A well insulated rigid tank contains 4 kg of argon gas at 450 kPa and 30 C. A valve is opened, allowing the argon to escape until the tank pressure drops to 200 kPa. Assuming that the argon remaining in the tank experiences a reversible adiabatic process, find the final mass of argon in the tank. Since you don't have argon gas tables, assume cp, cv, k as needed at some appropriate temperature(s).

Answer:

Hence the given statement is false.

Explanation:

For low-speed subsonic wind tunnels, the air density remains nearly constant decreasing the cross-section area cause the flow to extend velocity, and reduce pressure. Similarly increasing the world cause to decrease and therefore the pressure to extend.

The speed within the test section is decided by the planning of the tunnel.  

Thus by adjusting the pressure difference won't change the worth of test section velocity.

Answer:

The given statement is false .

The average power consumption for Eager WiFi, Lazy WiFi, Eager 4G, and Lazy 4G Split is maintained by Screen Mode.

Reducing such a need to move in between multiple tabs, the split-screen has been valuable for increasing wages. In the several instances running a two or more desktop system will allow different programs to run throughout multiple devices. That works with the same process on both PC and laptop monitors.

Just display them side by side, instead of the switching among both the apps that has been used frequently. In this phase, an app that the snap to either left or right occupies a third of the display, and yet another app holds the two-thirds remaining. It refers to Split-Screen Mode.

Similarly, assume that the communication radio for 4G has a power consumption of 190 mW when active and 25 mW when idle. The Idle Power of the CPU is 7 mW, whereas the Active Power of the CPU is 5 mW per unit utilization.

Therefore, The average power consumption for Eager WiFi, Lazy WiFi, Eager 4G, and Lazy 4G Split is maintained by Screen Mode.

Learn more about average power on:

https://brainly.com/question/14831024

#SPJ2

Answer:

1709.07 ft^3/s

Explanation:

Annual peak streamflow = Log10(Q [ft^3/s] )

mean = 1.835

standard deviation = 0.65

Probability of levee been overtopped in the next 15 years = 1/5

Determine the design flow ins ft^3/s

P₁₅ = 1 - ( q )^15 = 1 - ( 1 - 1/T )^15 = 0.2

                         ∴  T = 67.72 years

Q₁₅ = 1 - 0.2 = 0.8

Applying Lognormal distribution : Zt = mean + ( K₂ * std ) --- ( 1 )

K₂ = 2.054 + ( 67.72 - 50 ) / ( 100 - 50 ) * ( 2.326 - 2.054 )

    = 2.1504

back to equation 1

Zt = 1.835 + ( 2.1504 * 0.65 )  = 3.23276

hence:

Log₁₀ ( Qt(ft^3/s) ) = Zt  = 3.23276

hence ; Qt = 10^3.23276

                  = 1709.07 ft^3/s

Answer:

Each of the metal specimens HAS an indentation near the center to ensure that the fracture point would occur in this region. Tension tests WERE CONDUCTED as follows. Two pieces of reflective tape WERE PLACED approximately 1 inch apart in the center of the specimen where the indentation 4 WAS LOCATED. The width and the thickness of the specimen at this location WAS MEASURED using a Vernier caliper. Then the specimen WAS SECURED in the MTS Load Frame. A laser extensometer WAS PLACED into position to measure the deformation of the specimen. The laser extensometer WAS USED to measure the original distance between the pieces of reflective tape. The MTS WAS SET to elongate the specimen one tenth of an inch every minute.

Answer: hello your question is incomplete below is the complete question

answer:

N010 GO2 X7.0 Y2.0 15.0 J2.0  ( option 1 )

Explanation:

Given that the NC machining has to be moved from point ( 5,4 ) to point ( 7,2 ) along a circular path

GO2 = circular interpolation in a clockwise path

G91 = incremental dimension

hence the correct option is :

N010 GO2 X7.0 Y2.0 15.0 J2.0  

Answer:

[tex]a=45.31m/s^2[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=5.74[/tex]

Force [tex]F=317N[/tex]

Gravitational Acceleration [tex]g=9.81m/s^2[/tex]

Generally the equation for Force is mathematically given by

 [tex]F-mg=ma[/tex]

