the temperature is 28°C and the atmospheric pressure is 97 kPa.

Answer:

Đường dây siêu cao áp 500kV: Những chuyện giờ mới kể ... ​Ngày 27/5/1994, hệ thống đường dây điện siêu cao áp 500kV Bắc - Nam chính thức đưa ... Tại thời điểm đó, các nước như Pháp, Úc, Mỹ khi xây dựng đường dây dài nhất ... và chế ra các máy kéo dây theo đặc thù công việc của từng đơn vị.

Explanation:

Answer:

small forewing

pro : Can be used in place of tail plane configuration

con : Can be very complex and difficult to use

Explanation:

A canard is generally used to provide some form of stability to an unstable or semi stable system.

An example of a Canard is a small forewing placed in an aircraft that will help stabilize the aircraft when in motion( in air ). because an airplane is generally an unstable system on its own

pro : Can be used in place of tail plane configuration

con : Can be very complex and difficult to use

Answer:

a)  ∝  and β

   The phase compositions are :

    C[tex]_{\alpha }[/tex] = 5wt% Sn - 95 wt% Pb

    C[tex]_{\beta }[/tex] =  98 wt% Sn - 2wt% Pb

b)

The phase is; ∝  

The phase compositions is;   82 wt% Sn - 91.8 wt% Pb

Explanation:

a) 15 wt% Sn - 85 wt% Pb at 100⁰C.

The phases are ; ∝  and β

The phase compositions are :

C[tex]_{\alpha }[/tex] = 5wt% Sn - 95 wt% Pb

C[tex]_{\beta }[/tex] =  98 wt% Sn - 2wt% Pb

b) 1.25 kg of Sn and 14 kg Pb at 200⁰C

The phase is ; ∝  

The phase compositions is;  82 wt% Sn - 91.8 wt% Pb

Csn = 1.25 * 100 / 1.25 + 14 = 8.2 wt%

Cpb = 14 * 100 / 1.25 + 14 = 91.8 wt%

Answer: hello some data related to your question is missing attached below is the missing data

answer:

T2 = 265°C

Explanation:

First step : calculate sum of vertical forces

∑ y = 0

Fmg - 2(0.5 Fst ) = 0

∴Fmg = ( 12 * 10^6 ) ( 2 * π/4 (0.01)^2 )

          = 1884.96 N

Also determine the Compatibility equation in order to determine the change in Temperature

ΔT = 250°C

therefore Temperature at which average normal stress becomes 12.0 MPa

ΔT = T2 - T1

250°C = T2 - 15°C

T2 = 250 + 15 = 265°C

attached below is the detailed solution

Solution :

For R-134a, we are given :

[tex]$T_i = 25^\circ C$[/tex]

[tex]$P_i=750 \ kPa$[/tex]

[tex]$P_e=165 \ kPa$[/tex]

Now we have one inlet and one exit flow, no work and no heat transfer. The energy equation is :

[tex]$h_e+\frac{1}{2}.v_e^2= h_i+\frac{1}{2}.v_i^2 $[/tex]

We also know that the gas is throttled and there is no change in the kinetic energy.

So, [tex]$v_e=v_i$[/tex]

Now from the energy equation above, we can see that the inlet and the exit enthalpies are also the same. Therefore,

[tex]$h_i=h_e$[/tex]

From the saturated R-134a table, corresponding to [tex]P_e = 165 \ kPa[/tex], we can find the exit saturation temperature.

[tex]$T_e=-15^\circ C$[/tex]

From the saturated R-134a table, corresponding to [tex]P_e = 165 \ kPa[/tex], we can find the specific enthalpies :

[tex]$h_f = 180.19 \ kJ/kg$[/tex]

[tex]$h_{fg} = 209 \ kJ/kg$[/tex]

Calculating the exit flow quality factor,

[tex]$x_e=\frac{h_e-h_f}{h_{fg}}$[/tex]

    [tex]$=\frac{234.59-180.19}{209}$[/tex]

   = 0.26

From the saturated R-134a table, corresponding to [tex]P_e = 165 \ kPa[/tex], we can find the specific volumes :

[tex]$v_f = 0.00746 \ m^3/kg$[/tex]

[tex]$v_{fg} = 0.11932 \ m^3/kg$[/tex]

Calculating the exit specific volume :

[tex]$v_e=v_f+x_e(v_{fg})$[/tex]

   = 0.000746 + 0.26 (0.11932)

   = 0.0318 [tex]m^3/kg[/tex]

The mass flow is equal to :

[tex]$\dot{m} = A_i . \frac{v}{v_i}$[/tex]

  [tex]$=A_e . \frac{v}{v_e}$[/tex]

So, [tex]$\frac{A_e}{A_i}=\frac{v_e}{v_i}$[/tex]

Therefore, the ratio of the exit pipe and the inlet pipe diameter is equal to

[tex]$\frac{D_e}{D_i}=\sqrt{\frac{A_e}{A_i}}$[/tex]

[tex]$\frac{D_e}{D_i}=\sqrt{\frac{v_e}{v_i}}$[/tex]

[tex]$\frac{D_e}{D_i}=\sqrt{\frac{0.0318}{0.000829}}$[/tex]

[tex]$\frac{D_e}{D_i}=6.19$[/tex]

Answer:

Probability of no red ball from 4 balls = 0.1258 (Approx.)

Explanation:

Given:

Number of black ball = 3

Number of red ball = 5

Number of green ball = 6

Find:

Probability of no red ball from 4 balls

Computation:

Probability of no red ball from 4 balls = 9c4 / 14c4

Probability of no red ball from 4 balls = 126 / 1001

Probability of no red ball from 4 balls = 0.12587

Probability of no red ball from 4 balls = 0.1258 (Approx.)

A find the ganeral solution to this equation

An object moving through the water has an initial velocity of 40 m/s. Two seconds later, the velocity has decreased to 30 m/s. The velocity after ten seconds is 0.

Velocity is defined as the speed at which an object's position changes in relation to time and a frame of reference.  Speed is the rate at which an object travels along a path over time, whereas velocity is the speed and direction of an item's motion. In other words, speed is a scalar value, but velocity is a vector.

As given

Initial velocity = 40 m / sec

Velocity after 2 seconds = 30 m / sec

So the velocity after 10 seconds will be = 0 m / sec

That is the object will stop moving.

Thus. an object moving through the water has an initial velocity of 40 m/s. Two seconds later, the velocity has decreased to 30 m/s. The velocity after ten seconds is 0.

To learn more about velocity, refer to the link below:

https://brainly.com/question/28738284

#SPJ2

Your question is incomplete, but probably your complete question was

This question is incomplete, the complete question is;

Consider a venturi with a throat-to-inlet area ratio of 0.75, mounted on the side of an airplane fuselage. The airplane is in flight at standard sea level. If the static pressure at the throat is 2050 lb/ft2 , calculate the velocity of the airplane.

Note that standard sea level density and pressure are 1.23 kg/m3 (0.002377 slug/ft3) and 1.01 x 105 N/m2 (2116lb/ft3), respectively.

Answer:

the velocity of the airplane is 267.2 ft/s

Explanation:

Given the data in the question;

throat-to-inlet area ratio A₂/A₁ = 0.75

density of air ρ = 0.002377 slug/ft³

the pressure at inlet p₁ = 2116 lb/ft³

the pressure at the throat p₂ = 2050 lb/ft³

Now, for a venturi duct, the velocity of the airplane V is given as;

V = √[ (2( p₁ - p₂ )) / (ρ( [A₁/A₂]² - 1 )) ]

so we substitute in our values

V = √[ (2( 2116 - 2050 )) / (0.002377 ( [1/0.75]² - 1 )) ]

V = √[ ( 2 × 66 ) / (0.002377 ( 1.7778 - 1 )) ]

V = √[ ( 2 × 66 ) / (0.002377 × 0.7778 ) ]

V = √[ 132 / 0.0018488 ]

V = √[ 71397.663349 ]

V = 267.2 ft/s

Therefore, the velocity of the airplane is 267.2 ft/s

Answer:

Therefore the value of the line currents IR, IY, and IB are

[tex]I_{R}=I_{1}=83\angle 6.87^{o}A\\I_{B}=-I_{2}=-71.88\angle-23.13^{o}A\\I_{Y}=I_{2}-I_{1}\\I_{Y}=-I_{2}=41.50\angle113.3^{o}A[/tex]

Explanation:

Apply KVL for loop 1  

[tex]415\angle 120^{o}=\left ( 3+j4+3+j4 \right )I_{1}-\left ( 3+j4 \right )I_{2}\\415\angle 120^{o}=\left ( 6+j8 \right )I_{1}-\left ( 3+j4 \right )I_{2}\\415\angle 120^{o}=\left ( 10\angle 53.13^{o} \right )I_{1}-\left ( 5\angle 53.13^{o}\right )I_{2} \rightarrow \left ( 1 \right )[/tex]

Apply KVL for loop 2

[tex]415\angle 0^{o}=\left ( 3+j4+3+j4 \right )I_{2}-\left ( 3+j4 \right )I_{1}\\415\angle 0^{o}=\left ( 6+j8 \right )I_{2}-\left ( 3+j4 \right )I_{1}\\415\angle 0^{o}=\left ( 10\angle 53.13^{o} \right )I_{2}-\left ( 5\angle 53.13^{o}\right )I_{1} \rightarrow \left ( 2 \right )[/tex]

Solving the above equations,

[tex]415\angle 120^{o}+830\angle 0^{o}=\left ( 10\angle 53.13^{o} \right )I_{1}-\left ( 5\angle 53.13^{o}\right )I_{2} +\left ( 20\angle 53.13^{o}\right )I_{2}- \left ( 10\angle 53.13^{o} \right )I_{1}\\415\angle 120^{o}+830\angle 0^{o}=\left ( 20\angle 53.13^{o}\right )I_{2}- \left ( 10\angle 53.13^{o} \right )I_{2}\\415\angle 120^{o}+830\angle 0^{o}=\left ( 10\angle 53.13^{o} \right )I_{2}\\I_{2}= 71.88\angle -23.13^{o}A[/tex]&

[tex]415\angle 0^{o}=\left ( 10\angle 53.13^{o} \right ) \times 71.88\angle -23.13^{o}-\left ( 5\angle 53.13^{o} \right )I_{1}\\415\angle 0^{o}=718.8\angle 30^{o}-\left ( 5\angle 53.13^{o} \right )I_{1}\\\left ( 5\angle 53.13^{o} \right )I_{1}=415\angle 60^{o}\\I_{1}= 83\angle 6.87^{o}A[/tex]

Hence,

[tex]I_{R}=I_{1}=83\angle 6.87^{o}A\\I_{B}=-I_{2}=-71.88\angle-23.13^{o}A\\I_{Y}=I_{2}-I_{1}\\I_{Y}=-I_{2}=41.50\angle113.3^{o}A[/tex]

Answer:

 0.287

Explanation:

Design-stage uncertainty can be expressed as :

Ud = √ Uo^2 + Uc^2 ------ ( 1 )

where : Uo = 1/2( resolution value ) = 1/2 * 0.01 V = 0.005 V

Uc = √(0.10)^2 + (0.10)^2 + (0.15)^2 + (0.20)^2  = 0.287

back to equation 1

Ud = √ ( 0.005)^2 + ( 0.287 )^2 =  0.287

Answer:

0.287

Explanation:

Answer:

Video game designers typically have a bachelor's degree in game design, computer engineering, or computer science which takes four to five years to complete. Courses in a game design degree program may include project management, integrated video design and technology, game prototyping and level design.

Explanation:

उह्ह्नमजज्ल्ह्ह्बनुतनकुहक्जो

Answer: hello your question is poorly written below is the complete question

Ethanol blended with gasoline can be used to power a flex-fueled car. One particular blend that is gaining in popularity is E85 , which is 85% ethanol and 15% gasoline. E85 is 80% cleaner burning than gasoline alone, and it reduces our dependency on foreign oil. But a flex-fueled car costs $1,000 more than a conventional gasoline-fueled car. Additionally, E85 fuel gets 10% less miles per gallon than a conventional automobile. Consider a 100% gasoline -fueled car that averages 30 miles per gallon. The E85-fueled car will average about 27 miles per gallon. If either car will be driven 81000 miles before being traded in, how much will the E85 fuel have to cost (per gallon) to make the flex-fueled car as economically attractive as a conventional gasoline-fueled car? Gasoline costs $3.89 per gallon. Work this problem without considering the time value of money

answer :

$3.17

Explanation:

Determine how much E85 fuel have to cost

Fuel needed by 100% gasoline fueled car

= 81,000 miles / 30 = 2700 gallons

Fueled needed by E85 car

= 81,000 miles / 27 = 3000 gallons

next step : use the relation below

[ (Additional cost of flex fuel car) + (flex fuel consumption * cost of flex fuel per gallon( x ) )  ]   = [ Gasoline consumption * cost of gasoline per gallon ]