 [tex]317-5.74*9.81=5.74 a[/tex]

 [tex]a=\frac{260.7}{5.74}[/tex]

 [tex]a=45.31m/s^2[/tex]

Answer:

i did not known answer but anobody help you

Answer:

H(s) =  10 / [ 1 + s / (200*10^3π ) ]^2

Explanation:

Characteristics of the high pass filter

Cutoff frequency = 100 kHz

stopband attenuation rate = 40 dB/decade

nominal passband gain = 20dB  = 20logK = 20

Formula for H(s) satisfying the requirements above

given that the stopband attenuation = 40 dB/decade the formula for H(s) that will satisfy the requirements is a second order filter

H(s) = K / ( 1 + s/Wo ) ^2  ----- ( 1 )

Wo = 2πf = 2π ( 100 * 10^3 ) =  200 * 10^3 π

 K = 10

back to equation ( 1 )

H(s) =  10 / [ 1 + s / (200*10^3π ) ]^2

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Answer:

Explanation:

[tex]r_2=[/tex]∞

[tex]q=4\pi kT_1(T_2-T_1)\\[/tex]

[tex]q=2\pi kD.[/tex]ΔT--------(1)

[tex]q=hA[/tex] ΔT[tex]=4\pi r_1^2(T_2_s-T_1_s)\\[/tex]

[tex]N_u=\frac{hD}{k} = 2+\frac{0.589 R_a^\frac{1}{4} }{[1+(\frac{0.046}{p_r}\frac{9}{16} )^\frac{4}{9} } ------(3)[/tex]

By equation (1) and (2)

[tex]2\pi kD.[/tex]ΔT=h.4[tex]\pi r_1^2[/tex]ΔT

[tex]2kD=hD^2\\\frac{hD}{k} =2\\N_u=\frac{hD}{k}=2\\[/tex]-------(4)

From equation (3) and (4)

So for sphere [tex]R_a[/tex]→0

Answer:

Explanation:

Tech Social Media giant FB is one of those companies. Not long ago the ceo was brought to court to accusations that his company was selling user data. Turns out this is true and they are selling their users private data to companies all over the word. Once the news turned to something else, people focused on something new but the company still continues to sell it's users data the same as before. This is completely unethical as the information belongs to the user and they are not getting anything while the corporation is profiting.

Answer:

Interval:  x∈ ( 0, ∞ )

There are no transient terms

Explanation:

x (dy/dx) – y= x^2sinx

Attached below is the detailed solution of the Given problem

There are no transient terms found in the general solution

Interval:  x∈ ( 0, ∞ )

Solution :

Given :

[tex]$H(S) =\frac{2S-2}{S^2+\left(\frac{10}{3}\right) S+1}$[/tex]

Transfer function, [tex]$H(S) =\frac{Y(S)}{K(S)}= \frac{2S-2}{S^2+\left(\frac{10}{3}\right) S+1}$[/tex]

[tex]$Y(S) \left(S^2+\frac{10}{3}S+1\right) = (2S-2) \times (S)$[/tex]

[tex]$S^2Y(S) + \frac{10}{3}(SY(S)) + Y(S) = 2(S \times (S)) - 2 \times (S)$[/tex]

Apply Inverse Laplace Transforms,

[tex]$\frac{d^2y(t)}{dt^2} + \frac{10}{3} \frac{dy(t)}{dt} + y(t)=2 \frac{dx(t)}{dt} - 2x(t)$[/tex]

The above equation represents the differential equation of transfer function.

Given : [tex]$x(t)=e^{-t} u(t) \Rightarrow X(S) = \frac{1}{S+1}$[/tex]

We have : [tex]$H(S) =\frac{Y(S)}{K(S)}= \frac{2S-2}{S^2+\left(\frac{10}{3}\right) S+1}$[/tex]

[tex]$Y(S) = X(S) \times \frac{6S-6}{3S^2+10 S + 3} = \frac{6S-6}{(S+1)(3S+1)(S+3)}$[/tex]

[tex]$Y(S) = \frac{A}{S+1}+\frac{B}{3S+1} + \frac{C}{S+3}[/tex]