= 1000 + 3000x = 2700 * 3.89

= 1000 + 3000x = 10503

∴ x = 9503 / 3000 = $3.17 per gallon

Answer:

a) Solid solution strengthening and alloying,  Precipitation hardening, work hardening

b) Absence of enough  crystallographic misalignment in the grain boundary region for a small-angle

Explanation:

A) strengthening mechanism

i) Solid solution strengthening and alloying:

In solid solution strengthening and alloying mechanism there is an addition of one atom of solute to another during this process, there might be substitution of interstitial point defect in crystal

also the shear stress required can be represented as:  Δz = Gb√Ce^3/2

where : C = solute concentration , e = strain on material

ii) Precipitation hardening:

During precipitation hardening the alloying above the concentrate will lead to the formation of a second phase also under precipitation hardening a second phase can also be created via thermal treatments

particle bowing cab be written as :  Δz = Gb / L-2x

iii) work hardening :

Dislocation caused by stress fields been generated hardens metals under the work hardening mechanism

dislocation can be represented as ; Gb √ p

where : G = shear modulus , b = Burgess vector, p = dislocation density

B) The small angle grain boundaries are not effective enough because there is less crystallographic misalignment in the grain boundary region for a small-angle

Answer:

Una canica de masa 15g, con velocidad horizontal de 10 m/s impacta a otra de masa de 8g que se encuentra en el piso y en reposo. Suponga que el movimiento ocurre en la linea que une sus centros. Encuentra la velocidad de cada canica inmediatamente despues de la colisionExplanation:

Answer:

a g i

Explanation:

nnj

Answer:

a) 5.2 kPa

b) 49.3%

Explanation:

Given data:

Thermal efficiency ( л ) = 56.9% = 0.569

minimum pressure ( P1 ) = 100 kpa

a) Determine the pressure at inlet to expansion process

P2 = ?

r = 1.4

efficiency = 1 - [ 1 / (rp)[tex]\frac{r-1}{r}[/tex] ]

   0.569   = 1 - [ 1 / (rp)^0.4/1.4

1 - 0.569  = 1 / (rp)^0.285

∴ (rp)^0.285 = 0.431

rp = 0.0522

note : rp = P2 / P1

therefore P2 = rp * P1 = 0.0522 * 100 kpa

                                   = 5.2 kPa  

b) Thermal efficiency

Л = 1 - [ 1 / ( 10.9 )^0.285 ]

   = 0.493 = 49.3%

Answer:

Many techniques have been developed for the measurement of pressure and vacuum. Instruments used to measure and display pressure in an integral unit are called pressure meters or pressure gauges or vacuum gauges.

Answer:

A tunnel built in soft ground—such as clay, silt, sand, gravel or mud—requires specialized techniques compared to hard rock, to compensate for the shifting nature of the soil.

It's from web...

Soft ground tunneling describes the additional measures needed when Microtunneling through soil conditions that are vulnerable to collapse. ... This process ensures tunneling can happen effectively in soft grounds.

It's from me...

Answer:

a)  [tex]r_a=37.8m^3/h[/tex]

b)[tex]Q=2.8Kw[/tex]

Explanation:

Temperature of CH_4[tex]t=25C[/tex]

CH_4Flow rate of [tex]r=27m3/h[/tex]

Air Percentage [tex]=140\%=1.4[/tex]

Temperature of air [tex]t_a=127=>400K[/tex]

Temperature at exit[tex]t_e=427C=>700k[/tex]

Generally the equation for Air's flow Rate is mathematically given by

 [tex]r_a=air\%*r[/tex]

 [tex]r_a=1.4*27[/tex]

 [tex]r_a=37.8m^3/h[/tex]

Generally the equation for Ideal Gas is mathematically given by

 [tex]PV=mRT[/tex]

 [tex]m=\frac{PV}{RT}[/tex]

 [tex]m=\frac{1.01*10^5*37.8}{0.287*10^3*400}[/tex]

 [tex]m=33.35kg[/tex]

Therefore

The rate of heat transfer from the furnace, in kJ/h is

 [tex]Q=mC_p(T_e-T_a)[/tex]

 [tex]Q=33.35*1.005*(700-400)[/tex]

 [tex]Q=2.8Kw[/tex]

Answer:

a. the desired signal is 90 degrees out of phase with the intelligence signal.

Explanation:

The signal to noise ratio of FM detector is defined as function of modulation index for SSB FM signal plus narrow band Gaussian noise at input. The ratio is usually higher than 1:1 which indicates more signals than noise.

Answer: True

Explanation:

The statement that "a masonry chimney should be braced with horizontal metal straps every few feet against the structure and into reinforced points such as wall studs to stabilize the chimney from the shaking force of an earthquake" is true.

In a scenario whereby the chimney isn't braced with the horizontal metal straps every few feet, this can lead to its collapse in case of an earthquake. Therefore, the correct option is "true".

Answer:

Time required = 287.2 secs  

Explanation:

Volume  of room = 25 * 10 * 4 = 1000 m^3

power consumed by pump = 15 kW

T1 ( initial temperature ) = 12°C

P1 ( Initial pressure ) = 1 bar

COPhp = 3

Calculate time taken to raise room Temp to 27°C

average heat supplied ( ∅ ) = COPhp * power consumed by pump

                                             = 3 * 15 = 45 kW

Time required can be calculated using the relation below

∅t = P*V*Cv ( T2 - T1 )   [ p = 1.2 kg/m^3 , Cv = 0.718 KJ/kg ( air properties ) ]

45 * 10^3 ( t )  = 1.2*1000* 718 ( 27 - 12 )

∴ solving for t

t = 287.2 secs  ≈ 4.79 mins

Answer:

-Khái niệm:

Đo dòng điện là sử dụng các dụng cụ như ôm kê, vôn kế, ampe kế, tần số kế… để xác định các đại lượng vật lý của dòng điện

-Đo lường điện để làm gì?

Phát hiện hư hỏng sự cố trong mạch điện và các thiết bị vi mạch

Xác định các giá trị cần đo

Đánh giá chất lượng của các thiết bị sau sản xuất

Xác định thông số kỹ thuật của thiết bị

-Phân loại dụng cụ đo điện: Hiên nay có 2 phương pháp phân loại chính

a. Theo nguyên lý làm việc

Dụng cụ đo kiểu điện từ

Dụng cụ đo kiểu điện động

Dụng cụ đo kiểu cảm ứng

Dụng cụ đo kiểu từ điện

b. Theo đai lượng, giá trị cần đo

+Đo điện năng: Ví dụ công tơ điện

+Đo điện áp: Ví dụ: Vôn kế

+Đo dòng điện: Ví dụ: Ampe kế

+Đo công suất: Oát kế

+Đo điện trở: Ôm kế

Sai số khi đo: Khi đo lường luôn xảy ra các sai số

Sai số tương đối: Là tỷ lệ % của sự chênh lệch giữa giá trị đo được và giá trị thự

Answer:

Minimum city gas mileage is 15

Explanation:

Minimum city gas mileage among 13 different car sizes in 2008 is 15.

Answer:

0.2063

Explanation:

Given data:

packing factor = 0.5

percentage of reduction of powders = 70%

Calculate the final porosity

after sintering Bulk specific volume = 0.9 * 0.7 = 0.63

assuming true specific volume = 1

packing factor = 0.5 , bulk specific volume = 2

packing factor after pressing and sintering

= 1 / ( 2 * 0.63 ) = 0.7937

hence : porosity = 1 - packing factor

                            = 1 - 0.7937 = 0.2063

Answer:

The costs to run the dryer for one year are $ 9.03.

Explanation:

Given that the clothes dryer in my home has a power rating of 2250 Watts, and to dry one typical load of clothes the dryer will run for approximately 45 minutes, and in Ontario, the cost of electricity is $ 0.11 / kWh, to calculate the costs to run the dryer for one year the following calculation must be performed:

1 watt = 0.001 kilowatt

2250/45 = 50 watts per minute

45 x 365 = 16,425 / 60 = 273.75 hours of consumption

50 x 60 = 300 watt = 0.3 kw / h

0.3 x 273.75 = 82.125

82.125 x 0.11 = 9.03

Therefore, the costs to run the dryer for one year are $ 9.03.

Answer:

a. 67607.9psi

b. 123278.33

Explanation:

to get the yield strength of the material

= load/ cross sectional area

cross sectional area = π * 0.253²/4

= 0.0502927

The yield strenght

= 3400/0.0502927

= 67609.9 psi

b. the uts of the material

= maximum load/cross sectional area

= 6200/0.0502927

= 123278.33