[tex]$A = Lt_{S \to -1} (S+1)Y(S)=\frac{6S-6}{(3S+1)(S+3)} = \frac{-6-6}{(-3+1)(-1+3)} = 3$[/tex]

[tex]$B = Lt_{S \to -1/3} (3S+1)Y(S)=\frac{6S-6}{(S+1)(S+3)} = \frac{-6/3-6}{(1/3+1)(-1/3+3)} = \frac{-9}{2}$[/tex]

[tex]$C = Lt_{S \to -3} (S+3)Y(S)=\frac{6S-6}{(S+1)(3S+1)} = \frac{-18-6}{(-3+1)(-9+1)} = \frac{-3}{2}$[/tex]

So,

[tex]$Y(S) = \frac{3}{S+1} - \frac{9/2}{3S+1} - \frac{3/2}{S+3}$[/tex]

        [tex]$=\frac{3}{S+1} - \frac{3/2}{S+1/3} - \frac{3/2}{S+3}$[/tex]

Applying Inverse Laplace Transform,

[tex]$y(t) = 3e^{-t}u(t)-\frac{3}{2}e^{-t/3}u(t) - \frac{3}{2}e^{-3t} u(t)$[/tex]

       [tex]$=\frac{-3}{2}e^{-\frac{1}{3}t}u(t) + \frac{3}{1}e^{-t}u(t)-\frac{3}{2}e^{-3t} u(t)$[/tex]

where, a = 2

            b = 1

            c= 2

Answer:

False

Explanation:

Bainite is a type of steel. It is formed by the decomposition of austenite. This happens at a temperature above MS but given that MS temperature is below the one where fine pearlite is formed.

In other words, when iron goes through a metastable crystallization phase Bainite is formed. It takes the rapid cooling, or quenching, of austenite for that to happen.

Metastability is the phenomenon where matter remains in an apparent state of equilibrium or stability even though it can achieve a more stable state. It can also be referred to as the condition of remaining in that pseudo unstable state for a very protracted period of time.

Bainite steel is used for the construction of components that require frequent usage and where plasticity, or tendency for deformity, as well as wear, cannot be tolerated.

Cheers

Answer:

H(s) = 20 / [ 1 + s / 10^5 ]^2

Explanation:

Given data:

cutoff frequency = 100 kHz

stopband attenuation rate = 40 dB/decade

nominal passband gain = 20 dB

new nominal passband gain at cutoff = 14 dB

Represent the transfer function H(s)

The attenuation rate show that there are two(2) poles

H(s) = k / [ 1 + s/Wc ]^2  ----- ( 1 )

where : Wc = 100 kHz = 10^5 Hz , K = 20 log k = 20 dB ∴ k = 20

Input values into equation 1

H(s) = 20 / [ 1 + s / 10^5 ]^2

Answer:

The code is as follows:

<form name = "myForm">

       <div>

           <input type="radio" name="vehicle" value="D0" id="D0"/>

           <label for="D0">Car</label>

       </div>

       <div>

           <input type="radio" name="vehicle" value="D1" id="D1"/>

           <label for="D1">Bike</label>

       </div>

   </form>

Explanation:

This defines the first button

           <input type="radio" name="vehicle" value="D0" id="D0"/>

           <label for="D0">Car</label>

This defines the second button

           <input type="radio" name="vehicle" value="D1" id="D1"/>

           <label for="D1">Bike</label>

The code is self-explanatory, as it follows all the required details in the question

Answer:

a.  164 °F b. 91.11 °C c. 1439.54 kJ

Explanation:

a. [1 pts] How many degrees Fahrenheit (°F) must you raise the temperature?

Since the starting temperature is 48°F and the final temperature which water boils is 212°F, the number of degrees Fahrenheit we would need to raise the temperature is the difference between the final temperature and the initial temperature.

So, Δ°F = 212 °F - 48 °F = 164 °F

b. [2 pts] How many degrees Celsius (°C) must you raise the temperature?

To find the degree change in Celsius, we convert the initial and final temperature to Celsius.

°C = 5(°F - 32)/9

So, 48 °F in Celsius is

°C₁ = 5(48 - 32)/9

°C₁ = 5(16)/9

°C₁ = 80/9

°C₁ = 8.89 °C

Also, 212 °F in Celsius is

°C₂ = 5(212 - 32)/9

°C₂ = 5(180)/9

°C₂ = 5(20)

°C₂ = 100 °C

So, the number of degrees in Celsius you must raise the temperature is the temperature difference between the final and initial temperatures in Celsius.

So, Δ°C = °C₂ - °C₁ = 100 °C - 8.89 °C = 91.11 °C

c. [2 pts] How much energy is required to heat the four quarts of water from

48°F to 212°F (boiling)?

Since we require 15.8 kJ for every degree Celsius of temperature increase of the four quarts of water, that is 15.8 kJ/°C and it rises by 91.11 °C, then the amount of energy Q required is Q = amount of heat per temperature rise × temperature rise =  15.8 kJ/°C × 91.11 °C = 1439.54 kJ

Answer:

True

Explanation:

Yes it is true that the Temperature gradient would also decrease with magnitude just as the distances rise from the centre line.

We have this cylinder equation as

[T1-T2 / ln(r1-r2)]2πKL

The radial distance is r2-r1

The gradient of temperature is T1-T2

From the equation,

The temperature gradient has a direct and proportional relationship to radial distance

T1-T2 ∝ ln(r2-r1)

1/T1-T2 = k(r2-r1)

This inverse relationship above confirms that the statement is true

Answer:

you are right but then you ddnt ask a question

Solution :

Characteristic length  = thickness / 2

                                    [tex]$=\frac{0.04}{2}$[/tex]

                                    = 0.02 m

Thermal conductivity for steel is 42.5 W/m.K

[tex]$\text{Biot number} = \frac{\text{convective heat transfer coefficient} \times \text{characteristic length}}{\text{thermal conductivity}}$[/tex]

                  [tex]$=\frac{30 \times 0.02}{42.5}$[/tex]

                  = 0.014

Since the Biot number is less than 0.01, the lumped system analysis is applicable.

[tex]$\frac{T-T_{\infty}}{T_0-T_{\infty}} = e^{-b\times t}$[/tex]

Where,

T = temperature after t time

[tex]$T_{\infty}$[/tex] = surrounding temperature

[tex]$T_0$[/tex] = initial temperature

[tex]$b=\frac{\text{heat transfer coefficient}}{\text{density} \times {\text{specific heat } \times \text{characteristic length }}}$[/tex]

t = time

We calculate B:

[tex]$b=\frac{30}{7833 \times 460 \times 0.02}$[/tex]

  = 0.000416

Thus, [tex]$\frac{100-50}{500-50}=e^{-0.00416 \times t}$[/tex]

t = 5281.78 second

  = 88.02 minutes

Thus the time taken for reaching 100 degree Celsius is 88.02 minutes.

Answer:

-50 W

Explanation:

The heat transfer rate Q = kA(T₂ - T₁)/d where k = thermal conductivity of material = 0.15 W/m-K, A = surface area of tube = πdL where d = diameter of tube = 17 mm = 0.017 m and L = length of tube = 5 m, T₁ = inside temperature = 46 °C, T₂ = outside temperature = 43 °C and d = thickness of tube = 2.4 mm = 0.0024 m

Since Q = kA(T₂ - T₁)/d ,

Q = kπdL(T₂ - T₁)/d

substituting the values of the variables into the equation, we have

Q = 0.15 W/m-K × π × 0.017 m × 5 m(43 °C  - 46 °C )/0.0024 m

Q = 0.01275π Wm/K(-3 K )/0.0024 m

Q = -0.03825π Wm/0.0024 m

Q = -0.1202 Wm/0.0024 m

Q = -50.07 W

Q = -50 W

So, the heat transfer rate is -50 W meaning heat transfer out of the tube.

Answer: False

Explanation:

Reinforced concrete is simply a combination of the traditional cement concrete with the steel bars which are the reinforcements.

Reinforced concrete is utilized for construction purpose mostly on a large scale. The reinforced concrete was invented by French gardener Joseph in 1849 therefore, it has always been available and appreciated by architects before the 19th century.

Answer:

the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Explanation:

Given:

Diameter of the pipe = 100mm = 0.1m

Contraction ratio = 0.5

thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m

The formula for discharge through a venturimeter is given as:

Where,

is the coefficient of discharge = 0.97 (given)

A₁ = Area of the pipe

A₁ =  

A₂ = Area at the throat

A₂ =  

g = acceleration due to gravity = 9.8m/s²

Now,

The gauge pressure at throat = Absolute pressure - The atmospheric pressure

⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)

Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m

Substituting the values in the discharge formula we get

or

or

Q = 29.28 ×10⁻³ m³/s

Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s

Hope This Helps :D

Answer:

a) -505.229 kJ/Kg

b) -1.724 kJ/kg

Explanation:

T1 = 400°C

P1 = 3 MPa

P2 = 125 kPa

work output   = 530 kJ/kg

surrounding temperature = 20°C = 293 k

A) Calculate heat transfer from Turbine to surroundings

Q = h2 + w - h1

h ( enthalpy )

h1 = 3231.229 kj/kg

enthalpy at P2

h2 = hg = 2676 kj/kg

back to equation 1

Q = 2676 + 50 - 3231.229  = -505.229 kJ/Kg  ( i.e.  heat is lost )

b) Entropy generation

entropy generation = Δs ( surrounding )  + Δs(system)

                                =  - 505.229 / 293   + 0

                                = -1.724 kJ/kg  

Explanation:

ask your dad please and use your brain

Answer:

a. Rockwell              3. hardness

b. Instron                 2. stress vs strain

c. Charpy                 1. impact strength

d. Fatigue                4. Endurance Limit

e. Brinell                  3. hardness

f. Izod                      1. impact strength

Explanation:

Izod and Charpy are the impact strength testing procedure of a material in which a heavy hammer is attached to an arm is released to impact on the test specimen. In Izod test the specimen with v-notch is held vertical with the notch facing outward while in Charpy test the specimen is supported horizontally with notch facing inward to the impacting hammer.

Instron testing system does universal testing of the material which gradually applies the load recording all the stresses and the corresponding strains until the material fails.

Fatigue is the property of a material due to which it fails under the repeated cyclic loading by the initiation and propagation of cracks. The property of a material resist failure subjected to infinite number of repeated cyclic loads below a certain stress limit.

Rockwell and Brinell are the hardness testing methods. In Rockwell test an intender ball is firstly pressed against the specimen using minor load for a certain time and then a major load is pressed against it for a certain time. After the intender is removed the depth of impression on the surface is measured while in case of Brinell hardness we apply only one load against the intender ball for a certain time and after its removal the radius of impression is measured.

Answer:

D. Turn on the light source and the spectrophotometer.

Explanation:

A spectrophotometer is a machine used to measure the presence of any light-absorbing particle in a solution as well as its concentration. To prepare a spectrophotometer for use, the first step is to turn on the spectrophotometer and allow it to warm up for at least 15 minutes. After this is done, the next step will be to ensure that the samples and blank are ready. Next, an appropriate wavelength is set for the solute being determined. Finally, the absorbance is measured of both the blank and samples.

Answer:

[tex]v_2=549.2 m/s\\[/tex]

Explanation:

Given:

[tex]P_1=2500kPa\\T_1=1500 k\\V_1=5 m/s\\P_2=100 kPa\\T_2=1400 k\\A_2=10 cm^2[/tex]

Solution:

For [tex]Co_2[/tex] y=1.4

Since Nozzle is adiababic

So,

[tex]h_1+\frac{V_1^2}{2}=h_2+\frac{V_2^2}{2}\\\frac{v_2^2}{2}=(h_2-h_2)+\frac{r^2}{2}\\v_2^2=2(h_1-h_2)+v_1^2\\v_2=\sqrt{2(h_1-h_2)+v_1^2}[/tex]

Now,

[tex]h_1-h_2=Cp_1T_1-CP_2T_2\\h_1-h_2=(1989-1838.2)*10^3\\ =150.8 * 10^3\\Cp for co_2\\C_{p1}=1.326 kj/kg\\C_{p2}=1.313 kj/kg\\v_2=\sqrt{301600+25}\\ =549.2 m/s[/tex]